NCERT Solutions for Class 12 Maths Chapter 4 Determinants

Exercise 4.1: This exercise introduces the concept of determinants and their notation. Students will learn how to find the determinants of 2x2 and 3x3 matrices using various methods, including the rule of Sarrus and the cofactor expansion method.

Exercise 4.2: This exercise focuses on the application of determinants in solving systems of linear equations. Students will learn about Cramer's rule, which allows them to solve systems of linear equations using determinants.

Exercise 4.3: In this exercise, students will learn about the inverse of a matrix and how to find it using determinants. They will also practice using determinants to find the area of a triangle and the volume of a parallelepiped.

Exercise 4.4: This exercise covers the concept of minors and cofactors of a matrix. Students will learn how to find the minors and cofactors of a matrix and use them to find the inverse of a matrix.

Exercise 4.5: In this exercise, students will learn about the adjoint of a matrix and its properties. They will also practice finding the adjoint of a matrix and using it to find the inverse of a matrix.

Miscellaneous Exercise:  In this exercise all the concepts taught throughout the chapter are covered, giving students the opportunity to practice their skills. Additionally, students will learn about the adjoint of a matrix, which is a fundamental concept related to determinants.

NCERT Solutions for Class 12 Maths Chapter 4 - Determinants

Exercise (4.1)

1. Evaluate the determinants in Exercise $\mathbf{1}$ and $\mathbf{2}$. 

\[\mathbf{\left| \begin{matrix}   2 & 4  \\   -5 & -1  \\ \end{matrix} \right|}\]

Ans: Solving the determinant 

\[\left| \begin{matrix}  \text{2} & \text{4}  \\ \text{-5} & \text{-1}  \\\end{matrix} \right|\], 

we have:

\[\Rightarrow \left| \begin{matrix} \text{2} & \text{4}  \\ \text{-5} & \text{-1}  \\ \end{matrix} \right|\text{= -2+20}\]

\[\therefore \left| \begin{matrix} \text{2} & \text{4}  \\ \text{-5} & \text{-1}  \\ \end{matrix} \right|\text{=18}\]

2. Evaluate the determinants in Exercise $1$ and $2$.

1. \[\mathbf{\left| \begin{matrix}\cos \theta  & -\sin \theta   \\\sin \theta  & \cos \theta   \\ \end{matrix} \right|}\].

Ans: Solving the determinant 

\[\left| \begin{matrix} \cos \theta  & -\sin \theta   \\\sin \theta  & \cos \theta   \\ \end{matrix} \right|\], 

we have:

\[\Rightarrow \left| \begin{matrix}\cos \theta  & -\sin \theta   \\\sin \theta  & \cos \theta   \\\end{matrix} \right|=\left( \cos \theta  \right)\left( \cos \theta  \right)-\left( -\sin \theta  \right)\left( \sin \theta  \right)\]

\[\Rightarrow \left| \begin{matrix}\cos \theta  & -\sin \theta   \\\sin \theta  & \cos \theta   \\\end{matrix} \right|=\text{ }{{\cos }^{2}}\theta +{{\sin }^{2}}\theta \]

We know, \[{{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1\]

\[\therefore \left| \begin{matrix}\cos \theta  & -\sin \theta   \\\sin \theta  & \cos \theta   \\\end{matrix} \right|=1\]

2. \[\mathbf{\left| \begin{matrix}{{\text{x}}^{\text{2}}}\text{-x+1} & \text{x-1}  \\\text{x+1} & \text{x+1}  \\\end{matrix} \right|}\].

Ans: Solving the determinant 

\[\left| \begin{matrix}{{\text{x}}^{\text{2}}}\text{-x+1} & \text{x-1}  \\\text{x+1} &\text{x+1}  \\\end{matrix} \right|\], 

we have:

\[\Rightarrow \left| \begin{matrix}{{\text{x}}^{\text{2}}}\text{-x+1} & \text{x-1}  \\\text{x+1} & \text{x+1}  \\\end{matrix} \right|\text{=}\left( {{\text{x}}^{\text{2}}}\text{-x+1} \right)\left( \text{x+1} \right)\text{-}\left( \text{x-1} \right)\left( \text{x+1} \right)\]

\[\Rightarrow \left| \begin{matrix}{{\text{x}}^{\text{2}}}\text{-x+1} & \text{x-1}  \\\text{x+1} & \text{x+1}  \\\end{matrix} \right|\text{= }{{\text{x}}^{\text{3}}}\text{-}{{\text{x}}^{\text{2}}}\text{+x+}{{\text{x}}^{\text{2}}}\text{-x+1-}\left( {{\text{x}}^{\text{2}}}\text{-1} \right)\]

So, 

\[\left| \begin{matrix}{{\text{x}}^{\text{2}}}\text{-x+1} & \text{x-1}  \\\text{x+1} & \text{x+1}  \\\end{matrix} \right|\text{= }{{\text{x}}^{\text{3}}}\text{+1-}{{\text{x}}^{\text{2}}}\text{+1}\]

\[\therefore \left| \begin{matrix} {{\text{x}}^{\text{2}}}\text{-x+1} & \text{x-1}  \\ \text{x+1} & \text{x+1}  \\ \end{matrix} \right|\text{= }{{\text{x}}^{\text{3}}}\text{-}{{\text{x}}^{\text{2}}}\text{+2}\].

3. If \[\mathbf{\text{A=}\left[ \begin{matrix} \text{1} & \text{2}  \\ \text{4} & \text{2}  \\ \end{matrix} \right]}\]. , then show that \[\mathbf{\left| \text{2A} \right|\text{=4}\left| \text{A} \right|}\].

Ans: Given that, 

\[\text{A=}\left[ \begin{matrix} \text{1} & \text{2}  \\ \text{4} & \text{2}  \\ \end{matrix} \right]\]

Multiplying $\text{A}$ by $2$, we have:

\[\Rightarrow \text{2A= 2}\left[ \begin{matrix} \text{1} & \text{2}  \\ \text{4} & \text{2}  \\ \end{matrix} \right]\text{=}\left[ \begin{matrix} \text{2} & \text{4}  \\ \text{8} & \text{4}  \\ \end{matrix} \right]\]

\[\Rightarrow \text{2A=}\left[ \begin{matrix} \text{2} & \text{4}  \\ \text{8} & \text{4}  \\ \end{matrix} \right]\]

\[\therefore L.H.S \text{=}\left| \text{2A} \right|\text{=}\left| \begin{matrix}\text{2} & \text{4}  \\\text{8} & \text{4}  \\\end{matrix} \right|\]

\[\Rightarrow \left| \text{2A} \right|\text{=}2\times 4-4\times 8\]

\[\Rightarrow \left| \text{2A} \right|\text{=}8-32\]

\[\therefore \left| \text{2A} \right|\text{=}-24\]

The value of determinant $\text{A}$ is

\[\Rightarrow \left| \text{A} \right|\text{=}\left| \begin{matrix} \text{1} & \text{2}  \\ \text{4} & \text{2}  \\ \end{matrix} \right|\]

\[\Rightarrow \left| \text{A} \right|\text{=}2-8\]

\[\therefore \left| \text{A} \right|\text{=}-6\]

R.H.S is given as $\text{4}\left| \text{A} \right|$.

\[\therefore \text{4}\left| \text{A} \right|\text{=4 }\!\!\times\!\!\text{ }\left( \text{-6} \right)\text{=-24}\]

Hence, we have L.H.S $=$ R.H.S

\[\therefore \left| \text{2A} \right|\text{=4}\left| \text{A} \right|\].

4. If \[\mathbf{\text{A=}\left[ \begin{matrix} \text{1} & \text{0} & \text{1}  \\ \text{0} & \text{1} & \text{2}  \\ \text{0} & \text{0} & \text{4}  \\ \end{matrix} \right]}\]. , 

then show that $\mathbf{\left | 3A \right |=27\left | A \right |}$.

Ans: Given, 

\[\text{A=}\left[ \begin{matrix} \text{1} & \text{0} & \text{1}  \\ \text{0} & \text{1} & \text{2}  \\ \text{0} & \text{0} & \text{4}  \\ \end{matrix} \right]\]

Determining the value of determinant $\text{A}$, by expanding along the first column, i.e., \[C_1\], we get:

\[\Rightarrow \left| \text{A} \right|\text{=1}\left| \begin{matrix} \text{1} & \text{2}  \\ \text{0} & \text{4}  \\ \end{matrix} \right|\text{-0}\left| \begin{matrix} \text{0} & \text{1}  \\ \text{0} & \text{4}  \\ \end{matrix} \right|\text{+0}\left| \begin{matrix} \text{0} & \text{1}  \\ \text{1} & \text{2}  \\ \end{matrix} \right|\]

\[\Rightarrow \left| \text{A} \right|\text{=}1\left( 4-0 \right)-0+0\]

\[\therefore \left| \text{A} \right|\text{=4}\]

Hence, $27\left| A \right|=27\times 4$

$\Rightarrow 27\left| A \right|=108$   ……(1)

The value of $\left| 3\text{A} \right|$ is obtained as:

\[\Rightarrow \text{3A=3}\left[ \begin{matrix} \text{1} & \text{0} & \text{1}  \\ \text{0} & \text{1} & \text{2}  \\ \text{0} & \text{0} & \text{4}  \\ \end{matrix} \right]\] 

\[\Rightarrow \text{3A=}\left[ \begin{matrix} \text{3} & \text{0} & \text{3}  \\ \text{0} & \text{3} & \text{6}  \\ \text{0} & \text{0} & \text{12}  \\ \end{matrix} \right] \]

\[\therefore \left| \text{3A} \right|\text{=3}\left| \begin{matrix} \text{3} & \text{6}  \\ \text{0} & \text{12}  \\ \end{matrix} \right|\text{-0}\left| \begin{matrix} \text{0} & \text{3}  \\ \text{0} & \text{12}  \\ \end{matrix} \right|\text{+0}\left| \begin{matrix} \text{0} & \text{3}  \\ \text{3} & \text{6}  \\ \end{matrix} \right|\]

\[\Rightarrow \text{3}\left( \text{36-0} \right)\text{+0+0}\]

\[\Rightarrow \left| \text{3A} \right|\text{=3}\times \text{36}\]

Thus, \[\left| \text{3A} \right|\text{=}108\]   ……(2)  

From equations (1) and (2), we have:

\[\left| \text{3A} \right|\text{=27}\left| \text{A} \right|\]

Hence proved.

5. Evaluate the determinants

1. \[\mathbf{\left| \begin{matrix} \text{3} & \text{-1} & \text{-2}  \\ \text{0} & \text{0} & \text{-1}  \\ \text{3} & \text{-5} & \text{0}  \\ \end{matrix} \right|}\]

Ans: Let \[\text{A=}\left| \begin{matrix} \text{3} & \text{-1} & \text{-2}  \\ \text{0} & \text{0} & \text{-1}  \\ \text{3} & \text{-5} & \text{0}  \\ \end{matrix} \right|\]

Determining the value of $\text{A}$ by expanding along the third row, we have:

\[\Rightarrow \left| \text{A} \right|\text{=3}\left| \begin{matrix}  -1 & -2  \\  0 & -1  \\ \end{matrix} \right|-\left( -5 \right)\left| \begin{matrix}  3 & -2  \\ 0 & -1  \\ \end{matrix} \right|+0\left| \begin{matrix} 3 & -1  \\ 0 & 0  \\ \end{matrix} \right|\]

\[\Rightarrow \left| \text{A} \right|\text{=}\left( \text{3-15} \right)\]

\[\therefore \left| \text{A} \right|\text{=-12}\]

2. \[\mathbf{\left| \begin{matrix} \text{3} & \text{-4} & \text{5}  \\ \text{1} & \text{1} & \text{-2}  \\ \text{2} & \text{3} & \text{1}  \\ \end{matrix} \right|}\]

Ans: Let \[\text{A=}\left| \begin{matrix} \text{3} & \text{-4} & \text{5}  \\ \text{1} & \text{1} & \text{-2}  \\ \text{2} & \text{3} & \text{1}  \\ \end{matrix} \right|\]

Determining the value of $\text{A}$ by expanding along the first row, we have:

\[\Rightarrow \left| \text{A} \right|\text{=3}\left| \begin{matrix} \text{1} & \text{-2}  \\ \text{3} & \text{1}  \\ \end{matrix} \right|\text{+4}\left| \begin{matrix} \text{1} & \text{-2}  \\ \text{2} & \text{1}  \\ \end{matrix} \right|\text{+5}\left| \begin{matrix} \text{1} & \text{1}  \\ \text{2} & \text{3}  \\ \end{matrix} \right|\]

\[\Rightarrow \left| \text{A} \right|\text{=3}\left( \text{1+6} \right)\text{+4}\left( \text{1+4} \right)\text{+5}\left( \text{3-2} \right)\]

\[\Rightarrow \left| \text{A} \right|\text{=21+20+5}\]

\[\therefore \left| \text{A} \right|\text{=}46\]

3. \[\mathbf{\left| \begin{matrix} \text{0} & \text{1} & \text{2}  \\ \text{-1} & \text{0} & \text{-3}  \\ \text{-2} & \text{3} & \text{0}  \\ \end{matrix} \right|}\].

Ans: Let \[\text{A=}\left| \begin{matrix} \text{0} & \text{1} & \text{2}  \\ \text{-1} & \text{0} & \text{-3}  \\ \text{-2} & \text{3} & \text{0}  \\ \end{matrix} \right|\]

Determining the value of $\text{A}$ by expanding along the first row, we have:

\[\Rightarrow \left| \text{A} \right|\text{=0}\left| \begin{matrix} \text{0} & \text{-3}  \\ \text{3} & \text{0}  \\ \end{matrix} \right|\text{-1}\left| \begin{matrix} \text{-1} & \text{-3}  \\ \text{-2} & \text{0}  \\ \end{matrix} \right|\text{+2}\left| \begin{matrix} \text{-1} & \text{0}  \\ \text{-2} & \text{3}  \\ \end{matrix} \right|\]

\[\Rightarrow \left| \text{A} \right|\text{=0}\left( 9 \right)-\left( -6 \right)+2\left( -3 \right)\]

\[\therefore \left| \text{A} \right|\text{=0}\]

4. \[\mathbf{\left| \begin{matrix} \text{2} & \text{-1} & \text{-2}  \\ \text{0} & \text{2} & \text{-1}  \\ \text{3} & \text{-5} & \text{0}  \\ \end{matrix} \right|}\]

Ans: Let $\text{A=}\left| \begin{matrix} \text{2} & \text{-1} & \text{-2}  \\ \text{0} & \text{2} & \text{-1}  \\ \text{3} & \text{-5} & \text{0}  \\ \end{matrix} \right|$

Determining the value of $\text{A}$ by expanding along the first column, we have:

\[\Rightarrow \left| \text{A} \right|\text{=2}\left| \begin{matrix} \text{2} & \text{-1}  \\ \text{-5} & \text{0}  \\ \end{matrix} \right|\text{-0}\left| \begin{matrix} \text{-1} & \text{-2}  \\ \text{-5} & \text{0}  \\ \end{matrix} \right|\text{+3}\left| \begin{matrix} \text{-1} & \text{-2}  \\ \text{2} & \text{-1}  \\ \end{matrix} \right|\]

\[\Rightarrow \left| \text{A} \right|\text{=}2\left( -5 \right)-0+3\left( 5 \right)\]

\[\Rightarrow \left| \text{A} \right|\text{=}-\text{10+15}\]

\[\therefore \left| \text{A} \right|\text{=5}\]

6.  If \[\mathbf{\text{A=}\left[ \begin{matrix} \text{1} & \text{1} & \text{-2}  \\ \text{2} & \text{1} & \text{-3}  \\ \text{5} & \text{4} & \text{-9}  \\ \end{matrix} \right]}\] , 

find \[\mathbf{\left| \text{A} \right|}\].

Ans: Given, \[\text{A=}\left[ \begin{matrix} \text{1} & \text{1} & \text{-2}  \\ \text{2} & \text{1} & \text{-3}  \\ \text{5} & \text{4} & \text{-9}  \\ \end{matrix} \right]\]

Determining the value of $\text{A}$ by expanding along the first row, we have:

\[\Rightarrow \left| \text{A} \right|\text{=1}\left| \begin{matrix} \text{1} & \text{-3}  \\ \text{4} & \text{-9}  \\ \end{matrix} \right|\text{-1}\left| \begin{matrix} \text{2} & \text{-3}  \\ \text{5} & \text{-9}  \\ \end{matrix} \right|\text{-2}\left| \begin{matrix} \text{2} & \text{1}  \\ \text{5} & \text{4}  \\ \end{matrix} \right|\]

\[\Rightarrow \left| \text{A} \right|\text{=}1\left( -9+12 \right)-1\left( -18+15 \right)-2\left( 8-5 \right)\]

\[\Rightarrow \left| \text{A} \right|\text{=}3+3-6\]

\[\therefore \left| \text{A} \right|\text{=}0\]

7. Find values of \[x\] , if

1. \[\mathbf{\left| \begin{matrix} \text{2} & \text{4}  \\ \text{5} & \text{1}  \\ \end{matrix} \right|\text{=}\left| \begin{matrix} \text{2x} & \text{4}  \\ \text{6} & \text{x}  \\ \end{matrix} \right|}\]

Ans: Given, 

\[\left| \begin{matrix} \text{2} & \text{4}  \\ \text{5} & \text{1}  \\ \end{matrix} \right|\text{=}\left| \begin{matrix} \text{2x} & \text{4}  \\ \text{6} & \text{x}  \\ \end{matrix} \right|\]

 Solving it, we have:

\[\Rightarrow \left( 2\times 1 \right)-\left( 5\times 4 \right)=\left( 2x\times x \right)-\left( 6\times 4 \right)\]

\[\Rightarrow 2-20=2{{x}^{2}}-24\]

\[\Rightarrow -18+24=2{{x}^{2}}\]

\[\Rightarrow 3={{x}^{2}}\]

Applying square root on both the sides, we obtain:

\[\Rightarrow \text{x =  }\!\!\pm\!\!\text{ }\sqrt{\text{3}}\]

2. \[\mathbf{\left| \begin{matrix} \text{2} & \text{3}  \\ \text{4} & \text{5}  \\ \end{matrix} \right|\text{=}\left| \begin{matrix} \text{x} & \text{3}  \\ \text{2x} & \text{5}  \\ \end{matrix} \right|}\]

Ans: Given, 

\[\left| \begin{matrix} \text{2} & \text{3}  \\ \text{4} & \text{5}  \\n\end{matrix} \right|\text{=}\left| \begin{matrix} \text{x} & \text{3}  \\ \text{2x} & \text{5}  \\ \end{matrix} \right|\]

Solving it, we have:

\[\Rightarrow \left( \text{2 }\!\!\times\!\!\text{ 5} \right)-\left( \text{3 }\!\!\times\!\!\text{ 4} \right)\text{=}\left( \text{x }\!\!\times\!\!\text{ 5} \right)-\left( \text{3 }\!\!\times\!\!\text{ 2x} \right)\]

\[\Rightarrow \text{10}-\text{12=5x}-\text{6x}\]9

\[\Rightarrow -\text{2=}-\text{x}\]

Multiplying by $\left( -1 \right)$ on both the sides, we obtain:

\[\Rightarrow \text{x = 2}\]

8. If \[\mathbf{\left| \begin{matrix} \text{x} & \text{2}  \\ \text{18} & \text{x}  \\ \end{matrix} \right|\text{=}\left| \begin{matrix} \text{6} & \text{2}  \\ \text{18} & \text{6}  \\\end{matrix} \right|}\] , 

then \[\mathbf{x}\] is equal to

1. \[\mathbf{\text{6}}\]

2. \[\mathbf{\text{ }\!\!\pm\!\!\text{ 6}}\]

3. \[\mathbf{\text{-6}}\]

4. \[\mathbf{\text{0}}\]

Ans: Given, 

$\begin{vmatrix}x &2 \\ 18 &x \end{vmatrix}=\begin{vmatrix}6 &2 \\ 18 &6 \end{vmatrix}$

Solving it, we have:

\[\Rightarrow {{\text{x}}^{\text{2}}}-\text{36 = 36}-\text{36}\]

\[\Rightarrow {{\text{x}}^{\text{2}}}-\text{36=0}\]

\[\Rightarrow {{\text{x}}^{\text{2}}}\text{=36}\]

Applying square root on both the sides, we obtain:

\[\Rightarrow \text{x= }\!\!\pm\!\!\text{ 6}\]

Hence, B. \[\text{ }\!\!\pm\!\!\text{ 6}\] is the correct answer.

Exercise (4.2)

1. Find area of the triangle with vertices at the point given in each of the following:

1. \[\mathbf{\left( \text{1,0} \right)\text{,}\left( \text{6,0} \right)\text{,}\left( \text{4,3} \right)}\]

Ans: Given vertices, \[\left( \text{1,0} \right)\text{,}\left( \text{6,0} \right)\text{,}\left( \text{4,3} \right)\]

We know, if we have three points $\left( {{x}_{1}},{{y}_{1}} \right)$, $\left( {{x}_{2}},{{y}_{2}} \right)$ and $\left( {{x}_{3}},{{y}_{3}} \right)$, then the area of the triangle is given by,

$\Delta =\left| \begin{matrix} {{x}_{1}} & {{y}_{1}} & 1  \\ {{x}_{2}} & {{y}_{2}} & 1  \\ {{x}_{3}} & {{y}_{3}} & 1  \\ \end{matrix} \right|$

Thus, the area of the triangle is given by,

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}\dfrac{\text{1}}{\text{2}}\left| \begin{matrix} \text{1} & \text{0} & \text{1}  \\ \text{6} & \text{0} & \text{1}  \\ \text{4} & \text{3} & \text{1}  \\ \end{matrix} \right|\]

\[\Rightarrow \Delta =\dfrac{\text{1}}{\text{2}}\left[ \text{1}\left( \text{0-3} \right)\text{-0}\left( \text{6-4} \right)\text{+1}\left( \text{18-0} \right) \right]\]

\[\Rightarrow \Delta \text{=}\dfrac{\text{1}}{\text{2}}\left[ \text{-3+18} \right]\]

\[\Rightarrow \Delta \text{=}\dfrac{\text{15}}{\text{2}}\] square units

$\therefore $Area of the triangle with vertices \[\left( \text{1,0} \right)\text{,}\left( \text{6,0} \right)\text{,}\left( \text{4,3} \right)\] is \[\dfrac{\text{15}}{\text{2}}\] square units.

2. \[\mathbf{\left( \text{2,7} \right)\text{,}\left( \text{1,1} \right)\text{,}\left( \text{10,8} \right)}\]

Ans: Given vertices, \[\left( \text{2,7} \right)\text{,}\left( \text{1,1} \right)\text{,}\left( \text{10,8} \right)\]

We know, if we have three points $\left( {{x}_{1}},{{y}_{1}} \right)$, $\left( {{x}_{2}},{{y}_{2}} \right)$ and $\left( {{x}_{3}},{{y}_{3}} \right)$, then the area of the triangle is given by,

$\Delta =\left| \begin{matrix} {{x}_{1}} & {{y}_{1}} & 1  \\ {{x}_{2}} & {{y}_{2}} & 1  \\ {{x}_{3}} & {{y}_{3}} & 1  \\ \end{matrix} \right|$

Thus, the area of the triangle is given by,

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}\dfrac{\text{1}}{\text{2}}\left| \begin{matrix} \text{2} & \text{7} & \text{1}  \\ \text{1} & \text{1} & \text{1}  \\ \text{10} & \text{8} & \text{1}  \\ \end{matrix} \right|\]

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}\dfrac{\text{1}}{\text{2}}\left[ \text{2}\left( \text{1-8} \right)\text{-7}\left( \text{1-10} \right)\text{+1}\left( \text{8-10} \right) \right]\]

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}\dfrac{\text{1}}{\text{2}}\left[ \text{2}\left( \text{-7} \right)\text{-7}\left( \text{-9} \right)\text{+1}\left( \text{-2} \right) \right]\]

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}\dfrac{\text{1}}{\text{2}}\left[ \text{-16+63} \right]\]

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}\dfrac{\text{47}}{\text{2}}\] square units

$\therefore $Area of the triangle with vertices \[\left( \text{2,7} \right)\text{,}\left( \text{1,1} \right)\text{,}\left( \text{10,8} \right)\] is \[\dfrac{47}{\text{2}}\] square units.

3. \[\mathbf{\text{(-2,-3),}\left( \text{3,2} \right)\text{,(-1,-8)}}\]

Ans: Given vertices, \[\text{(-2,-3),}\left( \text{3,2} \right)\text{,(-1,-8)}\]

We know, if we have three points $\left( {{x}_{1}},{{y}_{1}} \right)$, $\left( {{x}_{2}},{{y}_{2}} \right)$ and $\left( {{x}_{3}},{{y}_{3}} \right)$, then the area of the triangle is given by,

$\Delta =\left| \begin{matrix} {{x}_{1}} & {{y}_{1}} & 1  \\ {{x}_{2}} & {{y}_{2}} & 1  \\ {{x}_{3}} & {{y}_{3}} & 1  \\ \end{matrix} \right|$

Thus, the area of the triangle with vertices \[\text{(-2,-3),}\left( \text{3,2} \right)\text{,(-1,-8)}\] is given by,

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}\dfrac{\text{1}}{\text{2}}\left| \begin{matrix} \text{-2} & \text{-3} & \text{1}  \\ \text{3} & \text{2} & \text{1}  \\ \text{-1} & \text{-8} & \text{1}  \\ \end{matrix} \right|\]

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}\dfrac{\text{1}}{\text{2}}\left[ \text{-2}\left( \text{2+8} \right)\text{+3}\left( \text{3+1} \right)\text{+1}\left( \text{-24+2} \right) \right]\]

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}\dfrac{\text{1}}{\text{2}}\left[ \text{-20+12-22} \right]\]

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ }=-\dfrac{\text{30}}{\text{2}}\]

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =-15}\]

$\therefore $The area of the triangle with vertices \[\left( \text{2,7} \right)\text{,}\left( \text{1,1} \right)\text{,}\left( \text{10,8} \right)\]is \[\left| \text{-15} \right|\text{=15}\] square units.

2. Show that points \[\mathbf{\text{A}\left( \text{a,b+c} \right)\text{,B}\left( \text{b,c +a} \right)\text{,C}\left( \text{c,a +b} \right)}\] are collinear.

Ans: To show that the points \[\text{A}\left( \text{a,b+c} \right)\text{,B}\left( \text{b,c +a} \right)\text{,C}\left( \text{c,a +b} \right)\] are collinear, the area of the triangle formed by these points as vertices should be zero.

$\therefore $ Area of \[\text{ }\!\!\Delta\!\!\text{ ABC}\] is given by,

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}\dfrac{\text{1}}{\text{2}}\left| \begin{matrix} \text{a} & \text{b+a} & \text{1}  \\ \text{b} & \text{c+a} & \text{1}  \\ \text{c} & \text{a+b} & \text{1}  \\ \end{matrix} \right|\]

Applying the row operations \[{{\text{R}}_{2}}\to {{\text{R}}_{2}}-{{\text{R}}_{1}}\] and \[{{\text{R}}_{3}}\to {{\text{R}}_{3}}-{{\text{R}}_{1}}\]

\[\Rightarrow \Delta \text{=}\dfrac{\text{1}}{\text{2}}\left| \begin{matrix} \text{a} & \text{b+c} & \text{1}  \\ \text{b-a} & \text{a-b} & \text{0}  \\ \text{c-a} & \text{a-c} & \text{0}  \\ \end{matrix} \right|\]

Taking out $\left( a-b \right)$ and $\left( c-a \right)$ common from \[R_2\] and \[R_3\] respectively,

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}\dfrac{\text{1}}{\text{2}}\left( \text{a-b} \right)\left( \text{c-a} \right)\left| \begin{matrix} \text{a} & \text{b+c} & \text{1}  \\ \text{-1} & \text{1} & \text{0}  \\ \text{1} & \text{-1} & \text{0}  \\ \end{matrix} \right|\]

Applying the row operation \[{{\text{R}}_{3}}\to {{\text{R}}_{3}}\text{+}{{\text{R}}_{2}}\]

\[\Rightarrow \Delta \text{=}\dfrac{\text{1}}{\text{2}}\left( \text{a-b} \right)\left( \text{c-a} \right)\left| \begin{matrix} \text{a} & \text{b+c} & \text{1}  \\ \text{-1} & \text{1} & \text{0}  \\ \text{0} & \text{0} & \text{0}  \\ \end{matrix} \right|\]

Since all the elements of the last row of the matrix are zero then the value of the determinant will be $0$.

\[\therefore \Delta \text{=0}\]                                           

Thus, the area of the triangle formed by points \[\text{A}\] , \[\text{B}\] and \[\text{C}\] is zero.

Hence, the points \[\text{A}\left( \text{a,b+c} \right)\text{,B}\left( \text{b,c +a} \right)\text{,C}\left( \text{c,a +b} \right)\] are collinear.

3. Find values of \[\mathbf{\text{k}}\] if area of triangle is \[\mathbf{\text{4}}\] square units and vertices are

1. \[\mathbf{\left( \text{k,0} \right)\text{,}\left( \text{4,0} \right)\text{,}\left( \text{0,2} \right)}\]

Ans: Given vertices are \[\left( \text{k,0} \right)\text{,}\left( \text{4,0} \right)\text{,}\left( \text{0,2} \right)\].

We know, if we have three points $\left( {{x}_{1}},{{y}_{1}} \right)$, $\left( {{x}_{2}},{{y}_{2}} \right)$ and $\left( {{x}_{3}},{{y}_{3}} \right)$, then the area of the triangle is given by,

$\Delta =\left| \begin{matrix} {{x}_{1}} & {{y}_{1}} & 1  \\ {{x}_{2}} & {{y}_{2}} & 1  \\ {{x}_{3}} & {{y}_{3}} & 1  \\ \end{matrix} \right|$

Thus, the area of the triangle is given by,

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}\dfrac{\text{1}}{\text{2}}\left| \begin{matrix} \text{k} & \text{0} & \text{1}  \\ \text{4} & \text{0} & \text{1}  \\ \text{0} & \text{2} & \text{1}  \\ \end{matrix} \right|\]

\[\Rightarrow \Delta \text{=}\dfrac{\text{1}}{\text{2}}\left[ \text{k}\left( \text{0-2} \right)\text{-0}\left( \text{4-0} \right)\text{+1}\left( \text{8-0} \right) \right]\]

\[\Rightarrow \Delta \text{=}\dfrac{\text{1}}{\text{2}}\left[ \text{-2k+8} \right]\]

\[\therefore \Delta =-k+4\]

Since the area is given to be \[\text{4}\] square units, thus

$-k+4=\pm 4$

When \[-k+4=-4\]

\[\therefore k=8\].

When \[-k+4=4\]

\[\therefore k=0\].

Hence, \[\text{k=0,8}\].

2. \[\mathbf{\text{(-2,0),}\left( \text{0,4} \right)\text{,}\left( \text{0,k} \right)}\]

Ans: Given vertices are \[\text{(-2,0),}\left( \text{0,4} \right)\text{,}\left( \text{0,k} \right)\].

We know, if we have three points $\left( {{x}_{1}},{{y}_{1}} \right)$, $\left( {{x}_{2}},{{y}_{2}} \right)$ and $\left( {{x}_{3}},{{y}_{3}} \right)$, then the area of the triangle is given by,

$\Delta =\left| \begin{matrix} {{x}_{1}} & {{y}_{1}} & 1  \\ {{x}_{2}} & {{y}_{2}} & 1  \\ {{x}_{3}} & {{y}_{3}} & 1  \\ \end{matrix} \right|$

The area of the triangle is given by,

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}\dfrac{\text{1}}{\text{2}}\left| \begin{matrix} -2 & \text{0} & \text{1}  \\ \text{0} & \text{4} & \text{1}  \\ \text{0} & \text{k} & \text{1}  \\ \end{matrix} \right|\]

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}\dfrac{\text{1}}{\text{2}}\left[ -2\left( 4-k \right) \right]\]

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ }=k-4\]

Since the area is given to be \[\text{4}\] square units, thus

\[k-4=\pm 4\]

When \[k-4=-4\]

\[\therefore k=0\].

When \[k-4=4\]

\[\therefore k=8\].

Hence, \[k=0,8\].

3. Determine the following:

1. Find equation of line joining (1,2) and (3,6) using determinants.

Ans: Let us assume a point, \[\text{P}\left( \text{x, y} \right)\] on the line joining points \[\text{A}\left( \text{1,2} \right)\] and \[\text{B}\left( \text{3,6} \right)\] .

Then, the point \[\text{A}\],$B$ and $P$ are collinear.

Thus, the area of triangle \[\text{ABP}\] will be zero.

\[\therefore \dfrac{\text{1}}{\text{2}}\left| \begin{matrix} \text{1} & \text{2} & \text{1}  \\ \text{3} & \text{6} & \text{1}  \\ \text{x} & \text{y} & \text{1}  \\ \end{matrix} \right|\text{=0}\]

\[\Rightarrow \dfrac{\text{1}}{\text{2}}\left[ \text{1}\left( \text{6-y} \right)\text{-2}\left( \text{3-x} \right)\text{+1}\left( \text{3y-6x} \right) \right]\text{=0}\]

\[\Rightarrow \text{6-y-6+2x+3y-6x=0}\]

\[\Rightarrow \text{2y-4x=0}\]

\[\Rightarrow \text{y=2x}\]

$\therefore $ The equation of the line joining the given points is \[y=2x\] .

2. Find equation of line joining (3,1) and (9,3) using determinants.

Ans: Let us assume a point, \[\text{P}\left( \text{x, y} \right)\] on the line joining points \[\text{A}\left( \text{3,1} \right)\] and \[\text{B}\left( \text{9,3} \right)\]. 

Then, the point \[\text{A}\],$B$ and $P$ are collinear.

Thus, the area of the triangle \[\text{ABP}\] will be zero.

\[\therefore \dfrac{1}{2}\left|\begin{matrix} \text{3}&\text{1}&\text{1}\\ \text{9}&\text{3}&\text{1}\\ \text{x}&\text{y}&\text{1}\\ \end{matrix}  \right|=0\]

\[\Rightarrow \dfrac{1}{2}\left[ 3(3-\text{y})-1(9-\text{x})+1(9\text{y}-3\text{x})\right]=0\]

\[\Rightarrow 9-3\text{y}-9+\text{x}+9\text{y}-3\text{x}=0\]

\[\Rightarrow 6\text{y}-2\text{x}=0\]

\[\Rightarrow \text{x}-3\text{y}=0\]

\[\therefore\] The equation of the line joining the given points is \[\text{x}-3\text{y}=0\].

4. If the area of triangle is \[\mathbf{\text{35}}\] square units with vertices \[\mathbf{\text{(2,-6)}}\] , \[\mathbf{\left( \text{5,4} \right)}\] and \[\mathbf{\left( \text{k,4} \right)}\] . Then \[\mathbf{\text{k}}\] is

1. \[\mathbf{\text{12}}\]

2. \[\mathbf{\text{-2}}\]

3. \[\mathbf{\text{-12,-2}}\]

4. \[\mathbf{\text{12,-2}}\]

Ans: Given vertices, \[\text{(2,-6)}\],\[\left( \text{5,4} \right)\] and \[\left( \text{k,4} \right)\]

We know, if we have three points $\left( {{x}_{1}},{{y}_{1}} \right)$, $\left( {{x}_{2}},{{y}_{2}} \right)$ and $\left( {{x}_{3}},{{y}_{3}} \right)$, then the area of the triangle is given by,

$\Delta =\left| \begin{matrix} {{x}_{1}} & {{y}_{1}} & 1  \\ {{x}_{2}} & {{y}_{2}} & 1  \\ {{x}_{3}} & {{y}_{3}} & 1  \\ \end{matrix} \right|$

The area of the triangle is given by,

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}\dfrac{\text{1}}{\text{2}}\left| \begin{matrix} \text{2} & \text{-6} & \text{1}  \\ \text{5} & \text{4} & \text{1}  \\ \text{k} & \text{4} & \text{1}  \\ \end{matrix} \right|\]

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}\dfrac{\text{1}}{\text{2}}\left[ \text{2}\left( \text{4-4} \right)\text{+6}\left( \text{5-k} \right)\text{+1}\left( \text{20-4k} \right) \right]\]

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}\dfrac{\text{1}}{\text{2}}\left[ \text{50-10k} \right]\]

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =25-5k}\]

Given, the area of the triangle is \[\text{35}\] square units .

Thus, we have:

\[\Rightarrow 25-5k=\pm 35\]

\[\Rightarrow 5\left( 5-k \right)=\pm 35\]

\[\Rightarrow 5-k=\pm 7\].

When \[5-k=7\]

\[\therefore k=-2\].

When \[5-k=-7\]

\[\therefore k=12\].

Hence, \[k=12,-2\] .

Thus, D. \[12,-2\] is the correct option.

Exercise (4.3)

1. Write Minors and Cofactors of the elements of following determinants:

1. \[\mathbf{\left| \begin{matrix} \text{2} & \text{-4}  \\ \text{0} & \text{3}  \\ \end{matrix} \right|}\]

Ans: Given, \[\left| \begin{matrix} \text{2} & \text{-4}  \\ \text{0} & \text{3}  \\ \end{matrix} \right|\]

Minor of an element is termed as the determinant obtained by removing the row and the column in which that element is present.

Minor of element \[{{\text{a}}_{ij}}\] is denoted by \[{{M}_{ij}}\],where

$i$ and $j$ denotes the row and the column of the determinant respectively.

\[\therefore {{\text{M}}_{11}}\text{=3}\]

\[{{\text{M}}_{12}}\text{=0}\]

\[{{\text{M}}_{21}}\text{=-4}\]

\[{{\text{M}}_{22}}\text{=2}\]

Cofactor of an element is termed as the determinant obtained by removing the row and the column in which that element is present preceded by a negative or a positive sign based on the position of the element.

Thus,

Cofactor of \[{{\text{a}}_{ij}}\] is \[{{\text{A}}_{ij}}\text{=}{{\left( \text{-1} \right)}^{\text{i+j}}}{{\text{M}}_{ij}}\]

\[\Rightarrow {{\text{A}}_{11}}\text{=}{{\left( \text{-1} \right)}^{\text{1+1}}}{{\text{M}}_{11}}\]

\[\Rightarrow {{\text{A}}_{11}}={{\left( \text{-1} \right)}^{\text{2}}}\left( \text{3} \right)\]

\[\therefore {{\text{A}}_{11}}\text{=3}\]

Similarly,

\[\Rightarrow {{\text{A}}_{12}}\text{=}{{\left( \text{-1} \right)}^{\text{1+2}}}{{\text{M}}_{12}}\]

\[\Rightarrow {{\text{A}}_{12}}\text{=}{{\left( \text{-1} \right)}^{\text{3}}}\left( \text{0} \right)\]

\[\therefore {{\text{A}}_{12}}\text{=0}\]

\[\Rightarrow {{\text{A}}_{21}}\text{=}{{\left( \text{-1} \right)}^{\text{2+1}}}{{M}_{21}}\]

\[\Rightarrow {{\text{A}}_{21}}\text{=}{{\left( \text{-1} \right)}^{\text{3}}}\left( \text{-4} \right)\]

\[\therefore {{\text{A}}_{21}}\text{=4}\]

\[\Rightarrow {{\text{A}}_{22}}\text{=}{{\left( \text{-1} \right)}^{\text{2+2}}}{{M}_{22}}\]

\[\Rightarrow {{\text{A}}_{22}}\text{=}{{\left( \text{-1} \right)}^{\text{4}}}\left( \text{2} \right)\]

\[\therefore {{\text{A}}_{22}}\text{=2}\]

2. \[\mathbf{\left| \begin{matrix} \text{a} & \text{c}  \\ \text{b} & \text{d}  \\ \end{matrix} \right|}\]

Ans: Given, \[\left| \begin{matrix} \text{a} & \text{c}  \\ \text{b} & \text{d}  \\ \end{matrix} \right|\]

Minor of element \[{{\text{a}}_{ij}}\] is denoted by \[{{M}_{ij}}\].

\[\therefore {{\text{M}}_{11}}\text{=d}\]

\[{{\text{M}}_{12}}\text{=b}\]

\[{{\text{M}}_{21}}\text{=c}\]

\[{{\text{M}}_{22}}\text{=a}\]

Cofactor of \[{{\text{a}}_{ij}}\] is \[{{\text{A}}_{ij}}\text{=}{{\left( \text{-1} \right)}^{\text{i+j}}}{{\text{M}}_{ij}}\]

\[\Rightarrow {{\text{A}}_{11}}\text{=}{{\left( \text{-1} \right)}^{\text{1+1}}}{{\text{M}}_{11}}\]

\[\Rightarrow {{\text{A}}_{11}}={{\left( \text{-1} \right)}^{\text{2}}}\left( d \right)\]

\[\therefore {{\text{A}}_{11}}\text{=d}\]

Similarly,

\[\Rightarrow {{\text{A}}_{12}}\text{=}{{\left( \text{-1} \right)}^{\text{1+2}}}{{\text{M}}_{12}}\]

\[\Rightarrow {{\text{A}}_{12}}\text{=}{{\left( \text{-1} \right)}^{\text{3}}}\left( b \right)\]

\[\therefore {{\text{A}}_{12}}\text{=}-b\]

\[\Rightarrow {{\text{A}}_{21}}\text{=}{{\left( \text{-1} \right)}^{\text{2+1}}}{{M}_{21}}\]

\[\Rightarrow {{\text{A}}_{21}}\text{=}{{\left( \text{-1} \right)}^{\text{3}}}\left( c \right)\]

\[\therefore {{\text{A}}_{21}}\text{=}-\text{c}\]

\[\Rightarrow {{\text{A}}_{22}}\text{=}{{\left( \text{-1} \right)}^{\text{2+2}}}{{M}_{22}}\]

\[\Rightarrow {{\text{A}}_{22}}\text{=}{{\left( \text{-1} \right)}^{\text{4}}}\left( a \right)\]

\[\therefore {{\text{A}}_{22}}\text{=a}\]

2. Write Minors and Cofactors of the elements of following determinants:

1. \[\mathbf{\left| \begin{matrix} \text{1} & \text{0} & \text{0}  \\ \text{0} & \text{1} & \text{0}  \\ \text{0} & \text{0} & \text{1}  \\ \end{matrix} \right|}\]

Ans: Given determinant, \[\left| \begin{matrix} \text{1} & \text{0} & \text{0}  \\ \text{0} & \text{1} & \text{0}  \\ \text{0} & \text{0} & \text{1}  \\ \end{matrix} \right|\].

Minor of element \[{{\text{a}}_{ij}}\] is denoted by \[{{M}_{ij}}\].

\[\therefore {{\text{M}}_{11}}\text{=}\left| \begin{matrix} \text{1} & \text{0}  \\ \text{0} & \text{1}  \\ \end{matrix} \right|\text{=1}\]

\[\Rightarrow {{\text{M}}_{12}}\text{=}\left| \begin{matrix} \text{0} & \text{0}  \\ \text{0} & \text{1}  \\ \end{matrix} \right|\text{=0}\]

\[\Rightarrow {{\text{M}}_{13}}\text{=}\left| \begin{matrix} \text{0} & \text{1}  \\ \text{0} & \text{0}  \\ \end{matrix} \right|\text{=0}\]

\[\Rightarrow {{\text{M}}_{21}}\text{=}\left| \begin{matrix} \text{0} & \text{0}  \\ \text{0} & \text{1}  \\ \end{matrix} \right|\text{=0}\]

\[\Rightarrow {{\text{M}}_{22}}\text{=}\left| \begin{matrix} \text{1} & \text{0}  \\ \text{0} & \text{1}  \\ \end{matrix} \right|\text{=1}\]

\[\Rightarrow {{\text{M}}_{23}}\text{=}\left| \begin{matrix} \text{1} & \text{0}  \\ \text{0} & \text{0}  \\ \end{matrix} \right|\text{=0}\]

\[\Rightarrow {{\text{M}}_{31}}\text{=}\left| \begin{matrix} \text{0} & \text{0}  \\ \text{1} & \text{0}  \\ \end{matrix} \right|\text{=0}\]

\[\Rightarrow {{\text{M}}_{32}}\text{=}\left| \begin{matrix} \text{1} & \text{0}  \\ \text{0} & \text{0}  \\ \end{matrix} \right|\text{=0}\]

\[\Rightarrow {{\text{M}}_{33}}\text{=}\left| \begin{matrix} \text{1} & \text{0}  \\ \text{0} & \text{1}  \\ \end{matrix} \right|\text{=1}\]

Cofactor of \[{{\text{a}}_{ij}}\] is \[{{A}_{ij}}\text{=}{{\left( \text{-1} \right)}^{\text{i+j}}}{{M}_{ij}}\]. 

\[\Rightarrow {{\text{A}}_{11}}\text{=}{{\left( \text{-1} \right)}^{1+1}}{{\text{M}}_{11}}=1\]

\[\Rightarrow {{\text{A}}_{12}}\text{=}{{\left( \text{-1} \right)}^{1+2}}{{\text{M}}_{12}}=0\]

\[\Rightarrow {{\text{A}}_{13}}\text{=}{{\left( \text{-1} \right)}^{1+3}}{{\text{M}}_{13}}=0\]

\[\Rightarrow {{\text{A}}_{21}}\text{=}{{\left( \text{-1} \right)}^{2+1}}{{\text{M}}_{21}}=0\]

\[\Rightarrow {{\text{A}}_{22}}\text{=}{{\left( \text{-1} \right)}^{2+2}}{{\text{M}}_{22}}=1\]

\[\Rightarrow {{\text{A}}_{23}}\text{=}{{\left( \text{-1} \right)}^{2+3}}{{\text{M}}_{23}}=0\]

\[\Rightarrow {{\text{A}}_{31}}\text{=}{{\left( \text{-1} \right)}^{3+1}}{{\text{M}}_{31}}=0\]

\[\Rightarrow {{\text{A}}_{32}}\text{=}{{\left( \text{-1} \right)}^{3+2}}{{\text{M}}_{32}}=0\]

\[\Rightarrow {{\text{A}}_{33}}\text{=}{{\left( \text{-1} \right)}^{3+3}}{{\text{M}}_{33}}=1\]

2. \[\mathbf{\left| \begin{matrix} \text{1} & \text{0} & \text{4}  \\ \text{3} & \text{5} & \text{-1}  \\ \text{0} & \text{1} & \text{2}  \\ \end{matrix} \right|}\]

Ans: Given determinant, \[\left| \begin{matrix} \text{1} & \text{0} & \text{4}  \\ \text{3} & \text{5} & \text{-1}  \\ \text{0} & \text{1} & \text{2}  \\ \end{matrix} \right|\]

Minor of element \[{{\text{a}}_{ij}}\] is denoted by \[{{M}_{ij}}\].

\[\Rightarrow {{\text{M}}_{11}}\text{=}\left| \begin{matrix} \text{5} & \text{-1}  \\ \text{1} & \text{2}  \\ \end{matrix} \right|\text{=10+1=11}\]

\[\Rightarrow {{\text{M}}_{12}}\text{=}\left| \begin{matrix} \text{3} & \text{-1}  \\ \text{0} & \text{2}  \\ \end{matrix} \right|\text{=6-0=6}\]

\[\Rightarrow {{\text{M}}_{13}}\text{=}\left| \begin{matrix} \text{3} & \text{5}  \\ \text{0} & \text{1}  \\ \end{matrix} \right|\text{=3-0=3}\]

\[\Rightarrow {{\text{M}}_{21}}\text{=}\left| \begin{matrix} \text{0} & \text{4}  \\ \text{1} & \text{2}  \\ \end{matrix} \right|\text{=0-4=-4}\]

\[\Rightarrow {{\text{M}}_{22}}\text{=}\left| \begin{matrix} \text{1} & \text{4}  \\ \text{0} & \text{2}  \\ \end{matrix} \right|\text{=2-0=2}\]

\[\Rightarrow {{\text{M}}_{23}}\text{=}\left| \begin{matrix} \text{1} & \text{0}  \\ \text{0} & \text{1}  \\ \end{matrix} \right|\text{=1-0=1}\]

\[\Rightarrow {{\text{M}}_{31}}\text{=}\left| \begin{matrix} \text{0} & \text{4}  \\ \text{5} & \text{-1}  \\ \end{matrix} \right|\text{=0-20=-20}\]

\[\Rightarrow {{\text{M}}_{32}}\text{=}\left| \begin{matrix} \text{1} & \text{4}  \\ \text{3} & \text{-1}  \\ \end{matrix} \right|\text{=-1-12=-13}\]

\[\Rightarrow {{\text{M}}_{33}}\text{=}\left| \begin{matrix} \text{1} & \text{0}  \\ \text{3} & \text{5}  \\ \end{matrix} \right|\text{=5-0=5}\]

Cofactor of \[{{\text{a}}_{ij}}\] is \[{{A}_{ij}}\text{=}{{\left( \text{-1} \right)}^{\text{i+j}}}{{M}_{ij}}\]. 

\[\Rightarrow {{\text{A}}_{11}}\text{=}{{\left( \text{-1} \right)}^{1+1}}{{\text{M}}_{11}}=11\]

\[\Rightarrow {{\text{A}}_{12}}\text{=}{{\left( \text{-1} \right)}^{1+2}}{{\text{M}}_{12}}=6\]

\[\Rightarrow {{\text{A}}_{13}}\text{=}{{\left( \text{-1} \right)}^{1+3}}{{\text{M}}_{13}}=3\]

\[\Rightarrow {{\text{A}}_{21}}\text{=}{{\left( \text{-1} \right)}^{2+1}}{{\text{M}}_{21}}=-4\]

\[\Rightarrow {{\text{A}}_{22}}\text{=}{{\left( \text{-1} \right)}^{2+2}}{{\text{M}}_{22}}=2\]

\[\Rightarrow {{\text{A}}_{23}}\text{=}{{\left( \text{-1} \right)}^{2+3}}{{\text{M}}_{23}}=1\]

\[\Rightarrow {{\text{A}}_{31}}\text{=}{{\left( \text{-1} \right)}^{3+1}}{{\text{M}}_{31}}=-20\]

\[\Rightarrow {{\text{A}}_{32}}\text{=}{{\left( \text{-1} \right)}^{3+2}}{{\text{M}}_{32}}=-13\]

\[\Rightarrow {{\text{A}}_{33}}\text{=}{{\left( \text{-1} \right)}^{3+3}}{{\text{M}}_{33}}=5\]

3. Using Cofactors of elements of second row, evaluate \[\mathbf{\text{ }\!\!\Delta\!\!\text{ =}\left| \begin{matrix} \text{5} & \text{3} & \text{8}  \\ \text{2} & \text{0} & \text{1}  \\ \text{1} & \text{2} & \text{3}  \\ \end{matrix} \right|}\].

Ans: Given determinant, \[\left| \begin{matrix} \text{5} & \text{3} & \text{8}  \\ \text{2} & \text{0} & \text{1}  \\ \text{1} & \text{2} & \text{3}  \\ \end{matrix} \right|\]

Determining the minors and cofactors, we get:

\[\Rightarrow {{\text{M}}_{21}}\text{=}\left| \begin{matrix} \text{3} & \text{8}  \\ \text{2} & \text{3}  \\ \end{matrix} \right|=9-16=-7\]

\[\therefore {{\text{A}}_{21}}\text{=}{{\left( \text{-1} \right)}^{\text{2+1}}}{{M}_{21}}\text{=7}\]

\[\Rightarrow {{\text{M}}_{22}}\text{=}\left| \begin{matrix} \text{5} & \text{8}  \\ \text{1} & \text{3}  \\ \end{matrix} \right|=15-8=7\]

\[\therefore {{\text{A}}_{22}}\text{=}{{\left( \text{-1} \right)}^{\text{2+2}}}{{M}_{22}}\text{=7}\]

\[\Rightarrow {{\text{M}}_{23}}\text{=}\left| \begin{matrix} \text{5} & \text{3}  \\ \text{1} & \text{2}  \\ \end{matrix} \right|=10-3=7\]

\[\therefore {{\text{A}}_{23}}\text{=}{{\left( \text{-1} \right)}^{\text{2+1}}}{{M}_{23}}=-7\]

Since, \[\text{ }\!\!\Delta\!\!\text{ }\] is equal to the sum of the product of the elements of the second row with their corresponding cofactors.

\[\therefore \Delta ={{a}_{21}}{{A}_{21}}+{{a}_{22}}{{A}_{22}}+{{a}_{23}}{{A}_{23}}\]

\[\Rightarrow \Delta \text{=2}\left( \text{7} \right)\text{+0}\left( \text{7} \right)\text{+1}\left( \text{7} \right)\]

Hence, \[\Delta \text{=21}\].

4. Using Cofactors of elements of third column, evaluate  \[\mathbf{\text{ }\!\!\Delta\!\!\text{ =}\left| \begin{matrix} \text{1} & \text{x} & \text{yz}  \\ \text{1} & \text{y} & \text{zx}  \\ \text{1} & \text{z} & \text{xy}  \\ \end{matrix} \right|}\].

Ans: Given determinant, \[\left| \begin{matrix} \text{1} & \text{x} & \text{yz}  \\ \text{1} & \text{y} & \text{zx}  \\ \text{1} & \text{z} & \text{xy}  \\ \end{matrix} \right|\]

Determining the minors and cofactors, we get:

\[\Rightarrow {{\text{M}}_{13}}\text{=}\left| \begin{matrix} 1 & y  \\ 1 & z  \\ \end{matrix} \right|=z-y\]

\[\therefore {{\text{A}}_{13}}\text{=}{{\left( \text{-1} \right)}^{\text{1+3}}}{{M}_{13}}\text{=}\left( z-y \right)\]

\[\Rightarrow {{\text{M}}_{23}}\text{=}\left| \begin{matrix} 1 & x  \\ \text{1} & z  \\ \end{matrix} \right|=z-x\]

\[\therefore {{\text{A}}_{23}}\text{=}{{\left( \text{-1} \right)}^{\text{2+3}}}{{M}_{23}}\text{=}\left( x-z \right)\]

\[\Rightarrow {{\text{M}}_{33}}\text{=}\left| \begin{matrix} 1 & x  \\ \text{1} & y  \\ \end{matrix} \right|=y-x\]

\[\therefore {{\text{A}}_{33}}\text{=}{{\left( \text{-1} \right)}^{\text{3+3}}}{{M}_{33}}=\left( y-x \right)\]

Since, \[\text{ }\!\!\Delta\!\!\text{ }\] is equal to the sum of the product of the elements of the first row with their corresponding cofactors. 

\[\therefore \Delta ={{a}_{13}}{{A}_{13}}+{{a}_{23}}{{A}_{23}}+{{a}_{33}}{{A}_{33}}\]

\[\Rightarrow \Delta =yz\left( z-y \right)+zx\left( x-z \right)+xy\left( y-x \right)\]

\[\Rightarrow \Delta =y{{z}^{2}}-{{y}^{2}}z+{{x}^{2}}z-x{{z}^{2}}+x{{y}^{2}}-{{x}^{2}}y\]

\[\Rightarrow \Delta =\left( {{x}^{2}}z-{{y}^{2}}z \right)+\left( y{{z}^{2}}-x{{z}^{2}} \right)+\left( x{{y}^{2}}-{{x}^{2}}y \right)\]

\[\Rightarrow \Delta =\left( x-y \right)\left[ zx+zy-{{z}^{2}}-xy \right]\]

\[\Rightarrow \Delta =\left( x-y \right)\left[ z\left( x-z \right)+y\left( z-x \right) \right]\]

Thus, \[\text{ }\!\!\Delta\!\!\text{ =}\left( \text{x-y} \right)\left( \text{y-z} \right)\left( \text{z-x} \right)\].

5. For the matrices \[\mathbf{\text{A}}\] and \[\mathbf{\text{B}}\] , verify that \[\mathbf{\left( \text{AB} \right)\text{ }\!\!'\!\!\text{ =B }\!\!'\!\!\text{ A }\!\!'\!\!\text{ }}\] where

If $\mathbf{\Delta =\left| \begin{matrix} {{a}_{11}} & {{a}_{12}} & {{a}_{13}}  \\ {{a}_{21}} & {{a}_{22}} & {{a}_{23}}  \\ {{a}_{31}} & {{a}_{32}} & {{a}_{33}}  \\ \end{matrix} \right|}$ and ${{A}_{ij}}$ is Cofactors of ${{a}_{ij}}$ , then value of $\Delta $ is given by

1. \[\mathbf{{{a}_{11}}{{A}_{31}}+{{a}_{12}}{{A}_{32}}+{{a}_{13}}{{A}_{33}}}\]

2. \[\mathbf{{{a}_{11}}{{A}_{11}}+{{a}_{12}}{{A}_{21}}+{{a}_{13}}{{A}_{31}}}\]

3. \[\mathbf{{{a}_{21}}{{A}_{11}}+{{a}_{22}}{{A}_{12}}+{{a}_{23}}{{A}_{13}}}\]

4. \[\mathbf{{{a}_{11}}{{A}_{11}}+{{a}_{21}}{{A}_{21}}+{{a}_{31}}{{A}_{31}}}\]

Ans: It is given that $\Delta =\left| \begin{matrix} {{a}_{11}} & {{a}_{12}} & {{a}_{13}}  \\ {{a}_{21}} & {{a}_{22}} & {{a}_{23}}  \\ {{a}_{31}} & {{a}_{32}} & {{a}_{33}}  \\ \end{matrix} \right|$.

The value of $\Delta $ by expanding along first column is obtained as,

\[{{a}_{11}}\left( {{a}_{22}}.{{a}_{33}}-{{a}_{23}}.{{a}_{32}} \right)-{{a}_{21}}\left( {{a}_{12}}.{{a}_{33}}-{{a}_{13}}.{{a}_{32}} \right)+{{a}_{31}}\left( {{a}_{12}}.{{a}_{23}}-{{a}_{13}}.{{a}_{22}} \right)\] ….. 1

Now, the cofactor ${{A}_{ij}}$ of element ${{a}_{ij}}$ is given by ${{\left( -1 \right)}^{i+j}}{{M}_{ij}}$, where ${{M}_{ij}}$ is the minor. Minor is the determinant obtained by cancelling the ith row and jth column of the original matrix.

Now for element ${{a}_{11}}$, the minor is ${{M}_{11}}=\left| \begin{matrix} {{a}_{22}} & {{a}_{23}}  \\ {{a}_{32}} & {{a}_{33}}  \\ \end{matrix}\right|={{a}_{22}}.{{a}_{33}}-{{a}_{23}}.{{a}_{32}}$and the cofactor is ${{A}_{11}}={{\left( -1 \right)}^{1+1}}\left( {{a}_{22}}.{{a}_{33}}-{{a}_{23}}.{{a}_{32}} \right)\Rightarrow {{A}_{11}}=\left( {{a}_{22}}.{{a}_{33}}-{{a}_{23}}.{{a}_{32}} \right)$.

Next for element ${{a}_{21}}$, the minor is ${{M}_{21}}=\left| \begin{matrix} {{a}_{12}} & {{a}_{13}}  \\ {{a}_{32}} & {{a}_{33}}  \\ \end{matrix}\right|={{a}_{12}}.{{a}_{33}}-{{a}_{13}}.{{a}_{32}}$ and the cofactor is ${{A}_{21}}={{\left( -1 \right)}^{2+1}}\left( {{a}_{12}}.{{a}_{33}}-{{a}_{13}}.{{a}_{32}} \right)\Rightarrow {{A}_{21}}=-\left( {{a}_{12}}.{{a}_{33}}-{{a}_{13}}.{{a}_{32}} \right)$.

Next for element ${{a}_{31}}$, the minor is ${{M}_{31}}=\left| \begin{matrix} {{a}_{12}} & {{a}_{13}}  \\ {{a}_{22}} & {{a}_{23}}  \\ \end{matrix}\right|={{a}_{12}}.{{a}_{23}}-{{a}_{13}}.{{a}_{22}}$ and the cofactor is ${{A}_{31}}={{\left( -1 \right)}^{3+1}}\left( {{a}_{12}}.{{a}_{23}}-{{a}_{13}}.{{a}_{22}} \right)\Rightarrow {{A}_{31}}=\left( {{a}_{12}}.{{a}_{23}}-{{a}_{13}}.{{a}_{22}} \right)$.

Now substituting the terms as obtained from above computation in equation 1,

\[{{a}_{11}}{{A}_{11}}-{{a}_{21}}\left( -{{A}_{21}} \right)+{{a}_{31}}{{A}_{31}}\]

\[{{a}_{11}}{{A}_{11}}+{{a}_{21}}{{A}_{21}}+{{a}_{31}}{{A}_{31}}\]

This matches with option d.

Exercise (4.4)

1. Find the adjoint of each of the matrices. \[\mathbf{\left[ \begin{matrix} \text{1} & \text{2}  \\ \text{3} & \text{4}  \\ \end{matrix} \right]}\]

Ans: Let \[\text{A=}\left[ \begin{matrix} \text{1} & \text{2}  \\ \text{3} & \text{4}  \\ \end{matrix} \right]\]

Since, Cofactor of \[{{\text{a}}_{ij}}\] is \[{{\text{A}}_{ij}}\text{=}{{\left( \text{-1} \right)}^{\text{i+j}}}{{\text{M}}_{ij}}\].

Thus,

\[\Rightarrow {{\text{A}}_{11}}\text{=}{{\left( \text{-1} \right)}^{1+1}}{{\text{M}}_{11}}={{\left( \text{-1} \right)}^{2}}{{\text{M}}_{11}}\]

\[\Rightarrow {{A}_{11}}\text{=4}\]

\[\Rightarrow {{\text{A}}_{12}}\text{=}{{\left( \text{-1} \right)}^{1+2}}{{\text{M}}_{12}}={{\left( \text{-1} \right)}^{3}}{{\text{M}}_{12}}\]

\[\Rightarrow {{A}_{12}}=-3\]

\[\Rightarrow {{\text{A}}_{21}}\text{=}{{\left( \text{-1} \right)}^{2+1}}{{\text{M}}_{21}}={{\left( \text{-1} \right)}^{3}}{{\text{M}}_{21}}\]

\[\Rightarrow {{\text{A}}_{21}}\text{=}-2\]

\[\Rightarrow {{\text{A}}_{22}}\text{=}{{\left( \text{-1} \right)}^{2+2}}{{\text{M}}_{22}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{22}}\]

\[\Rightarrow {{\text{A}}_{22}}=1\]

We know that adjoint of a matrix is the transpose of its cofactor matrix.

Thus,\[\text{adjA=}{{\left[ \begin{matrix} {{\text{A}}_{11}} & {{\text{A}}_{12}}  \\ {{\text{A}}_{21}} & {{\text{A}}_{22}}  \\ \end{matrix} \right]}^{T}}\]

\[\therefore adjA\text{=}\left[ \begin{matrix} \text{4} & \text{-2}  \\ \text{-3} & \text{1}  \\ \end{matrix} \right]\].

2. Find adjoint of each of the matrices \[\mathbf{\left[ \begin{matrix} \text{1} & \text{-1} & \text{2}  \\ \text{2} & \text{3} & \text{5}  \\ \text{-2} & \text{0} & \text{1}  \\ \end{matrix} \right]}\].

Ans: Let \[\text{A=}\left[ \begin{matrix} \text{1} & \text{-1} & \text{2}  \\ \text{2} & \text{3} & \text{5}  \\ \text{-2} & \text{0} & \text{1}  \\ \end{matrix} \right]\]

Since, Cofactor of \[{{\text{a}}_{ij}}\] is \[{{\text{A}}_{ij}}\text{=}{{\left( \text{-1} \right)}^{\text{i+j}}}{{\text{M}}_{ij}}\].

Thus,

\[\Rightarrow {{\text{A}}_{11}}\text{=}{{\left( \text{-1} \right)}^{1+1}}{{\text{M}}_{11}}={{\left( \text{-1} \right)}^{2}}{{\text{M}}_{11}}\]

\[\Rightarrow {{A}_{11}}\text{=}\left| \begin{matrix} \text{3} & \text{5}  \\ \text{0} & \text{1}  \\ \end{matrix} \right|=3-0=3\]

\[\Rightarrow {{\text{A}}_{12}}\text{=}{{\left( \text{-1} \right)}^{1+2}}{{\text{M}}_{12}}={{\left( \text{-1} \right)}^{3}}{{\text{M}}_{12}}\]

\[\Rightarrow {{A}_{12}}=\left| \begin{matrix} \text{2} & \text{5}  \\ \text{-2} & \text{1}  \\ \end{matrix} \right|=-\left( \text{2+10} \right)=-12\]

\[\Rightarrow {{\text{A}}_{13}}\text{=}{{\left( \text{-1} \right)}^{1+3}}{{\text{M}}_{13}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{13}}\]

\[\Rightarrow {{A}_{13}}=\left| \begin{matrix} 2 & 3  \\ -2 & 0  \\ \end{matrix} \right|\text{=0+6=6}\]

Similarly,

\[\Rightarrow {{\text{A}}_{21}}\text{=}{{\left( \text{-1} \right)}^{2+1}}{{\text{M}}_{21}}={{\left( \text{-1} \right)}^{3}}{{\text{M}}_{21}}\]

\[\Rightarrow {{\text{A}}_{21}}\text{=}\left| \begin{matrix} \text{-1} & \text{2}  \\ \text{0} & \text{1}  \\ \end{matrix} \right|=-\left( -1-0 \right)\text{=1}\]

\[\Rightarrow {{\text{A}}_{22}}\text{=}{{\left( \text{-1} \right)}^{2+2}}{{\text{M}}_{22}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{22}}\]

\[\Rightarrow {{\text{A}}_{22}}=\left| \begin{matrix} \text{1} & \text{2}  \\ -2 & \text{1}  \\ \end{matrix} \right|\text{=1+4=5}\]

\[\Rightarrow {{\text{A}}_{23}}\text{=}{{\left( \text{-1} \right)}^{2+3}}{{\text{M}}_{23}}={{\left( \text{-1} \right)}^{5}}{{\text{M}}_{23}}\]

\[\Rightarrow {{\text{A}}_{23}}\text{=}\left| \begin{matrix} \text{1} & \text{-1}  \\ \text{-2} & 0  \\ \end{matrix} \right|\text{=}\left( 0-2 \right)\text{=2}\]

and

\[\Rightarrow {{\text{A}}_{31}}\text{=}{{\left( \text{-1} \right)}^{3+1}}{{\text{M}}_{31}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{31}}\]

\[\Rightarrow {{\text{A}}_{31}}=\left| \begin{matrix} -1 & \text{2}  \\ \text{2} & \text{5}  \\ \end{matrix} \right|=-5-4=-9\]

\[\Rightarrow {{\text{A}}_{32}}\text{=}{{\left( \text{-1} \right)}^{3+2}}{{\text{M}}_{32}}={{\left( \text{-1} \right)}^{5}}{{\text{M}}_{32}}\]

\[\Rightarrow {{\text{A}}_{32}}=\left| \begin{matrix} \text{1} & \text{2}  \\ \text{2} & \text{5}  \\ \end{matrix} \right|=-\left( 5-4 \right)=-1\]

\[\Rightarrow {{\text{A}}_{33}}\text{=}{{\left( \text{-1} \right)}^{3+3}}{{\text{M}}_{33}}={{\left( \text{-1} \right)}^{6}}{{\text{M}}_{33}}\]

\[\Rightarrow {{\text{A}}_{33}}=\left| \begin{matrix} \text{1} & \text{-1}  \\ \text{2} & \text{3}  \\ \end{matrix} \right|\text{=3+2=5}\].

We know that adjoint of a matrix is the transpose of its cofactor matrix.

Thus, \[\text{adjA=}{{\left[ \begin{matrix} A11 & A\text{12} & A\text{13}  \\ A\text{21} & {{\text{A}}_{\text{22}}} & A23  \\ A\text{31} & {{\text{A}}_{\text{32}}} & A\text{33}  \\ \end{matrix} \right]}^{T}}\text{=}\left[ \begin{matrix} \text{3} & -12 & 6  \\ 1 & \text{5} & 2  \\ -9 & -1 & \text{5}  \\ \end{matrix} \right]\]

3. Verify \[\mathbf{\text{A}\left( \text{adjA} \right)\text{=}\left( \text{adjA} \right)\text{A=}\left| \text{A} \right|I}\]. \[\mathbf{\left[ \begin{matrix} \text{2} & \text{3}  \\ \text{-4} & \text{-6}  \\ \end{matrix} \right]}\]

Ans: Given,\[\text{A=}\,\left[ \begin{matrix} \text{2} & \text{3}  \\ \text{-4} & \text{-6}  \\ \end{matrix} \right]\]

\[\therefore \left| \text{A} \right|=-12-\left( -12 \right)\]

\[\Rightarrow \left| \text{A} \right|=0\]

Hence, \[\left| A \right|I=0\left[ \begin{matrix} 1 & 0  \\ 0 & 1  \\ \end{matrix} \right]\]

\[\Rightarrow \left| A \right|I=\left[ \begin{matrix} 0 & 0  \\ 0 & 0  \\ \end{matrix} \right]\]

Since, Cofactor of \[{{\text{a}}_{ij}}\] is \[{{\text{A}}_{ij}}\text{=}{{\left( \text{-1} \right)}^{\text{i+j}}}{{\text{M}}_{ij}}\].

Then,

\[\Rightarrow {{\text{A}}_{11}}\text{=}{{\left( \text{-1} \right)}^{1+1}}{{\text{M}}_{11}}={{\left( \text{-1} \right)}^{2}}{{\text{M}}_{11}}\]

\[\Rightarrow {{A}_{11}}\text{=}-6\]

\[\Rightarrow {{\text{A}}_{12}}\text{=}{{\left( \text{-1} \right)}^{1+2}}{{\text{M}}_{12}}={{\left( \text{-1} \right)}^{3}}{{\text{M}}_{12}}\]

\[\Rightarrow {{A}_{12}}=4\]

\[\Rightarrow {{\text{A}}_{21}}\text{=}{{\left( \text{-1} \right)}^{2+1}}{{\text{M}}_{21}}={{\left( \text{-1} \right)}^{3}}{{\text{M}}_{21}}\]

\[\Rightarrow {{\text{A}}_{21}}\text{=}-3\]

\[\Rightarrow {{\text{A}}_{22}}\text{=}{{\left( \text{-1} \right)}^{2+2}}{{\text{M}}_{22}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{22}}\]

\[\Rightarrow {{\text{A}}_{22}}=2\]

Cofactor matrix is $\left[ \begin{matrix} -6 & 4  \\ -3 & 2  \\ \end{matrix} \right]$.

We know that adjoint of a matrix is the transpose of its cofactor matrix.

Thus, \[\text{adj}\,\text{A=}\,\left[ \begin{matrix} -6 & -3  \\ 4 & \text{2}  \\ \end{matrix} \right]\]

Now, multiplying $A$ with its adjoint, we have:

\[\Rightarrow \text{A}\left( \text{adjA} \right)\text{=}\left[ \begin{matrix} \text{2} & \text{3}  \\ \text{-4} & \text{-6}  \\ \end{matrix} \right]\left[ \begin{matrix} -6 & -3  \\ 4 & \text{2}  \\ \end{matrix} \right]\]

\[\Rightarrow \text{A}\left( \text{adjA} \right)\text{=}\left[ \begin{matrix} -12+12 & -6+6  \\ 24-24 & 12-12  \\ \end{matrix} \right]\]

\[\therefore \text{A}\left( \text{adjA} \right)\text{=}\left[ \begin{matrix} \text{0} & \text{0}  \\ \text{0} & \text{0}  \\ \end{matrix} \right]\]

Similarly, multiplying \[\left( adjA \right)\] with \[A\], we get:

\[\Rightarrow \left( \text{adjA} \right)\text{A=}\left[ \begin{matrix} -6 & -3  \\ 4 & \text{2}  \\ \end{matrix} \right]\left[ \begin{matrix} \text{2} & \text{3}  \\ -4 & -6  \\ \end{matrix} \right]\]

\[\Rightarrow \left( \text{adjA} \right)\text{A=}\left[ \begin{matrix} -12+12 & -18+18  \\ 8-8 & 12-12  \\ \end{matrix} \right]\]

\[\therefore \left( \text{adjA} \right)\text{A=}\left[ \begin{matrix} \text{0} & \text{0}  \\ \text{0} & \text{0}  \\ \end{matrix} \right]\]

Thus, \[\text{A}\left( \text{adjA} \right)\text{=}\left( \text{adjA} \right)\text{A=}\left| \text{A} \right|I\]

Hence verified.

4. Verify \[\mathbf{\text{A}\left( \text{adjA} \right)\text{=}\left( \text{adjA} \right)\text{A=}\left| \text{A} \right|I}\]. \[\mathbf{\left[ \begin{matrix} \text{1} & -1 & \text{2}  \\ \text{3} & \text{0} & -2  \\ \text{1} & \text{0} & \text{3}  \\ \end{matrix} \right]}\].

Ans: Let \[\text{A=}\left[ \begin{matrix} \text{1} & \text{-1} & \text{2}  \\ \text{3} & \text{0} & \text{-2}  \\ \text{1} & \text{0} & \text{3}  \\ \end{matrix} \right]\]

\[\Rightarrow \left| \text{A} \right|\text{=1}\left( \text{0-0} \right)\text{+1}\left( \text{9+2} \right)\text{+2}\left( \text{0-0} \right)\]

\[\therefore \left| \text{A} \right|\text{=11}\]

\[\left| \text{A} \right|\text{I=11}\left[ \begin{matrix} \text{1} & \text{0} & \text{0}  \\ \text{0} & \text{1} & \text{0}  \\ \text{0} & \text{0} & \text{1}  \\ \end{matrix} \right]\text{=}\left[ \begin{matrix} \text{11} & \text{0} & \text{0}  \\ \text{0} & \text{11} & \text{0}  \\ \text{0} & \text{0} & \text{11}  \\ \end{matrix} \right]\]

Since, Cofactor of \[{{\text{a}}_{ij}}\] is \[{{\text{A}}_{ij}}\text{=}{{\left( \text{-1} \right)}^{\text{i+j}}}{{\text{M}}_{ij}}\].

Thus,

\[\Rightarrow {{\text{A}}_{11}}\text{=}{{\left( \text{-1} \right)}^{1+1}}{{\text{M}}_{11}}={{\left( \text{-1} \right)}^{2}}{{\text{M}}_{11}}\]

\[\Rightarrow {{A}_{11}}\text{=0}\]

\[\Rightarrow {{\text{A}}_{12}}\text{=}{{\left( \text{-1} \right)}^{1+2}}{{\text{M}}_{12}}={{\left( \text{-1} \right)}^{3}}{{\text{M}}_{12}}\]

\[\Rightarrow {{A}_{12}}=-\left( 9+2 \right)=-11\]

\[\Rightarrow {{\text{A}}_{13}}\text{=}{{\left( \text{-1} \right)}^{1+3}}{{\text{M}}_{13}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{13}}\]

\[\Rightarrow {{A}_{13}}=0\]

Similarly,

\[\Rightarrow {{\text{A}}_{21}}\text{=}{{\left( \text{-1} \right)}^{2+1}}{{\text{M}}_{21}}={{\left( \text{-1} \right)}^{3}}{{\text{M}}_{21}}\]

\[\Rightarrow {{\text{A}}_{21}}\text{=}-\left( -3+0 \right)=3\]

\[\Rightarrow {{\text{A}}_{22}}\text{=}{{\left( \text{-1} \right)}^{2+2}}{{\text{M}}_{22}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{22}}\]

\[\Rightarrow {{\text{A}}_{22}}=3-2=1\]

\[\Rightarrow {{\text{A}}_{23}}\text{=}{{\left( \text{-1} \right)}^{2+3}}{{\text{M}}_{23}}={{\left( \text{-1} \right)}^{5}}{{\text{M}}_{23}}\]

\[\Rightarrow {{\text{A}}_{23}}\text{=}-\left( 0+1 \right)=-1\]

and

\[\Rightarrow {{\text{A}}_{31}}\text{=}{{\left( \text{-1} \right)}^{3+1}}{{\text{M}}_{31}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{31}}\]

\[\Rightarrow {{\text{A}}_{31}}=2-0=2\]

\[\Rightarrow {{\text{A}}_{32}}\text{=}{{\left( \text{-1} \right)}^{3+2}}{{\text{M}}_{32}}={{\left( \text{-1} \right)}^{5}}{{\text{M}}_{32}}\]

\[\Rightarrow {{\text{A}}_{32}}=-\left( -2-6 \right)=8\]

\[\Rightarrow {{\text{A}}_{33}}\text{=}{{\left( \text{-1} \right)}^{3+3}}{{\text{M}}_{33}}={{\left( \text{-1} \right)}^{6}}{{\text{M}}_{33}}\]

\[\Rightarrow {{\text{A}}_{33}}=0+3=3\].

Cofactor matrix is \[\left[ \begin{matrix} 0 & -11 & 0  \\ 3 & 1 & -1  \\ 2 & 8 & 3  \\ \end{matrix} \right]\].

We know that adjoint of a matrix is the transpose of its cofactor matrix.

\[\Rightarrow adjA={{\left[ \begin{matrix} 0 & -11 & 0  \\ 3 & 1 & -1  \\ 2 & 8 & 3  \\ \end{matrix} \right]}^{T}}\]

\[\therefore \text{adj}\,\text{A=}\left[ \begin{matrix} \text{0} & 3 & 2  \\ -11 & \text{1} & \text{8}  \\ \text{0} & -1 & \text{3}  \\ \end{matrix} \right]\]

Now, multiplying $A$ with its adjoint, we have:

\[\Rightarrow \text{A}\left( \text{adj}\,\text{A} \right)\text{=}\left[ \begin{matrix} \text{1} & \text{-1} & \text{2}  \\ \text{3} & \text{0} & \text{-2}  \\ \text{1} & \text{0} & \text{3}  \\ \end{matrix} \right]\left[ \begin{matrix} \text{0} & \text{3} & \text{2}  \\ \text{-11} & \text{1} & \text{8}  \\ \text{0} & \text{-1} & \text{3}  \\ \end{matrix} \right]\]

\[\Rightarrow \text{A}\left( \text{adj}\,\text{A} \right)\text{=}\left[ \begin{matrix} \text{0+11+0} & \text{3-1-2} & \text{2-8+6}  \\ \text{0+0+0} & \text{9+0+2} & \text{6+0-6}  \\ \text{0+0+0} & \text{3+0-3} & \text{2+0+9}  \\ \end{matrix} \right]\]

\[\therefore \text{A}\left( \text{adj}\,\text{A} \right)\text{=}\left[ \begin{matrix} \text{11} & \text{0} & \text{0}  \\ \text{0} & \text{11} & \text{0}  \\ \text{0} & \text{0} & \text{11}  \\ \end{matrix} \right]\]

Similarly, multiplying \[\left( adjA \right)\] with \[A\], we get:

\[\Rightarrow \left( \text{adj}\,\text{A} \right)\text{A=}\left[ \begin{matrix} \text{0} & \text{3} & \text{2}  \\ \text{-11} & \text{1} & \text{8}  \\ \text{0} & \text{-1} & \text{3}  \\ \end{matrix} \right]\left[ \begin{matrix} \text{1} & \text{-1} & \text{2}  \\ \text{3} & \text{0} & \text{-2}  \\ \text{1} & \text{0} & \text{3}  \\ \end{matrix} \right]\]

\[\Rightarrow \left( \text{adj}\,\text{A} \right)\text{A=}\left[ \begin{matrix} \text{0+9+2} & \text{0+0+0} & \text{0-6+6}  \\ \text{-11+3+8} & \text{11+0+0} & \text{-22-2+24}  \\ \text{0-3+3} & \text{0+0+0} & \text{0+2+9}  \\ \end{matrix} \right]\]

\[\therefore \left( \text{adj}\,\text{A} \right)\text{A=}\left[ \begin{matrix} \text{11} & \text{0} & \text{0}  \\ \text{0} & \text{11} & \text{0}  \\ \text{0} & \text{0} & \text{11}  \\ \end{matrix} \right]\]

Thus, \[\text{A}\left( \text{adjA} \right)\text{=}\left( \text{adjA} \right)\text{A=}\left| \text{A} \right|I\]

Hence verified.

5. Find the inverse of each of the matrices (if it exists). $\mathbf{\left[ \begin{matrix} 2 & -2  \\ 4 & 3  \\ \end{matrix} \right]}$

Ans: Let \[\text{A=}\left[ \begin{matrix} 2 & -2  \\ 4 & 3  \\ \end{matrix} \right]\]

\[\Rightarrow \left| \text{A} \right|=6+8\]

\[\therefore \left| \text{A} \right|=14\]

Since, Cofactor of \[{{\text{a}}_{ij}}\] is \[{{\text{A}}_{ij}}\text{=}{{\left( \text{-1} \right)}^{\text{i+j}}}{{\text{M}}_{ij}}\].

Then,

\[\Rightarrow {{\text{A}}_{11}}\text{=}{{\left( \text{-1} \right)}^{1+1}}{{\text{M}}_{11}}={{\left( \text{-1} \right)}^{2}}{{\text{M}}_{11}}\]

\[\Rightarrow {{\text{A}}_{11}}\text{=3}\]

\[\Rightarrow {{\text{A}}_{12}}\text{=}{{\left( \text{-1} \right)}^{1+2}}{{\text{M}}_{12}}={{\left( \text{-1} \right)}^{3}}{{\text{M}}_{12}}\]

\[\Rightarrow {{\text{A}}_{12}}\text{=-4}\]

\[\Rightarrow {{\text{A}}_{21}}\text{=}{{\left( \text{-1} \right)}^{2+1}}{{\text{M}}_{21}}={{\left( \text{-1} \right)}^{3}}{{\text{M}}_{21}}\]

\[\Rightarrow {{\text{A}}_{21}}=2\]

\[\Rightarrow {{\text{A}}_{22}}\text{=}{{\left( \text{-1} \right)}^{2+2}}{{\text{M}}_{22}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{22}}\]

\[\Rightarrow {{\text{A}}_{22}}=2\]

Cofactor matrix is $\left[ \begin{matrix} 3 & -4  \\ 2 & 2  \\ \end{matrix} \right]$.

We know that adjoint of a matrix is the transpose of its cofactor matrix.

\[\therefore \text{adjA=}\left[ \begin{matrix} 3 & 2  \\ -4 & 2  \\ \end{matrix} \right]\]

Hence, the inverse of the matrix $A$ is given by,

\[\Rightarrow {{\text{A}}^{\text{-1}}}\text{=}\dfrac{\text{1}}{\left| \text{A} \right|}\text{adjA}\]

\[\therefore {{\text{A}}^{\text{-1}}}\text{=}\dfrac{\text{1}}{\text{14}}\left[ \begin{matrix} 3 & 2  \\ \text{-4} & 2  \\ \end{matrix} \right]\].

6. Find the inverse of each of the matrices (if it exists). \[\mathbf{\left[ \begin{matrix} \text{-1} & \text{5}  \\ \text{-3} & \text{2}  \\ \end{matrix} \right]}\]

Ans: Let \[\text{A=}\left[ \begin{matrix} \text{-1} & \text{5}  \\ \text{-3} & \text{2}  \\ \end{matrix} \right]\]

\[\Rightarrow \left| \text{A} \right|=-2+15\]

\[\therefore \left| \text{A} \right|=13\]

Since, Cofactor of \[{{\text{a}}_{ij}}\] is \[{{\text{A}}_{ij}}\text{=}{{\left( \text{-1} \right)}^{\text{i+j}}}{{\text{M}}_{ij}}\].

Then,

\[\Rightarrow {{\text{A}}_{11}}\text{=}{{\left( \text{-1} \right)}^{1+1}}{{\text{M}}_{11}}={{\left( \text{-1} \right)}^{2}}{{\text{M}}_{11}}\]

\[\Rightarrow {{\text{A}}_{11}}\text{=2}\]

\[\Rightarrow {{\text{A}}_{12}}\text{=}{{\left( \text{-1} \right)}^{1+2}}{{\text{M}}_{12}}={{\left( \text{-1} \right)}^{3}}{{\text{M}}_{12}}\]

\[\Rightarrow {{\text{A}}_{12}}\text{=3}\]

\[\Rightarrow {{\text{A}}_{21}}\text{=}{{\left( \text{-1} \right)}^{2+1}}{{\text{M}}_{21}}={{\left( \text{-1} \right)}^{3}}{{\text{M}}_{21}}\]

\[\Rightarrow {{\text{A}}_{21}}=-5\]

\[\Rightarrow {{\text{A}}_{22}}\text{=}{{\left( \text{-1} \right)}^{2+2}}{{\text{M}}_{22}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{22}}\]

\[\Rightarrow {{\text{A}}_{22}}=-1\]

Cofactor matrix is $\left[ \begin{matrix} 2 & 3  \\ -5 & -1  \\ \end{matrix} \right]$.

We know that adjoint of a matrix is the transpose of its cofactor matrix.

\[\therefore \text{adjA=}\left[ \begin{matrix} \text{2} & \text{-5}  \\ \text{3} & \text{-1}  \\ \end{matrix} \right]\]

Hence, the inverse of the matrix $A$ is given by,

\[\Rightarrow {{\text{A}}^{\text{-1}}}\text{=}\dfrac{\text{1}}{\left| \text{A} \right|}\text{adjA}\]

\[\therefore {{\text{A}}^{\text{-1}}}\text{=}\dfrac{\text{1}}{\text{13}}\left[ \begin{matrix} \text{2} & \text{-5}  \\ \text{3} & \text{-1}  \\ \end{matrix} \right]\].

7. Find the inverse of each of the matrices (if it exists). \[\mathbf{\left[ \begin{matrix} \text{1} & \text{2} & \text{3}  \\ \text{0} & \text{2} & \text{4}  \\ \text{0} & \text{0} & \text{5}  \\ \end{matrix} \right]}\]

Ans: Let \[\text{A=}\left[ \begin{matrix} \text{1} & \text{2} & \text{3}  \\ \text{0} & \text{2} & \text{4}  \\ \text{0} & \text{0} & \text{5}  \\ \end{matrix} \right]\]

Then,

\[\Rightarrow \left| \text{A} \right|\text{=1}\left( 10-0 \right)-2\left( 0-0 \right)\text{+3}\left( 0-0 \right)\]

\[\therefore \left| \text{A} \right|\text{=10}\]

Since, Cofactor of \[{{\text{a}}_{ij}}\] is \[{{\text{A}}_{ij}}\text{=}{{\left( \text{-1} \right)}^{\text{i+j}}}{{\text{M}}_{ij}}\].

Thus,

\[\Rightarrow {{\text{A}}_{11}}\text{=}{{\left( \text{-1} \right)}^{1+1}}{{\text{M}}_{11}}={{\left( \text{-1} \right)}^{2}}{{\text{M}}_{11}}\]

\[\Rightarrow {{A}_{11}}\text{=}10-0=10\]

\[\Rightarrow {{\text{A}}_{12}}\text{=}{{\left( \text{-1} \right)}^{1+2}}{{\text{M}}_{12}}={{\left( \text{-1} \right)}^{3}}{{\text{M}}_{12}}\]

\[\Rightarrow {{A}_{12}}=-\left( 0+0 \right)=0\]

\[\Rightarrow {{\text{A}}_{13}}\text{=}{{\left( \text{-1} \right)}^{1+3}}{{\text{M}}_{13}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{13}}\]

\[\Rightarrow {{A}_{13}}\text{=0}\]

Similarly,

\[\Rightarrow {{\text{A}}_{21}}\text{=}{{\left( \text{-1} \right)}^{2+1}}{{\text{M}}_{21}}={{\left( \text{-1} \right)}^{3}}{{\text{M}}_{21}}\]

\[\Rightarrow {{\text{A}}_{21}}\text{=}-\left( 10-0 \right)\text{=}-10\]

\[\Rightarrow {{\text{A}}_{22}}\text{=}{{\left( \text{-1} \right)}^{2+2}}{{\text{M}}_{22}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{22}}\]

\[\Rightarrow {{\text{A}}_{22}}=5-0=5\]

\[\Rightarrow {{\text{A}}_{23}}\text{=}{{\left( \text{-1} \right)}^{2+3}}{{\text{M}}_{23}}={{\left( \text{-1} \right)}^{5}}{{\text{M}}_{23}}\]

\[\Rightarrow {{\text{A}}_{23}}\text{=}-\left( 0-0 \right)=0\]

And

\[\Rightarrow {{\text{A}}_{31}}\text{=}{{\left( \text{-1} \right)}^{3+1}}{{\text{M}}_{31}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{31}}\]

\[\Rightarrow {{\text{A}}_{31}}=8-6=2\]

\[\Rightarrow {{\text{A}}_{32}}\text{=}{{\left( \text{-1} \right)}^{3+2}}{{\text{M}}_{32}}={{\left( \text{-1} \right)}^{5}}{{\text{M}}_{32}}\]

\[\Rightarrow {{\text{A}}_{32}}=-\left( 4-0 \right)=-4\]

\[\Rightarrow {{\text{A}}_{33}}\text{=}{{\left( \text{-1} \right)}^{3+3}}{{\text{M}}_{33}}={{\left( \text{-1} \right)}^{6}}{{\text{M}}_{33}}\]

\[\Rightarrow {{\text{A}}_{33}}=2-0=2\].

Cofactor matrix is $\left[ \begin{matrix} 10 & 0 & 0  \\ -10 & 5 & 0  \\ 2 & -4 & 2  \\ \end{matrix} \right]$.

We know that adjoint of a matrix is the transpose of its cofactor matrix.

\[\therefore \text{adjA=}\left[ \begin{matrix} \text{10} & -10 & \text{2}  \\ \text{0} & \text{5} & -4  \\ \text{0} & \text{0} & \text{2}  \\ \end{matrix} \right]\]

Hence, the inverse of the matrix $A$ is given by,

\[\Rightarrow {{\text{A}}^{\text{-1}}}\text{=}\dfrac{\text{1}}{\left| \text{A} \right|}\text{adjA}\]

\[\therefore {{\text{A}}^{\text{-1}}}=\dfrac{\text{1}}{\text{10}}\left[ \begin{matrix} \text{10} & \text{-10} & \text{2}  \\ \text{0} & \text{5} & \text{-4}  \\ \text{0} & \text{0} & \text{2}  \\ \end{matrix} \right]\]

8. Find the inverse of each of the matrices (if it exists). \[\mathbf{\left[ \begin{matrix} \text{1} & \text{0} & \text{0}  \\ \text{3} & \text{3} & \text{0}  \\ \text{5} & \text{2} & \text{-1}  \\ \end{matrix} \right]}\]

Ans: Let \[\text{A=}\left[ \begin{matrix} \text{1} & \text{0} & \text{0}  \\ \text{3} & \text{3} & \text{0}  \\ \text{5} & \text{2} & \text{-1}  \\ \end{matrix} \right]\]

Then,

\[\Rightarrow \left| \text{A} \right|\text{=1}\left( -3-0 \right)\text{-0+0}\]

\[\therefore \left| \text{A} \right|=-3\]

Since, Cofactor of \[{{\text{a}}_{ij}}\] is \[{{\text{A}}_{ij}}\text{=}{{\left( \text{-1} \right)}^{\text{i+j}}}{{\text{M}}_{ij}}\].

Thus,

\[\Rightarrow {{\text{A}}_{11}}\text{=}{{\left( \text{-1} \right)}^{1+1}}{{\text{M}}_{11}}={{\left( \text{-1} \right)}^{2}}{{\text{M}}_{11}}\]

\[\Rightarrow {{A}_{11}}\text{=}-3-0=-3\]

\[\Rightarrow {{\text{A}}_{12}}\text{=}{{\left( \text{-1} \right)}^{1+2}}{{\text{M}}_{12}}={{\left( \text{-1} \right)}^{3}}{{\text{M}}_{12}}\]

\[\Rightarrow {{A}_{12}}=-\left( -3-0 \right)=3\]

\[\Rightarrow {{\text{A}}_{13}}\text{=}{{\left( \text{-1} \right)}^{1+3}}{{\text{M}}_{13}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{13}}\]

\[\Rightarrow {{A}_{13}}=6-15=-9\]

Similarly,

\[\Rightarrow {{\text{A}}_{21}}\text{=}{{\left( \text{-1} \right)}^{2+1}}{{\text{M}}_{21}}={{\left( \text{-1} \right)}^{3}}{{\text{M}}_{21}}\]

\[\Rightarrow {{\text{A}}_{21}}\text{=}-\left( 0+0 \right)=0\]

\[\Rightarrow {{\text{A}}_{22}}\text{=}{{\left( \text{-1} \right)}^{2+2}}{{\text{M}}_{22}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{22}}\]

\[\Rightarrow {{\text{A}}_{22}}=-1-0=-1\]

\[\Rightarrow {{\text{A}}_{23}}\text{=}{{\left( \text{-1} \right)}^{2+3}}{{\text{M}}_{23}}={{\left( \text{-1} \right)}^{5}}{{\text{M}}_{23}}\]

\[\Rightarrow {{\text{A}}_{23}}\text{=}-\left( 2-0 \right)=-2\]

and

\[\Rightarrow {{\text{A}}_{31}}\text{=}{{\left( \text{-1} \right)}^{3+1}}{{\text{M}}_{31}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{31}}\]

\[\Rightarrow {{\text{A}}_{31}}=0-0=0\]

\[\Rightarrow {{\text{A}}_{32}}\text{=}{{\left( \text{-1} \right)}^{3+2}}{{\text{M}}_{32}}={{\left( \text{-1} \right)}^{5}}{{\text{M}}_{32}}\]

\[\Rightarrow {{\text{A}}_{32}}=-\left( 0-0 \right)=0\]

\[\Rightarrow {{\text{A}}_{33}}\text{=}{{\left( \text{-1} \right)}^{3+3}}{{\text{M}}_{33}}={{\left( \text{-1} \right)}^{6}}{{\text{M}}_{33}}\]

\[\Rightarrow {{\text{A}}_{33}}=3-0=3\].

Cofactor matrix is $\left[ \begin{matrix} -3 & 3 & -9  \\ 0 & -1 & -2  \\ 0 & 0 & 3  \\ \end{matrix} \right]$.

We know that adjoint of a matrix is the transpose of its cofactor matrix.

\[\Rightarrow adjA={{\left[ \begin{matrix} -3 & 3 & -9  \\ 0 & -1 & -2  \\ 0 & 0 & 3  \\ \end{matrix} \right]}^{T}}\]

\[\therefore \text{adjA=}\left[ \begin{matrix} \text{-3} & \text{0} & \text{0}  \\ \text{3} & \text{-1} & \text{0}  \\ \text{-9} & \text{-2} & \text{3}  \\ \end{matrix} \right]\]

Hence, the inverse of the matrix $A$ is given by,

\[\Rightarrow {{\text{A}}^{\text{-1}}}\text{=}\dfrac{\text{1}}{\left| \text{A} \right|}\text{adjA}\]

\[\therefore {{\text{A}}^{\text{-1}}}\text{=}\dfrac{\text{1}}{\text{10}}\left[ \begin{matrix} \text{-3} & \text{0} & \text{0}  \\ \text{3} & \text{-1} & \text{0}  \\ \text{-9} & \text{-2} & \text{3}  \\ \end{matrix} \right]\]

9. Find the inverse of each of the matrices (if it exists). \[\mathbf{\left[ \begin{matrix} \text{2} & \text{1} & \text{3}  \\ \text{4} & \text{-1} & \text{0}  \\ \text{-7} & \text{2} & \text{1}  \\ \end{matrix} \right]}\]

Ans: Let \[\text{A=}\left[ \begin{matrix} \text{2} & \text{1} & \text{3}  \\ \text{4} & \text{-1} & \text{0}  \\ \text{-7} & \text{2} & \text{1}  \\ \end{matrix} \right]\]

Thus,

\[\Rightarrow \left| \text{A} \right|\text{=2}\left( \text{-1-0} \right)\text{-1}\left( \text{4-0} \right)\text{+3}\left( \text{8-7} \right)\]

\[\Rightarrow \left| A \right|=2\left( -1 \right)-1\left( 4 \right)+3\left( 1 \right)\]

\[\therefore \left| \text{A} \right|=-3\]

Since, Cofactor of \[{{\text{a}}_{ij}}\] is \[{{\text{A}}_{ij}}\text{=}{{\left( \text{-1} \right)}^{\text{i+j}}}{{\text{M}}_{ij}}\].

Thus,

\[\Rightarrow {{\text{A}}_{11}}\text{=}{{\left( \text{-1} \right)}^{1+1}}{{\text{M}}_{11}}={{\left( \text{-1} \right)}^{2}}{{\text{M}}_{11}}\]

\[\Rightarrow {{A}_{11}}\text{=}-1-0=-1\]

\[\Rightarrow {{\text{A}}_{12}}\text{=}{{\left( \text{-1} \right)}^{1+2}}{{\text{M}}_{12}}={{\left( \text{-1} \right)}^{3}}{{\text{M}}_{12}}\]

\[\Rightarrow {{A}_{12}}=-\left( 4-0 \right)=-4\]

\[\Rightarrow {{\text{A}}_{13}}\text{=}{{\left( \text{-1} \right)}^{1+3}}{{\text{M}}_{13}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{13}}\]

\[\Rightarrow {{A}_{13}}=8-7=1\]

Similarly,

\[\Rightarrow {{\text{A}}_{21}}\text{=}{{\left( \text{-1} \right)}^{2+1}}{{\text{M}}_{21}}={{\left( \text{-1} \right)}^{3}}{{\text{M}}_{21}}\]

\[\Rightarrow {{\text{A}}_{21}}\text{=}-\left( 1-6 \right)=5\]

\[\Rightarrow {{\text{A}}_{22}}\text{=}{{\left( \text{-1} \right)}^{2+2}}{{\text{M}}_{22}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{22}}\]

\[\Rightarrow {{\text{A}}_{22}}=2+21=23\]

\[\Rightarrow {{\text{A}}_{23}}\text{=}{{\left( \text{-1} \right)}^{2+3}}{{\text{M}}_{23}}={{\left( \text{-1} \right)}^{5}}{{\text{M}}_{23}}\]

\[\Rightarrow {{\text{A}}_{23}}\text{=}-\left( 4+7 \right)=-11\]

and

\[\Rightarrow {{\text{A}}_{31}}\text{=}{{\left( \text{-1} \right)}^{3+1}}{{\text{M}}_{31}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{31}}\]

\[\Rightarrow {{\text{A}}_{31}}=0+3=3\]

\[\Rightarrow {{\text{A}}_{32}}\text{=}{{\left( \text{-1} \right)}^{3+2}}{{\text{M}}_{32}}={{\left( \text{-1} \right)}^{5}}{{\text{M}}_{32}}\]

\[\Rightarrow {{\text{A}}_{32}}=-\left( 0-12 \right)=12\]

\[\Rightarrow {{\text{A}}_{33}}\text{=}{{\left( \text{-1} \right)}^{3+3}}{{\text{M}}_{33}}={{\left( \text{-1} \right)}^{6}}{{\text{M}}_{33}}\]

\[\Rightarrow {{\text{A}}_{33}}=-2-4=-6\].

Cofactor matrix is $\left[ \begin{matrix} -1 & -4 & 1  \\ 5 & 23 & -11  \\ 3 & 12 & -6  \\ \end{matrix} \right]$.

We know that adjoint of a matrix is the transpose of its cofactor matrix.

$\Rightarrow adjA={{\left[ \begin{matrix} -1 & -4 & 1  \\ 5 & 23 & -11  \\ 3 & 12 & -6  \\ \end{matrix} \right]}^{T}}$

\[\therefore \text{adjA=}\left[ \begin{matrix} \text{-1} & \text{5} & \text{3}  \\ \text{-4} & \text{23} & \text{12}  \\ \text{1} & \text{-11} & \text{-6}  \\ \end{matrix} \right]\]

Hence, the inverse of the matrix $A$ is given by,

\[\Rightarrow {{\text{A}}^{\text{-1}}}\text{=}\dfrac{\text{1}}{\left| \text{A} \right|}\text{adjA}\]

\[\therefore {{\text{A}}^{\text{-1}}}\text{=}\dfrac{\text{1}}{\text{10}}\left[ \begin{matrix} \text{-1} & \text{5} & \text{3}  \\ \text{-4} & \text{23} & \text{12}  \\ \text{1} & \text{-11} & \text{-6}  \\ \end{matrix} \right]\]

10. Find the inverse of each of the matrices (if it exists). \[\mathbf{\left[ \begin{matrix} \text{1} & \text{-1} & \text{2}  \\ \text{0} & \text{2} & \text{-3}  \\ \text{3} & \text{-2} & \text{4}  \\ \end{matrix} \right]}\]

Ans: Let \[\text{A=}\left[ \begin{matrix} \text{1} & \text{-1} & \text{2}  \\ \text{0} & \text{2} & \text{-3}  \\ \text{3} & \text{-2} & \text{4}  \\ \end{matrix} \right]\]

Expanding along column \[{{\text{C}}_{1}}\],

\[\Rightarrow \left| \text{A} \right|\text{=1}\left( 8-6 \right)\text{-0+3}\left( 3-4 \right)\]

\[\therefore \left| \text{A} \right|=-1\]

Since, Cofactor of \[{{\text{a}}_{ij}}\] is \[{{\text{A}}_{ij}}\text{=}{{\left( \text{-1} \right)}^{\text{i+j}}}{{\text{M}}_{ij}}\].

Thus,

\[\Rightarrow {{\text{A}}_{11}}\text{=}{{\left( \text{-1} \right)}^{1+1}}{{\text{M}}_{11}}={{\left( \text{-1} \right)}^{2}}{{\text{M}}_{11}}\]

\[\Rightarrow {{A}_{11}}\text{=8-6=2}\]

\[\Rightarrow {{\text{A}}_{12}}\text{=}{{\left( \text{-1} \right)}^{1+2}}{{\text{M}}_{12}}={{\left( \text{-1} \right)}^{3}}{{\text{M}}_{12}}\]

\[\Rightarrow {{A}_{12}}=-\left( 0+9 \right)=-9\]

\[\Rightarrow {{\text{A}}_{13}}\text{=}{{\left( \text{-1} \right)}^{1+3}}{{\text{M}}_{13}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{13}}\]

\[\Rightarrow {{A}_{13}}=0-6=-6\]

Similarly,

\[\Rightarrow {{\text{A}}_{21}}\text{=}{{\left( \text{-1} \right)}^{2+1}}{{\text{M}}_{21}}={{\left( \text{-1} \right)}^{3}}{{\text{M}}_{21}}\]

\[\Rightarrow {{\text{A}}_{21}}\text{=}-\left( -4+4 \right)=0\]

\[\Rightarrow {{\text{A}}_{22}}\text{=}{{\left( \text{-1} \right)}^{2+2}}{{\text{M}}_{22}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{22}}\]

\[\Rightarrow {{\text{A}}_{22}}=4-6=-2\]

\[\Rightarrow {{\text{A}}_{23}}\text{=}{{\left( \text{-1} \right)}^{2+3}}{{\text{M}}_{23}}={{\left( \text{-1} \right)}^{5}}{{\text{M}}_{23}}\]

\[\Rightarrow {{\text{A}}_{23}}\text{=}-\left( -2+3 \right)=-1\]

and

\[\Rightarrow {{\text{A}}_{31}}\text{=}{{\left( \text{-1} \right)}^{3+1}}{{\text{M}}_{31}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{31}}\]

\[\Rightarrow {{\text{A}}_{31}}=3-4=-1\]

\[\Rightarrow {{\text{A}}_{32}}\text{=}{{\left( \text{-1} \right)}^{3+2}}{{\text{M}}_{32}}={{\left( \text{-1} \right)}^{5}}{{\text{M}}_{32}}\]

\[\Rightarrow {{\text{A}}_{32}}=-\left( -3-0 \right)=3\]

\[\Rightarrow {{\text{A}}_{33}}\text{=}{{\left( \text{-1} \right)}^{3+3}}{{\text{M}}_{33}}={{\left( \text{-1} \right)}^{6}}{{\text{M}}_{33}}\]

\[\Rightarrow {{\text{A}}_{33}}=2-0=2\].

Cofactor matrix is \[\left[ \begin{matrix} 2 & -9 & -6  \\ 0 & -2 & -1  \\ -1 & 3 & 2  \\ \end{matrix} \right]\].

We know that adjoint of a matrix is the transpose of its cofactor matrix.

\[\Rightarrow adjA={{\left[ \begin{matrix} 2 & -9 & -6  \\ 0 & -2 & -1  \\ -1 & 3 & 2  \\ \end{matrix} \right]}^{T}}\]

\[\Rightarrow adjA=\left[ \begin{matrix} 2 & 0 & -1  \\ -9 & -2 & 3  \\ -6 & -1 & 2  \\ \end{matrix} \right]\]

The inverse of the matrix $A$ is given by,

\[\Rightarrow {{\text{A}}^{\text{-1}}}\text{=}\dfrac{\text{1}}{\left| \text{A} \right|}\text{adjA}\]

$\therefore A^{-1}=-1\begin{bmatrix}2 &0  &-1 \\ -9 &-2  &3 \\ -6 &-1  &2 \end{bmatrix}$

$\text{Hence},A^{-1}=\begin{bmatrix}-2 &0  &1 \\ 9 &2  &-3 \\ 6 &1  &-2 \end{bmatrix}$

11. Find the inverse of each of the matrices (if it exists). \[\mathbf{\left[ \begin{matrix} \text{1} & \text{0} & \text{0}  \\ \text{0} & \text{cos}\,\text{a} & \text{sin}\,\text{a}  \\ \text{0} & \text{sin}\,\text{a} & \text{-cos}\,\text{a}  \\ \end{matrix} \right]}\]

Ans: Let \[\text{A=}\left[ \begin{matrix} \text{1} & \text{0} & \text{0}  \\ \text{0} & \text{cos a} & \text{sin a}  \\ \text{0} & \text{sin a} & \text{-cos a}  \\ \end{matrix} \right]\]

Expanding along column, \[{{C}_{1}}\]

\[\Rightarrow \left| \text{A} \right|\text{=1}\left( \text{-co}{{\text{s}}^{\text{2}}}\text{a-si}{{\text{n}}^{\text{2}}}\text{a} \right)\]

\[\Rightarrow \left| \text{A} \right|=-\left( \text{co}{{\text{s}}^{\text{2}}}\text{a+si}{{\text{n}}^{\text{2}}}\text{a} \right)\]

\[\therefore \left| \text{A} \right|=-1\]

Since, Cofactor of \[{{\text{a}}_{ij}}\] is \[{{\text{A}}_{ij}}\text{=}{{\left( \text{-1} \right)}^{\text{i+j}}}{{\text{M}}_{ij}}\].

Thus,

\[\Rightarrow {{\text{A}}_{11}}\text{=}{{\left( \text{-1} \right)}^{1+1}}{{\text{M}}_{11}}={{\left( \text{-1} \right)}^{2}}{{\text{M}}_{11}}\]

\[\Rightarrow {{A}_{11}}\text{=}-\text{co}{{\text{s}}^{\text{2}}}\text{a}-\text{si}{{\text{n}}^{\text{2}}}\text{a=}-1\]

\[\Rightarrow {{\text{A}}_{12}}\text{=}{{\left( \text{-1} \right)}^{1+2}}{{\text{M}}_{12}}={{\left( \text{-1} \right)}^{3}}{{\text{M}}_{12}}\]

\[\Rightarrow {{A}_{12}}=0\]

\[\Rightarrow {{\text{A}}_{13}}\text{=}{{\left( \text{-1} \right)}^{1+3}}{{\text{M}}_{13}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{13}}\]

\[\Rightarrow {{A}_{13}}=0\]

Similarly,

\[\Rightarrow {{\text{A}}_{21}}\text{=}{{\left( \text{-1} \right)}^{2+1}}{{\text{M}}_{21}}={{\left( \text{-1} \right)}^{3}}{{\text{M}}_{21}}\]

\[\Rightarrow {{\text{A}}_{21}}\text{=0}\]

\[\Rightarrow {{\text{A}}_{22}}\text{=}{{\left( \text{-1} \right)}^{2+2}}{{\text{M}}_{22}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{22}}\]

\[\Rightarrow {{\text{A}}_{22}}=-\cos a\]

\[\Rightarrow {{\text{A}}_{23}}\text{=}{{\left( \text{-1} \right)}^{2+3}}{{\text{M}}_{23}}={{\left( \text{-1} \right)}^{5}}{{\text{M}}_{23}}\]

\[\Rightarrow {{\text{A}}_{23}}\text{=}-\sin a\]

and

\[\Rightarrow {{\text{A}}_{31}}\text{=}{{\left( \text{-1} \right)}^{3+1}}{{\text{M}}_{31}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{31}}\]

\[\Rightarrow {{\text{A}}_{31}}=0\]

\[\Rightarrow {{\text{A}}_{32}}\text{=}{{\left( \text{-1} \right)}^{3+2}}{{\text{M}}_{32}}={{\left( \text{-1} \right)}^{5}}{{\text{M}}_{32}}\]

\[\Rightarrow {{\text{A}}_{32}}=-\sin a\]

\[\Rightarrow {{\text{A}}_{33}}\text{=}{{\left( \text{-1} \right)}^{3+3}}{{\text{M}}_{33}}={{\left( \text{-1} \right)}^{6}}{{\text{M}}_{33}}\]

\[\Rightarrow {{\text{A}}_{33}}=\cos a\].

Cofactor matrix is \[\left[ \begin{matrix} -1 & 0 & 0  \\ 0 & -\cos a & -\sin a  \\ 0 & -\sin a & \cos a  \\ \end{matrix} \right]\].

We know that adjoint of a matrix is the transpose of its cofactor matrix.

\[\Rightarrow adjA={{\left[ \begin{matrix} -1 & 0 & 0  \\ 0 & -\cos a & -\sin a  \\ 0 & -\sin a & \cos a  \\ \end{matrix} \right]}^{T}}\]

\[\therefore \text{adjA=}\left[ \begin{matrix} \text{-1} & \text{0} & \text{0}  \\ \text{0} & \text{-cos}\,\text{a} & \text{-sin}\,\text{a}  \\ \text{0} & \text{-sin}\,\text{a} & \text{cos}\,\text{a}  \\ \end{matrix} \right]\]

The inverse of the matrix $A$ is given by,

\[\Rightarrow {{\text{A}}^{\text{-1}}}\text{=}\dfrac{\text{1}}{\left| \text{A} \right|}\text{adjA}\]

\[\therefore {{\text{A}}^{\text{-1}}}\text{=}-\text{1}\left[ \begin{matrix} \text{-1} & \text{0} & \text{0}  \\ \text{0} & \text{-cos}\,\text{a} & \text{-sin}\,\text{a}  \\ \text{0} & \text{-sin}\,\text{a} & \text{cos}\,\text{a}  \\ \end{matrix} \right]\]

Hence, \[{{\text{A}}^{-1}}\text{=}\left[ \begin{matrix} \text{1} & \text{0} & \text{0}  \\ \text{0} & \text{cos}\,\text{a} & \text{sin}\,\text{a}  \\ \text{0} & \text{sin}\,\text{a} & \text{-cos}\,\text{a}  \\ \end{matrix} \right]\].

12. Let \[\mathbf{\text{A=}\left[ \begin{matrix} \text{3} & \text{7}  \\ \text{2} & \text{5}  \\ \end{matrix} \right]}\] and \[\mathbf{\text{B=}\left[ \begin{matrix} \text{6} & \text{8}  \\ \text{7} & \text{9}  \\ \end{matrix} \right]}\] . Verify that \[\mathbf{{{\left( \text{AB} \right)}^{\text{-1}}}\text{=}{{\text{B}}^{\text{-1}}}{{\text{A}}^{\text{-1}}}}\].

Ans: Let \[\text{A=}\left[ \begin{matrix} \text{3} & \text{7}  \\ \text{2} & \text{5}  \\ \end{matrix} \right]\]

Thus, determining the value of \[\left| \text{A} \right|\],

\[\Rightarrow \left| \text{A} \right|\text{=}15-14\]

\[\therefore \left| \text{A} \right|=1\]

Since, Cofactor of \[{{\text{a}}_{ij}}\] is \[{{\text{A}}_{ij}}\text{=}{{\left( \text{-1} \right)}^{\text{i+j}}}{{\text{M}}_{ij}}\].

Thus,

\[\Rightarrow {{\text{A}}_{11}}\text{=}{{\left( \text{-1} \right)}^{1+1}}{{\text{M}}_{11}}={{\left( \text{-1} \right)}^{2}}{{\text{M}}_{11}}\]

\[\Rightarrow {{\text{A}}_{11}}\text{=5}\]

\[\Rightarrow {{\text{A}}_{12}}\text{=}{{\left( \text{-1} \right)}^{1+2}}{{\text{M}}_{12}}={{\left( \text{-1} \right)}^{3}}{{\text{M}}_{12}}\]

\[\Rightarrow {{\text{A}}_{12}}\text{=}-2\]

\[\Rightarrow {{\text{A}}_{21}}\text{=}{{\left( \text{-1} \right)}^{2+1}}{{\text{M}}_{21}}={{\left( \text{-1} \right)}^{3}}{{\text{M}}_{21}}\]

\[\Rightarrow {{\text{A}}_{21}}=-7\]

\[\Rightarrow {{\text{A}}_{22}}\text{=}{{\left( \text{-1} \right)}^{2+2}}{{\text{M}}_{22}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{22}}\]

\[\Rightarrow {{\text{A}}_{22}}=3\]

We know that adjoint of a matrix is the transpose of its cofactor matrix.

\[\therefore \text{adjA=}{{\left[ \begin{matrix} \text{5} & -2  \\ -7 & \text{3}  \\ \end{matrix} \right]}^{T}}\]

\[\Rightarrow \text{adjA=}\left[ \begin{matrix} \text{5} & -7  \\ -2 & \text{3}  \\ \end{matrix} \right]\]

The inverse of a matrix is given by, \[{{\text{A}}^{\text{-1}}}\text{=}\dfrac{\text{1}}{\left| \text{A} \right|}\text{adjA}\]

Hence, \[{{\text{A}}^{\text{-1}}}\text{=}\left[ \begin{matrix} \text{5} & \text{-7}  \\ \text{-2} & \text{3}  \\ \end{matrix} \right]\]

For \[\text{B=}\left[ \begin{matrix} \text{6} & \text{8}  \\ \text{7} & \text{9}  \\ \end{matrix} \right]\]

\[\Rightarrow \left| \text{B} \right|\text{=54}-\text{56}\]

\[\therefore \left| \text{B} \right|\text{=}-\text{2}\]

Thus,

\[\Rightarrow {{\text{A}}_{11}}\text{=}{{\left( \text{-1} \right)}^{1+1}}{{\text{M}}_{11}}={{\left( \text{-1} \right)}^{2}}{{\text{M}}_{11}}\]

\[\Rightarrow {{\text{A}}_{11}}\text{=9}\]

\[\Rightarrow {{\text{A}}_{12}}\text{=}{{\left( \text{-1} \right)}^{1+2}}{{\text{M}}_{12}}={{\left( \text{-1} \right)}^{3}}{{\text{M}}_{12}}\]

\[\Rightarrow {{\text{A}}_{12}}\text{=}-7\]

\[\Rightarrow {{\text{A}}_{21}}\text{=}{{\left( \text{-1} \right)}^{2+1}}{{\text{M}}_{21}}={{\left( \text{-1} \right)}^{3}}{{\text{M}}_{21}}\]

\[\Rightarrow {{\text{A}}_{21}}=-8\]

\[\Rightarrow {{\text{A}}_{22}}\text{=}{{\left( \text{-1} \right)}^{2+2}}{{\text{M}}_{22}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{22}}\]

\[\Rightarrow {{\text{A}}_{22}}=6\]

We know that adjoint of a matrix is the transpose of its cofactor matrix.

\[\therefore \text{adjA=}{{\left[ \begin{matrix} 9 & -7  \\ -8 & 6  \\ \end{matrix} \right]}^{T}}\]

\[\Rightarrow \text{adjA=}\left[ \begin{matrix} 9 & -8  \\ -7 & 6  \\ \end{matrix} \right]\]

Hence, \[\text{adjB=}\left[ \begin{matrix} \text{9} & \text{-8}  \\ \text{-7} & \text{6}  \\ \end{matrix} \right]\]

\[\therefore {{\text{B}}^{\text{-1}}}\text{=}\dfrac{\text{1}}{\left| \text{B} \right|}\text{adjB=}-\dfrac{\text{1}}{\text{2}}\left[ \begin{matrix} \text{9} & -\text{8}  \\ -\text{7} & \text{6}  \\ \end{matrix} \right]\]

Thus, \[{{\text{B}}^{\text{-1}}}=\left[ \begin{matrix} -\dfrac{\text{9}}{\text{2}} & \text{4}  \\ \dfrac{\text{7}}{\text{2}} & -\text{3}  \\ \end{matrix} \right]\].

Now, multiplying ${{B}^{-1}}$ and ${{A}^{-1}}$, we get:

\[\Rightarrow {{\text{B}}^{\text{-1}}}{{\text{A}}^{\text{-1}}}\text{=}\left[ \begin{matrix} \text{-}\dfrac{\text{9}}{\text{2}} & \text{4}  \\ \dfrac{\text{7}}{\text{2}} & \text{-3}  \\ \end{matrix} \right]\left[ \begin{matrix} \text{5} & \text{-7}  \\ \text{-2} & \text{3}  \\ \end{matrix} \right]\]

\[\Rightarrow {{\text{B}}^{\text{-1}}}{{\text{A}}^{\text{-1}}}\text{=}\left[ \begin{matrix} \text{-}\dfrac{\text{45}}{\text{2}}\text{-8} & \dfrac{\text{63}}{\text{2}}\text{+12}  \\ \dfrac{\text{35}}{\text{2}}\text{+6} & \text{-}\dfrac{\text{49}}{\text{2}}\text{-9}  \\ \end{matrix} \right]\]

\[\therefore {{\text{B}}^{\text{-1}}}{{\text{A}}^{\text{-1}}}\text{=}\left[ \begin{matrix} \text{-}\dfrac{\text{61}}{\text{2}} & \dfrac{\text{87}}{\text{2}}  \\ \dfrac{\text{47}}{\text{2}} & \text{-}\dfrac{\text{67}}{\text{2}}  \\ \end{matrix} \right]\]   ……(1)

Similarly, multiplying the matrices $A$ and $B$, we get:

\[\Rightarrow \text{AB=}\left[ \begin{matrix} \text{3} & \text{7}  \\ \text{2} & \text{5}  \\ \end{matrix} \right]\left[ \begin{matrix} \text{6} & \text{8}  \\ \text{7} & \text{9}  \\ \end{matrix} \right]\]

\[\Rightarrow \text{AB=}\left[ \begin{matrix} \text{18+49} & \text{24+63}  \\ \text{12+35} & \text{16+45}  \\ \end{matrix} \right]\]

\[\therefore \text{AB=}\left[ \begin{matrix} \text{67} & \text{87}  \\ \text{47} & \text{61}  \\ \end{matrix} \right]\]

The value of \[\left| \text{AB} \right|\] is 

\[\Rightarrow \left| \text{AB} \right|\text{=67 }\!\!\times\!\!\text{ 61-87 }\!\!\times\!\!\text{ 47}\]

\[\Rightarrow \left| \text{AB} \right|\text{=4087-4089}\]

\[\therefore \left| \text{AB} \right|=-2\]

The adjoint of $\left( AB \right)$ is given by,

\[\Rightarrow \text{adj}\left( \text{AB} \right)\text{=}\left[ \begin{matrix} \text{61} & \text{-87} \\\text{-47} & \text{67} \\\end{matrix} \right]\]

Thus, the inverse is,

\[\Rightarrow {{\left( \text{AB} \right)}^{\text{-1}}}\text{=}\dfrac{\text{1}}{\left| \text{AB} \right|}\text{adj}\left( \text{AB} \right)\]

\[\Rightarrow {{\left( \text{AB} \right)}^{\text{-1}}}-\dfrac{\text{1}}{\text{2}}\left[ \begin{matrix} \text{61} & \text{-87}  \\ \text{-47} & \text{67}  \\ \end{matrix} \right]\]

\[\therefore {{\left( \text{AB} \right)}^{\text{-1}}}\text{=}\left[ \begin{matrix} \text{-}\dfrac{\text{61}}{\text{2}} & \dfrac{\text{87}}{\text{2}}  \\ \dfrac{\text{47}}{\text{2}} & \text{-}\dfrac{\text{67}}{\text{2}}  \\ \end{matrix} \right]\]   ……. (2)

From (1) and (2), we have:

\[{{\left( \text{AB} \right)}^{\text{-1}}}\text{=}{{\text{B}}^{\text{-1}}}{{\text{A}}^{\text{-1}}}\]

Hence proved.

13. If $A = \begin{bmatrix} 3&1 \\ -1 &2 \end{bmatrix} , \text{show that}\;\; A^2-5A+7I=0. \text{Hence find} A^{-1}$.

Ans: Given, \[\text{A=}\left[ \begin{matrix} \text{3} & \text{1}  \\ \text{-1} & \text{2}  \\ \end{matrix} \right]\]

We can write, \[{{\text{A}}^{\text{2}}}\text{=A}\text{.A}\]

\[\Rightarrow {{\text{A}}^{\text{2}}}\text{=}\left[ \begin{matrix} \text{3} & \text{1}  \\ \text{-1} & \text{2}  \\ \end{matrix} \right]\left[ \begin{matrix} \text{3} & \text{1}  \\ \text{-1} & \text{2}  \\ \end{matrix} \right]\]

\[\Rightarrow {{\text{A}}^{\text{2}}}\text{=}\left[ \begin{matrix} \text{9-1} & \text{3+2}  \\ \text{-3-2} & \text{-1+4}  \\ \end{matrix} \right]\]

\[\Rightarrow {{\text{A}}^{\text{2}}}\text{=}\left[ \begin{matrix} \text{8} & \text{5}  \\ \text{-5} & \text{3}  \\ \end{matrix} \right]\]

\[\therefore \] The value of \[{{\text{A}}^{\text{2}}}\text{-5A+7I}\] is:

\[\Rightarrow {{\text{A}}^{\text{2}}}\text{-5A+7I=}\left[ \begin{matrix} \text{8} & \text{5}  \\ \text{-5} & \text{3}  \\ \end{matrix} \right]\text{-5}\left[ \begin{matrix} \text{3} & \text{1}  \\ \text{-1} & \text{2}  \\ \end{matrix} \right]\text{+7}\left[ \begin{matrix} \text{1} & \text{0}  \\ \text{0} & \text{1}  \\ \end{matrix} \right]\]

\[\Rightarrow {{\text{A}}^{\text{2}}}\text{-5A+7I=}\left[ \begin{matrix} \text{-7} & \text{0}  \\ \text{0} & \text{-7}  \\ \end{matrix} \right]\text{+}\left[ \begin{matrix} \text{7} & \text{0}  \\ \text{0} & \text{7}  \\ \end{matrix} \right]\]

\[\therefore {{\text{A}}^{\text{2}}}\text{-5A+7I=}\left[ \begin{matrix} \text{0} & \text{0}  \\ \text{0} & \text{0}  \\ \end{matrix} \right]\]

Hence, \[{{\text{A}}^{\text{2}}}\text{-5A+7I=0}\]

\[\Rightarrow {{\text{A}}^{\text{2}}}\text{-5A=-7I}\]

Multiplying by \[{{\text{A}}^{\text{-1}}}\] on both the sides, we have:

\[\Rightarrow \text{AA}\left( {{\text{A}}^{\text{-1}}} \right)-\text{5A}{{\text{A}}^{\text{-1}}}\text{=}-\text{7I}{{\text{A}}^{\text{-1}}}\]    

\[\Rightarrow \text{A}\left( \text{A}{{\text{A}}^{\text{-1}}} \right)-\text{5I=}-\text{7I}{{\text{A}}^{\text{-1}}}\]

\[\Rightarrow \text{AI}-\text{5I=}-\text{7I}{{\text{A}}^{\text{-1}}}\]

\[\Rightarrow {{\text{A}}^{\text{-1}}}\text{=}-\dfrac{\text{1}}{\text{7}}\left( \text{A}-\text{5I} \right)\]

\[\Rightarrow {{\text{A}}^{\text{-1}}}\text{=}\dfrac{\text{1}}{\text{7}}\left( \text{5I}-\text{A} \right)\]

\[\Rightarrow {{\text{A}}^{\text{-1}}}\text{=}\dfrac{\text{1}}{\text{7}}\left( \left[ \begin{matrix} \text{5} & \text{0}  \\ \text{0} & \text{5}  \\ \end{matrix} \right]-\left[ \begin{matrix} \text{3} & \text{1}  \\ \text{-1} & \text{2}  \\ \end{matrix} \right] \right)\]

\[\therefore {{\text{A}}^{\text{-1}}}\text{=}\dfrac{\text{1}}{\text{7}}\left[ \begin{matrix} \text{2} & \text{-1}  \\ \text{1} & \text{3}  \\ \end{matrix} \right]\]

14. For the matrix\[\mathbf{\text{A=}\left[ \begin{matrix} \text{3} & \text{2}  \\ \text{1} & \text{1}  \\ \end{matrix} \right]}\] .find the number \[\mathbf{\text{a}}\] and \[\mathbf{\text{b}}\] such that. \[\mathbf{{{\text{A}}^{\text{2}}}\text{+aA+bI=0}}\].

Ans: Given \[\text{A=}\left[ \begin{matrix} \text{3} & \text{2}  \\ \text{1} & \text{1}  \\ \end{matrix} \right]\]

We can write, \[{{\text{A}}^{\text{2}}}\text{=A}\text{.A}\]

\[\therefore {{\text{A}}^{\text{2}}}\text{=}\left[ \begin{matrix} \text{3} & \text{2}  \\ \text{1} & \text{1}  \\ \end{matrix} \right]\left[ \begin{matrix} \text{3} & \text{2}  \\ \text{1} & \text{1}  \\ \end{matrix} \right]\]

\[\Rightarrow {{\text{A}}^{\text{2}}}\text{=}\left[ \begin{matrix} \text{9+2} & \text{6+2}  \\ \text{3+1} & \text{2+1}  \\ \end{matrix} \right]\]

\[\Rightarrow {{\text{A}}^{\text{2}}}\text{=}\left[ \begin{matrix} \text{11} & \text{8}  \\ \text{4} & \text{3}  \\ \end{matrix} \right]\]

Solving \[{{\text{A}}^{\text{2}}}\text{+aA+bI=0}\] by multiplying the whole equation by \[{{A}^{-1}}\].

\[\Rightarrow \left( \text{AA} \right){{\text{A}}^{\text{-1}}}\text{+aA}{{\text{A}}^{\text{-1}}}\text{+bI}{{\text{A}}^{\text{-1}}}\text{=0}\]

\[\Rightarrow \text{A}\left( \text{A}{{\text{A}}^{\text{-1}}} \right)\text{+aI+b}\left( \text{I}{{\text{A}}^{\text{-1}}} \right)\text{=0}\]

\[\Rightarrow \text{AI+aI+b}{{\text{A}}^{\text{-1}}}\text{=0}\]

\[\Rightarrow \text{A+aI=-b}{{\text{A}}^{\text{-1}}}\]

\[\Rightarrow {{\text{A}}^{\text{-1}}}\text{=}\dfrac{\text{1}}{\text{b}}\left( \text{A+aI} \right)\]

Now, determining the value of  \[{{\text{A}}^{\text{-1}}}\].

We know that the adjoint of a square matrix is the transpose of its cofactor matrix.

Hence, the adjoint of matrix $A$ is:

$\therefore adjA=\left[ \begin{matrix} 1 & -2  \\ -1 & 3  \\ \end{matrix} \right]$

The inverse is given by, \[{{\text{A}}^{\text{-1}}}\text{=}\dfrac{\text{1}}{\left| \text{A} \right|}adjA\].

\[\Rightarrow {{\text{A}}^{\text{-1}}}\text{=}\dfrac{\text{1}}{\text{1}}\left[ \begin{matrix} \text{1} & \text{-2}  \\ \text{-1} & \text{3}  \\ \end{matrix} \right]\]

\[\therefore {{\text{A}}^{\text{-1}}}\text{=}\left[ \begin{matrix} \text{1} & \text{-2}  \\ \text{-1} & \text{3}  \\ \end{matrix} \right]\]

Thus,

\[\Rightarrow \left[ \begin{matrix} \text{1} & \text{-2}  \\ \text{-1} & \text{3}  \\ \end{matrix} \right]\text{=-}\dfrac{\text{1}}{\text{b}}\left( \left[ \begin{matrix} \text{3} & \text{2}  \\ \text{1} & \text{1}  \\ \end{matrix} \right]\text{+}\left[ \begin{matrix} \text{a} & \text{0}  \\ \text{0} & \text{a}  \\ \end{matrix} \right] \right)\]

\[\Rightarrow \left[ \begin{matrix} \text{1} & \text{-2}  \\ \text{-1} & \text{3}  \\ \end{matrix} \right]\text{=-}\dfrac{\text{1}}{\text{b}}\left[ \begin{matrix} \text{3+a} & \text{2}  \\ \text{1} & \text{1+a}  \\ \end{matrix} \right]\]

\[\Rightarrow \left[ \begin{matrix} \text{1} & \text{-2}  \\ \text{-1} & \text{3}  \\ \end{matrix} \right]\text{=}\left[ \begin{matrix} \dfrac{\text{-3-a}}{\text{b}} & \text{-}\dfrac{\text{2}}{\text{b}}  \\ \text{-}\dfrac{\text{1}}{\text{b}} & \dfrac{\text{-1-a}}{\text{b}}  \\ \end{matrix} \right]\]

Equating the corresponding elements of the two matrices, we get:

\[\Rightarrow \text{-}\dfrac{\text{1}}{\text{b}}\text{=-1}\]

\[\therefore \text{b=1}\]

\[\Rightarrow \dfrac{\text{-3-a}}{\text{b}}=1\]

\[\therefore \text{a=}-4\]

Thus, \[-4\] and \[1\] are the required values of \[\text{a}\] and \[\text{b}\] respectively.

15. For the matrix \[\mathbf{\text{A=}\left[ \begin{matrix} \text{1} & \text{1} & \text{1}  \\ \text{1} & \text{2} & \text{-3}  \\ \text{2} & \text{-1} & \text{3}  \\ \end{matrix} \right]}\] show that \[\mathbf{{{\text{A}}^{\text{3}}}\text{-6}{{\text{A}}^{\text{2}}}\text{+5A+11I=0}}\] . Hence, \[\mathbf{{{\text{A}}^{\text{-1}}}}\].

Ans: Given, \[\text{A=}\left[ \begin{matrix} \text{1} & \text{1} & \text{1}  \\ \text{1} & \text{2} & \text{-3}  \\ \text{2} & \text{-1} & \text{3}  \\ \end{matrix} \right]\]

\[\Rightarrow {{\text{A}}^{\text{2}}}\text{=}\left[ \begin{matrix} \text{1} & \text{1} & \text{1}  \\ \text{1} & \text{2} & \text{-3}  \\ \text{2} & \text{-1} & \text{3}  \\ \end{matrix} \right]\left[ \begin{matrix} \text{1} & \text{1} & \text{1}  \\ \text{1} & \text{2} & \text{-3}  \\ \text{2} & \text{-1} & \text{3}  \\ \end{matrix} \right]\]

\[\Rightarrow {{\text{A}}^{\text{2}}}\text{=}\left[ \begin{matrix} \text{1+1+1} & \text{1+2-1} & \text{1-3+3}  \\ \text{1+2-6} & \text{1+4+3} & \text{1-6-9}  \\ \text{2-1+6} & \text{2-2-3} & \text{2+3+9}  \\ \end{matrix} \right]\]

\[\therefore {{\text{A}}^{\text{2}}}\text{=}\left[ \begin{matrix} \text{4} & \text{2} & \text{1}  \\ \text{-3} & \text{8} & \text{-14}  \\ \text{7} & \text{-3} & \text{14}  \\ \end{matrix} \right]\]

\[\Rightarrow {{\text{A}}^{\text{3}}}\text{=}{{\text{A}}^{\text{2}}}\text{.A=}\left[ \begin{matrix} \text{4} & \text{2} & \text{1}  \\ \text{-3} & \text{8} & \text{-14}  \\ \text{7} & \text{-3} & \text{14}  \\ \end{matrix} \right]\left[ \begin{matrix} \text{1} & \text{1} & \text{1}  \\ \text{1} & \text{2} & \text{-3}  \\ \text{2} & \text{-1} & \text{3}  \\ \end{matrix} \right]\]

\[\Rightarrow {{\text{A}}^{\text{3}}}\text{=}\left[ \begin{matrix} \text{4+2+2} & \text{4+4-1} & \text{4-6+3}  \\ \text{-3+8-28} & \text{-3+16+14} & \text{-3-24-42}  \\ \text{7-3+28} & \text{7-6-14} & \text{7+9+42}  \\ \end{matrix} \right]\]

\[\therefore {{\text{A}}^{\text{3}}}\text{=}\left[ \begin{matrix} \text{8} & \text{7} & \text{1}  \\ \text{-23} & \text{27} & \text{-69}  \\ \text{32} & \text{-13} & \text{58}  \\ \end{matrix} \right]\]

Substituting the values for \[{{\text{A}}^{\text{3}}}\], \[{{\text{A}}^{2}}\] and \[\text{A}\] in \[{{\text{A}}^{\text{3}}}\text{-6}{{\text{A}}^{\text{2}}}\text{+5A+11I}\], we have:

\[\Rightarrow {{\text{A}}^{\text{3}}}\text{-6}{{\text{A}}^{\text{2}}}\text{+5A+11I=}\left[ \begin{matrix} \text{8} & \text{7} & \text{1}  \\ \text{-23} & \text{27} & \text{-69}  \\ \text{32} & \text{-13} & \text{58}  \\ \end{matrix} \right]\text{-6}\left[ \begin{matrix} \text{4} & \text{2} & \text{1}  \\ \text{-3} & \text{8} & \text{-14}  \\ \text{7} & \text{-3} & \text{14}  \\ \end{matrix} \right]\text{+5}\left[ \begin{matrix} \text{1} & \text{1} & \text{1}  \\ \text{1} & \text{2} & \text{-3}  \\ \text{2} & \text{-1} & \text{3}  \\ \end{matrix} \right]\text{+11}\left[ \begin{matrix} \text{1} & \text{0} & \text{0}  \\ \text{0} & \text{1} & \text{0}  \\ \text{0} & \text{0} & \text{1}  \\ \end{matrix} \right]\]

\[\Rightarrow {{\text{A}}^{\text{3}}}\text{-6}{{\text{A}}^{\text{2}}}\text{+5A+11I=}\left[ \begin{matrix} \text{8} & \text{7} & \text{1}  \\ \text{-23} & \text{27} & \text{-69}  \\ \text{32} & \text{-13} & \text{58}  \\ \end{matrix} \right]\text{-}\left[ \begin{matrix} \text{24} & \text{12} & \text{6}  \\ \text{-18} & \text{48} & \text{-84}  \\ \text{42} & \text{-18} & \text{84}  \\ \end{matrix} \right]\text{+}\left[ \begin{matrix} \text{5} & \text{5} & \text{5}  \\ \text{5} & \text{10} & \text{-15}  \\ \text{2} & \text{-5} & \text{15}  \\ \end{matrix} \right]\text{+11}\left[ \begin{matrix} \text{11} & \text{0} & \text{0}  \\ \text{0} & \text{11} & \text{0}  \\ \text{0} & \text{0} & \text{11}  \\ \end{matrix} \right]\]

\[\Rightarrow {{\text{A}}^{\text{3}}}\text{-6}{{\text{A}}^{\text{2}}}\text{+5A+11I=}\left[ \begin{matrix} \text{24} & \text{12} & \text{6}  \\ \text{-18} & \text{48} & \text{-84}  \\ \text{42} & \text{-18} & \text{84}  \\ \end{matrix} \right]\text{-}\left[ \begin{matrix} \text{24} & \text{12} & \text{6}  \\ \text{-18} & \text{48} & \text{-84}  \\ \text{42} & \text{-18} & \text{84}  \\ \end{matrix} \right]\]

\[\therefore {{\text{A}}^{\text{3}}}\text{-6}{{\text{A}}^{\text{2}}}\text{+5A+11I=}\left[ \begin{matrix} \text{0} & \text{0} & \text{0}  \\ \text{0} & \text{0} & \text{0}  \\ \text{0} & \text{0} & \text{0}  \\ \end{matrix} \right]\text{=0}\]

Thus, \[{{\text{A}}^{\text{3}}}\text{-6}{{\text{A}}^{\text{2}}}\text{+5A+11I=0}\]

Since, \[{{\text{A}}^{\text{3}}}\text{-6}{{\text{A}}^{\text{2}}}\text{+5A+11I=0}\]. 

Multiplying the whole equation by \[{{A}^{-1}}\], we have:

\[\Rightarrow \left( \text{AAA} \right){{\text{A}}^{\text{-1}}}\text{-6}\left( \text{AA} \right){{\text{A}}^{\text{-1}}}\text{+5A}{{\text{A}}^{\text{-1}}}\text{+11I}{{\text{A}}^{\text{-1}}}\text{=0}\] 

\[\Rightarrow \text{AA}\left( \text{A}{{\text{A}}^{\text{-1}}} \right)\text{-6A}\left( \text{A}{{\text{A}}^{\text{-1}}} \right)\text{+5}\left( \text{A}{{\text{A}}^{\text{-1}}} \right)\text{=11}\left( \text{I}{{\text{A}}^{\text{-1}}} \right)\]

\[\Rightarrow {{\text{A}}^{\text{2}}}\text{-6A+5I=-11}{{\text{A}}^{\text{-1}}}\]

\[\Rightarrow {{\text{A}}^{\text{-1}}}\text{=-}\dfrac{\text{1}}{\text{11}}\left( {{\text{A}}^{\text{2}}}\text{-6A+5I} \right)\]   …. (1)

Now, \[{{\text{A}}^{\text{2}}}\text{-6A+5I}\] is given by:

\[\Rightarrow {{\text{A}}^{\text{2}}}\text{-6A+5I=}\left[ \begin{matrix} \text{4} & \text{2} & \text{1}  \\ \text{-3} & \text{8} & \text{-14}  \\ \text{7} & \text{-3} & \text{14}  \\ \end{matrix} \right]\text{-6}\left[ \begin{matrix} \text{1} & \text{1} & \text{1}  \\ \text{1} & \text{2} & \text{-3}  \\ \text{2} & \text{-1} & \text{3}  \\ \end{matrix} \right]\text{+5}\left[ \begin{matrix} \text{1} & \text{0} & \text{0}  \\ \text{0} & \text{1} & \text{0}  \\ \text{0} & \text{0} & \text{1}  \\ \end{matrix} \right]\]

\[\Rightarrow {{\text{A}}^{\text{2}}}\text{-6A+5I=}\left[ \begin{matrix} \text{4} & \text{2} & \text{1}  \\ \text{-3} & \text{8} & \text{-14}  \\ \text{7} & \text{-3} & \text{14}  \\ \end{matrix} \right]\text{-}\left[ \begin{matrix} \text{6} & \text{6} & \text{6}  \\ \text{6} & \text{12} & \text{-18}  \\ \text{12} & \text{6} & \text{18}  \\ \end{matrix} \right]\text{+}\left[ \begin{matrix} \text{5} & \text{0} & \text{0}  \\ \text{0} & \text{5} & \text{0}  \\ \text{0} & \text{0} & \text{5}  \\ \end{matrix} \right]\]

\[\Rightarrow {{\text{A}}^{\text{2}}}\text{-6A+5I=}\left[ \begin{matrix} \text{4} & \text{2} & \text{1}  \\ \text{-3} & \text{13} & \text{-14}  \\ \text{7} & \text{-3} & \text{19}  \\ \end{matrix} \right]\text{-}\left[ \begin{matrix} \text{6} & \text{6} & \text{6}  \\ \text{6} & \text{12} & \text{-18}  \\ \text{12} & \text{-6} & \text{18}  \\ \end{matrix} \right]\]

\[\therefore {{\text{A}}^{\text{2}}}\text{-6A+5I=}\left[ \begin{matrix} \text{3} & \text{-4} & \text{-5}  \\ \text{-9} & \text{1} & \text{4}  \\ \text{-5} & \text{3} & \text{1}  \\ \end{matrix} \right]\]

Substituting for \[{{\text{A}}^{\text{2}}}\text{-6A+5I}\] equation (1), we get

\[\Rightarrow {{\text{A}}^{\text{-1}}}\text{=-}\dfrac{\text{1}}{\text{11}}\left[ \begin{matrix} \text{3} & \text{-4} & \text{-5}  \\ \text{-9} & \text{1} & \text{4}  \\ \text{-5} & \text{3} & \text{1}  \\ \end{matrix} \right]\]

\[\therefore {{\text{A}}^{\text{-1}}}\text{=}\dfrac{\text{1}}{\text{11}}\left[ \begin{matrix} \text{-3} & \text{4} & \text{5}  \\ \text{9} & \text{-1} & \text{-4}  \\ \text{5} & \text{-3} & \text{-1}  \\ \end{matrix} \right]\]

16. If \[\mathbf{\text{A=}\left[ \begin{matrix} \text{2} & \text{-1} & \text{1}  \\ \text{-1} & \text{2} & \text{-1}  \\ \text{1} & \text{-1} & \text{2}  \\ \end{matrix} \right]}\] verify that \[\mathbf{{{\text{A}}^{\text{3}}}\text{-6}{{\text{A}}^{\text{2}}}\text{+9A+4I=0}}\] and hence find \[\mathbf{{{\text{A}}^{\text{-1}}}}\].

Ans: Given,\[\text{A=}\left[ \begin{matrix} \text{2} & \text{-1} & \text{1}  \\ \text{-1} & \text{2} & \text{-1}  \\ \text{1} & \text{-1} & \text{2}  \\ \end{matrix} \right]\]

\[\Rightarrow {{\text{A}}^{\text{2}}}\text{=}\left[ \begin{matrix} \text{2} & \text{-1} & \text{1}  \\ \text{-1} & \text{2} & \text{-1}  \\ \text{1} & \text{-1} & \text{2}  \\ \end{matrix} \right]\left[ \begin{matrix} \text{2} & \text{-1} & \text{1}  \\ \text{-1} & \text{2} & \text{-1}  \\ \text{1} & \text{-1} & \text{2}  \\ \end{matrix} \right]\]

\[\Rightarrow {{\text{A}}^{\text{2}}}\text{=}\left[ \begin{matrix} \text{4+1+1} & \text{-2-2-1} & \text{2+1+2}  \\ \text{-2-2-1} & \text{1+4+1} & \text{-1-2-2}  \\ \text{2+1+2} & \text{-1-2-2} & \text{1+1+4}  \\ \end{matrix} \right]\]

\[\therefore {{\text{A}}^{\text{2}}}\text{=}\left[ \begin{matrix} \text{6} & \text{-5} & \text{5}  \\ \text{-5} & \text{6} & \text{-5}  \\ \text{5} & \text{-5} & \text{6}  \\ \end{matrix} \right]\]

Similarly,

\[\Rightarrow {{\text{A}}^{\text{3}}}\text{=}{{\text{A}}^{\text{2}}}\text{A=}\left[ \begin{matrix} \text{6} & \text{-5} & \text{5}  \\ \text{-5} & \text{6} & \text{-5}  \\ \text{5} & \text{-5} & \text{6}  \\ \end{matrix} \right]\left[ \begin{matrix} \text{2} & \text{-1} & \text{1}  \\ \text{-1} & \text{2} & \text{-1}  \\ \text{1} & \text{-1} & \text{2}  \\ \end{matrix} \right]\]

\[\Rightarrow {{\text{A}}^{\text{3}}}\text{=}\left[ \begin{matrix} \text{12+5+5} & \text{-6-10-5} & \text{6+5+10}  \\ \text{-10-6-5} & \text{5+12+5} & \text{-5-6-10}  \\ \text{10+5+6} & \text{-5-10-6} & \text{5+5+12}  \\ \end{matrix} \right]\]

\[\therefore {{\text{A}}^{\text{3}}}\text{=}\left[ \begin{matrix} \text{22} & \text{-21} & \text{21}  \\ \text{-21} & \text{22} & \text{-21}  \\ \text{21} & \text{-21} & \text{22}  \\ \end{matrix} \right]\]

Now, \[{{\text{A}}^{\text{3}}}\text{-6}{{\text{A}}^{\text{2}}}\text{+9A-4I}\] is given by:

\[\Rightarrow {{\text{A}}^{\text{3}}}\text{-6}{{\text{A}}^{\text{2}}}\text{+9A-4I=}\left[ \begin{matrix} \text{22} & \text{-21} & \text{21}  \\ \text{-21} & \text{22} & \text{-21}  \\ \text{21} & \text{-21} & \text{22}  \\ \end{matrix} \right]\text{-6}\left[ \begin{matrix} \text{6} & \text{-5} & \text{5}  \\ \text{-5} & \text{6} & \text{-5}  \\ \text{5} & \text{-5} & \text{6}  \\ \end{matrix} \right]\text{+9}\left[ \begin{matrix} \text{2} & \text{-1} & \text{1}  \\ \text{-1} & \text{2} & \text{-1}  \\ \text{1} & \text{-1} & \text{2}  \\ \end{matrix} \right]\text{-4}\left[ \begin{matrix} \text{1} & \text{0} & \text{0}  \\ \text{0} & \text{1} & \text{0}  \\ \text{0} & \text{0} & \text{1}  \\ \end{matrix} \right]\]

\[\Rightarrow {{\text{A}}^{\text{3}}}\text{-6}{{\text{A}}^{\text{2}}}\text{+9A-4I=}\left[ \begin{matrix} \text{22} & \text{-21} & \text{21}  \\ \text{-21} & \text{22} & \text{-21}  \\ \text{21} & \text{-21} & \text{22}  \\ \end{matrix} \right]\text{-}\left[ \begin{matrix} \text{36} & \text{-30} & \text{30}  \\ \text{-30} & \text{36} & \text{-30}  \\ \text{30} & \text{-30} & \text{36}  \\ \end{matrix} \right]\text{+}\left[ \begin{matrix} \text{18} & \text{-9} & \text{9}  \\ \text{-9} & \text{18} & \text{-9}  \\ \text{9} & \text{-9} & \text{18}  \\ \end{matrix} \right]\text{-}\left[ \begin{matrix} \text{4} & \text{0} & \text{0}  \\ \text{0} & \text{4} & \text{0}  \\ \text{0} & \text{0} & \text{4}  \\ \end{matrix} \right]\]

\[\Rightarrow {{\text{A}}^{3}}\text{-6}{{\text{A}}^{\text{2}}}\text{+9A-4I=}\left[ \begin{matrix} \text{40} & \text{-30} & \text{30}  \\ \text{-30} & \text{40} & \text{-30}  \\ \text{30} & \text{-30} & \text{40}  \\ \end{matrix} \right]\text{-}\left[ \begin{matrix} \text{40} & \text{-30} & \text{30}  \\ \text{-30} & \text{40} & \text{-30}  \\ \text{30} & \text{-30} & \text{40}  \\ \end{matrix} \right]\text{=}\left[ \begin{matrix} \text{0} & \text{0} & \text{0}  \\ \text{0} & \text{0} & \text{0}  \\ \text{0} & \text{0} & \text{0}  \\ \end{matrix} \right]\]

\[\therefore {{\text{A}}^{\text{3}}}\text{-6}{{\text{A}}^{\text{2}}}\text{+9A-4I=0}\]

Since, \[{{\text{A}}^{\text{3}}}\text{-6}{{\text{A}}^{\text{2}}}\text{+9A-4I=0}\]. 

Multiplying the whole equation by \[{{A}^{-1}}\], we have:

\[\Rightarrow \left( \text{AAA} \right){{\text{A}}^{\text{-1}}}\text{-6}\left( \text{AA} \right){{\text{A}}^{\text{-1}}}\text{+9A}{{\text{A}}^{\text{-1}}}\text{-4I}{{\text{A}}^{\text{-1}}}\text{=0}\] 

\[\Rightarrow \text{AA}\left( \text{A}{{\text{A}}^{\text{-1}}} \right)\text{-6A}\left( \text{A}{{\text{A}}^{\text{-1}}} \right)\text{+9}\left( \text{A}{{\text{A}}^{\text{-1}}} \right)\text{=4}\left( \text{I}{{\text{A}}^{\text{-1}}} \right)\]

\[\Rightarrow \text{AAI-6AI+9I=4}{{\text{A}}^{\text{-1}}}\]

\[\Rightarrow {{\text{A}}^{\text{2}}}\text{-6A+9I=4}{{\text{A}}^{\text{-1}}}\]

\[\Rightarrow {{\text{A}}^{\text{-1}}}\text{=}\dfrac{\text{1}}{\text{4}}\left( {{\text{A}}^{\text{2}}}\text{-6A+9I} \right)\]   …... (1)

Now, \[{{\text{A}}^{\text{2}}}\text{-6A+9I}\] is given by:

\[{{\text{A}}^{\text{2}}}\text{-6A+9I=}\left[ \begin{matrix} \text{6} & \text{-5} & \text{5}  \\ \text{-5} & \text{6} & \text{-5}  \\ \text{5} & \text{-5} & \text{6}  \\ \end{matrix} \right]\text{-6}\left[ \begin{matrix} \text{2} & \text{-1} & \text{1}  \\ \text{-1} & \text{2} & \text{-1}  \\ \text{1} & \text{-1} & \text{2}  \\ \end{matrix} \right]\text{+9}\left[ \begin{matrix} \text{0} & \text{0} & \text{0}  \\ \text{0} & \text{0} & \text{0}  \\ \text{0} & \text{0} & \text{0}  \\ \end{matrix} \right]\]

\[{{\text{A}}^{\text{2}}}\text{-6A+9I=}\left[ \begin{matrix} \text{6} & \text{-5} & \text{5}  \\ \text{-5} & \text{6} & \text{-5}  \\ \text{5} & \text{-5} & \text{6}  \\ \end{matrix} \right]\text{-}\left[ \begin{matrix} \text{12} & \text{-6} & \text{6}  \\ \text{-6} & \text{12} & \text{-6}  \\ \text{6} & \text{-6} & \text{12}  \\ \end{matrix} \right]\text{+}\left[ \begin{matrix} \text{9} & \text{0} & \text{0}  \\ \text{0} & \text{9} & \text{0}  \\ \text{0} & \text{0} & \text{9}  \\ \end{matrix} \right]\]

\[\therefore {{\text{A}}^{\text{2}}}\text{-6A+9I=}\left[ \begin{matrix} \text{3} & \text{1} & \text{-1}  \\ \text{1} & \text{3} & \text{1}  \\ \text{-1} & \text{3} & \text{3}  \\ \end{matrix} \right]\]

Substituting for \[{{\text{A}}^{\text{2}}}\text{-6A+9I}\] equation (1), we get

\[{{\text{A}}^{\text{-1}}}\text{=}\dfrac{\text{1}}{\text{4}}\left[ \begin{matrix} \text{3} & \text{1} & \text{-1}  \\ \text{1} & \text{3} & \text{1}  \\ \text{-1} & \text{3} & \text{3}  \\ \end{matrix} \right]\].

17. Let \[\mathbf{\text{A}}\] be nonsingular square matrix of order \[\mathbf{\text{3 }\!\!\times\!\!\text{ 3}}\] . Then \[\mathbf{\left| \text{adjA} \right|}\] is equal to

1. \[\mathbf{\left| \text{A} \right|}\]

2. \[\mathbf{{{\left| \text{A} \right|}^{\text{2}}}}\]

3. \[\mathbf{{{\left| \text{A} \right|}^{\text{3}}}}\]

4. \[\mathbf{\text{3}\left| \text{A} \right|}\]

Ans: Given $A$ is a nonsingular square matrix, i.e., it is a square matrix whose determinant is not equal to zero.

The inverse of a matrix is given as \[{{\text{A}}^{\text{-1}}}\text{=}\dfrac{\text{1}}{\left| \text{A} \right|}adjA\].

$ \Rightarrow {{A}^{-1}}A=\dfrac{1}{\left| A \right|}adjA$ 

$\Rightarrow \left| A \right|I=adjA$

The adjoint of matrix $A$ is given by,

\[\Rightarrow \left( \text{adjA} \right)\text{=A=}\left| \text{A} \right|\text{I=}\left[ \begin{matrix} \left| \text{A} \right| & \text{0} & \text{0}  \\ \text{0} & \left| \text{A} \right| & \text{0}  \\ \text{0} & \text{0} & \left| \text{A} \right|  \\ \end{matrix} \right]\]

\[\Rightarrow \left| \left( \text{adjA} \right)\text{A} \right|\text{=}\left[ \begin{matrix} \left| \text{A} \right| & \text{0} & \text{0}  \\ \text{0} & \left| \text{A} \right| & \text{0}  \\ \text{0} & \text{0} & \left| \text{A} \right|  \\ \end{matrix} \right]\]

\[\Rightarrow \left| \text{adjA} \right|\left| \text{A} \right|\text{=}{{\left| \text{A} \right|}^{\text{3}}}\left| \begin{matrix} \text{1} & \text{0} & \text{0}  \\ \text{0} & \text{1} & \text{0}  \\ \text{0} & \text{0} & \text{1}  \\ \end{matrix} \right|\text{=}{{\left| \text{A} \right|}^{\text{3}}}\left( \text{I} \right)\]

\[\therefore \left| \text{adjA} \right|\text{=}{{\left| \text{A} \right|}^{\text{3}}}\]

Hence, B. \[{{\left| \text{A} \right|}^{\text{2}}}\] is the correct answer.

18. If \[\mathbf{\text{A}}\] is an invertible matrix of order \[\mathbf{\text{2}}\] , then \[\mathbf{\text{det}\left( {{\text{A}}^{\text{-1}}} \right)}\] is equal to

1. \[\mathbf{\text{det}\left( \text{A} \right)}\]

2. \[\mathbf{\dfrac{\text{1}}{\text{det}\left( \text{A} \right)}}\]

3. \[\mathbf{\text{1}}\]

4. \[\mathbf{\text{0}}\]

Ans: Since \[\text{A}\] is an invertible matrix, thus \[{{\text{A}}^{\text{-1}}}\] exists and it is given by: \[{{\text{A}}^{\text{-1}}}\text{=}\dfrac{\text{1}}{\left| \text{A} \right|}\text{adjA}\].

As matrix \[\text{A}\] is of order \[\text{2}\], 

$\therefore $ Let \[\text{A=}\left[ \begin{matrix} \text{a} & \text{b}  \\ \text{c} & \text{d}  \\ \end{matrix} \right]\] .

Hence, \[\left| \text{A} \right|\text{=ad-bc}\].

The adjoint of \[\text{A}\] would be, \[\text{adjA=}\left[ \begin{matrix} \text{d} & \text{-b}  \\ \text{-c} & \text{a}  \\ \end{matrix} \right]\] .

Now, the inverse of the matrix is given by:

\[\Rightarrow {{\text{A}}^{\text{-1}}}\text{=}\dfrac{\text{1}}{\left| \text{A} \right|}\text{adjA}\]

\[\Rightarrow {{\text{A}}^{\text{-1}}}\text{=}\left[ \begin{matrix} \dfrac{\text{d}}{\left| \text{A} \right|} & \dfrac{\text{-b}}{\left| \text{A} \right|}  \\ \dfrac{\text{-c}}{\left| \text{A} \right|} & \dfrac{\text{a}}{\left| \text{A} \right|}  \\ \end{matrix} \right]\]

\[\Rightarrow {{\text{A}}^{\text{-1}}}\text{=}\left[ \begin{matrix} \dfrac{\text{d}}{\left| \text{A} \right|} & \dfrac{\text{-b}}{\left| \text{A} \right|}  \\ \dfrac{\text{-c}}{\left| \text{A} \right|} & \dfrac{\text{a}}{\left| \text{A} \right|}  \\ \end{matrix} \right]\]

\[\Rightarrow \left| {{\text{A}}^{\text{-1}}} \right|\text{=}\dfrac{\text{1}}{\left| {{\text{A}}^{\text{2}}} \right|}\left| \begin{matrix} \text{d} & \text{-b}  \\ \text{-c} & \text{a}  \\ \end{matrix} \right|\]

\[\Rightarrow \left| {{\text{A}}^{\text{-1}}} \right|=\dfrac{\text{1}}{\left| {{\text{A}}^{\text{2}}} \right|}\left( \text{ad-bc} \right)\]

\[\Rightarrow \left| {{\text{A}}^{\text{-1}}} \right|\text{=}\dfrac{\text{1}}{\left| {{\text{A}}^{\text{2}}} \right|}\text{.}\left| \text{A} \right|\]

\[\therefore \left| {{\text{A}}^{\text{-1}}} \right|\text{=}\dfrac{\text{1}}{\left| \text{A} \right|}\]

Thus, \[\text{det}\left( {{\text{A}}^{\text{-1}}} \right)\text{=}\dfrac{\text{1}}{\text{det}\left( \text{A} \right)}\].

Hence, B. \[\dfrac{\text{1}}{\text{det}\left( \text{A} \right)}\] is the correct answer.

Exercise (4.5)

1. Examine the consistency of the system of equations.

\[\mathbf{\text{x+2y=2}}\]

\[\mathbf{\text{2x+3y=3}}\]

Ans: Given equations,

\[\text{x+2y=2}\]

\[\text{2x+3y=3}\]

Let us suppose \[\text{A=}\left[ \begin{matrix} \text{1} & \text{2}  \\ \text{2} & \text{3}  \\ \end{matrix} \right]\], \[\text{X=}\left[ \begin{matrix} x  \\ y  \\ \end{matrix} \right]\] and \[\text{B=}\left[ \begin{matrix} \text{2}  \\ \text{3}  \\ \end{matrix} \right]\] such that, the given system of equations can be written in the form of \[\text{AX=B}\].

Determining the value of $A$, we have:

\[\Rightarrow \left| \text{A} \right|\text{=1}\left( \text{3} \right)-\text{2}\left( \text{2} \right)\text{=3}-\text{4}\]

\[\therefore \left| \text{A} \right|\text{=}-1\ne 0\]

Hence, \[\text{A}\] is non-singular.

Thus, the inverse of $A$, i.e., \[{{\text{A}}^{\text{-1}}}\] exists.

$\therefore $ The given system of equations is consistent.

2. Examine the consistency of the system of equations.

\[\mathbf{\text{2x-y=5}}\]

\[\mathbf{\text{x+y=4}}\]

Ans: Given equations,

\[\text{2x-y=5}\]

\[\text{x+y=4}\]

Let us suppose\[\text{A=}\left[ \begin{matrix} \text{2} & \text{-1}  \\ \text{1} & \text{1}  \\ \end{matrix} \right]\], \[\text{X=}\left[ \begin{matrix} \text{x}  \\ y  \\ \end{matrix} \right]\] and \[\text{B=}\left[ \begin{matrix} \text{5}  \\ \text{4}  \\ \end{matrix} \right]\] such that, the given system of equation can be written in the form of \[\text{AX=B}\].

Determining the value of $A$, we have:

\[\Rightarrow \left| \text{A} \right|\text{=2}\left( \text{1} \right)\text{-}\left( \text{-1} \right)\left( \text{1} \right)\text{=2+1}\]

\[\therefore \left| \text{A} \right|=3\ne 0\]

Hence, \[\text{A}\] is non-singular.

Thus, \[{{\text{A}}^{\text{-1}}}\] exists.

$\therefore $ The given system of equations is consistent.

3. Examine the consistency of the system of equations.

\[\mathbf{\text{x+3y=5}}\]

\[\mathbf{\text{2x+6y=8}}\]

Ans: Given equations,

\[\text{x+3y=5}\]

\[\text{2x+6y=8}\]

We know, that a given system of equations is consistent if it has at least one solution.

Let \[\text{A=}\left[ \begin{matrix} \text{1} & \text{3}  \\ \text{2} & \text{6}  \\ \end{matrix} \right]\] , \[\text{X=}\left[ \begin{matrix} \text{x}  \\ \text{y}  \\ \end{matrix} \right]\] and \[\text{B=}\left[ \begin{matrix} \text{5}  \\ \text{8}  \\ \end{matrix} \right]\] such that, the given system of equation can be written in the form of \[\text{AX=B}\].

Determining the value of $A$, we have:

\[\Rightarrow \left| \text{A} \right|\text{=1}\left( \text{6} \right)\text{-3}\left( \text{2} \right)\]

\[\therefore \left| \text{A} \right|\text{=}0\]

Hence, \[\text{A}\] is a singular matrix.

We know that the adjoint of a square matrix is the transpose of its cofactor matrix.

Determining the adjoint of the matrix $A$, we have:

\[\Rightarrow \left( \text{adjA} \right)\text{=}\left[ \begin{matrix} \text{6} & \text{-3}  \\ \text{-2} & \text{1}  \\ \end{matrix} \right]\]

\[\Rightarrow \left( \text{adjA} \right)\text{B=}\left[ \begin{matrix} \text{6} & \text{-3}  \\ \text{-2} & \text{1}  \\ \end{matrix} \right]\left[ \begin{matrix} \text{5}  \\ \text{8}  \\ \end{matrix} \right]\text{=}\left[ \begin{matrix} \text{30-24}  \\ \text{-10+8}  \\ \end{matrix} \right]\] \[\therefore \left( \text{adjA} \right)\text{B=}\left[ \begin{matrix} \text{6}  \\ \text{-2}  \\ \end{matrix} \right]\ne 0\]

Thus, the solution of the given system of equations does not exists.

$\therefore $ The given system of equations is inconsistent.

4. Examine the consistency of the system of equations.

\[\mathbf{\text{x+y+z=1}}\]

\[\mathbf{\text{2x+3y+2z=2}}\]

\[\mathbf{\text{ax+ay+2az=4}}\]

Ans: Given equations,

\[\text{x+y+z=1}\]

\[\text{2x+3y+2z=2}\]

\[\text{ax+ay+2az=4}\]

We know, that a given system of equations is consistent if it has at least one solution.

Let \[\text{A=}\left[ \begin{matrix} \text{1} & \text{1} & \text{1}  \\ \text{2} & \text{3} & \text{2}  \\ \text{a} & \text{a} & \text{2a}  \\ \end{matrix} \right]\] , \[\text{X=}\left[ \begin{matrix} \text{x}  \\ \text{y}  \\ \text{z}  \\ \end{matrix} \right]\] and \[\text{B=}\left[ \begin{matrix} \text{1}  \\ \text{2}  \\ \text{4}  \\ \end{matrix} \right]\] such that, the given system of equation can be written in the form of \[\text{AX=B}\].

Determining the value of $A$, we have:

\[\Rightarrow \left| \text{A} \right|\text{=1}\left( \text{6a}-\text{2a} \right)-\text{1}\left( \text{4a}-\text{2a} \right)\text{+1}\left( \text{2a}-\text{3a} \right)\]

\[\Rightarrow \left| A \right|\text{=4a}-\text{2a}-a\]

\[\therefore \left| A \right|\text{=}a\ne 0\]

Hence \[\text{A}\] is non-singular matrix.

Thus, \[{{\text{A}}^{\text{-1}}}\] exists.

$\therefore $ The given system of equation is consistent.

5. Examine the consistency of the system of equations.

\[\mathbf{\text{3x-y-2z=2}}\]

\[\mathbf{\text{2y-z=-1}}\]

\[\mathbf{\text{3x-5y=3}}\]

Ans: Given equations,

\[\text{3x-y-2z=2}\]

\[\text{2y-z=-1}\]

\[\text{3x-5y=3}\]

We know, that a given system of equations is consistent if it has at least one solution.

Let \[\text{A=}\left[ \begin{matrix} \text{3} & \text{-1} & \text{-2}  \\ \text{0} & \text{2} & \text{-1}  \\ \text{3} & \text{-5} & \text{0}  \\ \end{matrix} \right]\] , \[\text{X=}\left[ \begin{matrix} \text{x}  \\ \text{y}  \\ \text{z}  \\ \end{matrix} \right]\] and \[\text{B=}\left[ \begin{matrix} \text{2}  \\ \text{-1}  \\ \text{3}  \\ \end{matrix} \right]\] such that, this system of equations can be written in the form of \[\text{AX=B}\].

Determining the value of $A$, we have:

\[\Rightarrow \left| \text{A} \right|\text{=3}\left( \text{-5} \right)\text{-0+3}\left( \text{1+4} \right)\]

\[\Rightarrow \left| \text{A} \right|\text{=-15+15}\]

\[\Rightarrow \left| \text{A} \right|\text{=0}\]

\[\therefore \text{A}\] is a singular matrix.

Determining the adjoint of matrix $A$.

Writing the cofactors,

${{A}_{11}}={{(-1)}^{1+1}}\left| \begin{matrix} 2 & -1  \\ -5 & 0  \\ \end{matrix} \right|=-5$

${{A}_{12}}={{(-1)}^{1+2}}\left| \begin{matrix} 0 & -1  \\ 3 & 0  \\ \end{matrix} \right|=-3$

${{A}_{13}}={{(-1)}^{1+3}}\left| \begin{matrix} 0 & 2  \\ 3 & -5  \\ \end{matrix} \right|=-6$

${{A}_{21}}={{(-1)}^{2+1}}\left| \begin{matrix} -1 & -2  \\ -5 & 0  \\ \end{matrix} \right|=10$

${{A}_{22}}={{(-1)}^{2+2}}\left| \begin{matrix} 3 & -2  \\ 3 & 0  \\ \end{matrix} \right|=6$

${{A}_{23}}={{(-1)}^{2+3}}\left| \begin{matrix} 3 & -1  \\ 3 & -5  \\ \end{matrix} \right|=-[-15+3]=12$

${{A}_{31}}={{(-1)}^{3+1}}\left| \begin{matrix} -1 & -2  \\ 2 & -1  \\ \end{matrix} \right|=[1+4]=5$

${{A}_{32}}={{(-1)}^{3+2}}\left| \begin{matrix} 3 & -2  \\ 0 & -1  \\ \end{matrix} \right|=3$

${{A}_{33}}={{(-1)}^{3+3}}\left| \begin{matrix} 3 & -1  \\ 0 & 2  \\ \end{matrix} \right|=6$

Cofactor matrix is \[\left[ \begin{matrix} \text{-5} & \text{-3} & -6  \\ 10 & \text{6} & 12  \\ 5 & 3 & \text{6}  \\ \end{matrix} \right]\]

Taking transpose,

\[\Rightarrow \left( \text{adjA} \right)\text{=}\left[ \begin{matrix} \text{-5} & \text{10} & \text{5}  \\ \text{-3} & \text{6} & \text{3}  \\ \text{-6} & \text{12} & \text{6}  \\ \end{matrix} \right]\]

\[\Rightarrow \left( \text{adjA} \right)\text{B=}\left[ \begin{matrix} \text{-5} & \text{10} & \text{5}  \\ \text{-3} & \text{6} & \text{3}  \\ \text{-6} & \text{12} & \text{6}  \\ \end{matrix} \right]\left[ \begin{matrix} \text{2}  \\ \text{-1}  \\ \text{3}  \\ \end{matrix} \right]\text{=}\left[ \begin{matrix} \text{-10-10+15}  \\ \text{-6-6+9}  \\ \text{-12-12+18}  \\ \end{matrix} \right]\]

\[\therefore \left( \text{adjA} \right)\text{B=}\left[ \begin{matrix} \text{-5}  \\ \text{-3}  \\ \text{-6}  \\ \end{matrix} \right]\ne 0\]

Thus, the solution of the given system of equation does not exist.

$\therefore $ The system of equations is inconsistent.

6. Examine the consistency of the system of equations.

\[\mathbf{\text{5x-y+4z=5}}\]

\[\mathbf{\text{2x+3y+5z=2}}\]

\[\mathbf{\text{5x-2y+6z=-1}}\]

Ans: Given equations,

\[\text{5x-y+4z=5}\]

\[\text{2x+3y+5z=2}\]

\[\text{5x-2y+6z=-1}\]

Let \[\text{A=}\left[ \begin{matrix} \text{5} & \text{-1} & \text{4}  \\ \text{2} & \text{3} & \text{5}  \\ \text{3} & \text{-2} & \text{6}  \\ \end{matrix} \right]\] , \[\text{X=}\left[ \begin{matrix} \text{x}  \\ \text{y}  \\ \text{z}  \\ \end{matrix} \right]\] and \[\text{B=}\left[ \begin{matrix} \text{5}  \\ \text{2}  \\ \text{-1}  \\ \end{matrix} \right]\] such that, the system of equations can be written in the form of \[\text{AX=B}\].

Determining the value of $A$, we have:

\[\Rightarrow \left| \text{A} \right|\text{=5}\left( \text{18+10} \right)\text{+1}\left( \text{12-25} \right)\text{+4}\left( \text{-4-15} \right)\]

\[\Rightarrow \left| \text{A} \right|\text{=5}\left( \text{28} \right)\text{+1}\left( \text{-13} \right)\text{+4}\left( \text{-19} \right)\]

\[\Rightarrow \left| \text{A} \right|\text{=140-13-76}\]

\[\therefore \left| \text{A} \right|\text{=51}\ne \text{0}\]

Hence, \[\text{A}\] is a non-singular matrix.

Thus, \[{{\text{A}}^{\text{-1}}}\] exists.

$\therefore $ The given system of equations is consistent.

7. Solve system of linear equations, using matrix method.

\[\mathbf{\text{5x+2y=4}}\]

\[\mathbf{\text{7x+3y=5}}\]

Ans: Given equations,

\[\text{5x+2y=4}\]

\[\text{7x+3y=5}\]

Let \[\text{A=}\left[ \begin{matrix} \text{5} & \text{2}  \\ \text{7} & \text{3}  \\ \end{matrix} \right]\] , \[\text{X=}\left[ \begin{matrix} \text{x}  \\ \text{y}  \\ \end{matrix} \right]\] and \[\text{B=}\left[ \begin{matrix} \text{4}  \\ \text{5}  \\ \end{matrix} \right]\] such that, this system of equations can be written in the form of \[\text{AX=B}\].

Determining the value of $A$, we have:

\[\Rightarrow \left| \text{A} \right|\text{=15-14}\]

\[\therefore \left| \text{A} \right|=1\ne 0\]

Thus, \[\text{A}\] is a non-singular matrix.

Hence, its inverse exists.

We know that the adjoint of a square matrix is the transpose of its cofactor matrix.

$\therefore adjA=\left[ \begin{matrix} 3 & -2  \\ -7 & 5  \\ \end{matrix} \right]$

Thus,

\[\Rightarrow {{\text{A}}^{\text{-1}}}=\left[ \begin{matrix} \text{3} & \text{-2}  \\ \text{-7} & \text{5}  \\ \end{matrix} \right]\]

\[\Rightarrow \text{X=}{{\text{A}}^{\text{-1}}}\text{B=}\left[ \begin{matrix} \text{3} & \text{-2}  \\ \text{-7} & \text{5}  \\ \end{matrix} \right]\left[ \begin{matrix} \text{4}  \\ \text{5}  \\ \end{matrix} \right]\]

\[\Rightarrow \left[ \begin{matrix} \text{x}  \\ \text{y}  \\ \end{matrix} \right]\text{=}\left[ \begin{matrix} \text{12+10}  \\ \text{-28+52}  \\ \end{matrix} \right]\text{=}\left[ \begin{matrix} \text{2}  \\ \text{-3}  \\ \end{matrix} \right]\]

Thus, \[\text{x=2}\] and \[\text{y=}-3\].

8. Solve system of linear equations, using matrix method.

\[\mathbf{\text{2x-y=-2}}\]

\[\mathbf{\text{3x+4y=3}}\]

Ans: Given equations,

\[\text{2x-y=-2}\]

\[\text{3x+4y=3}\]

Let \[\text{A=}\left[ \begin{matrix} \text{2} & \text{-1}  \\ \text{3} & \text{4}  \\ \end{matrix} \right]\] , \[\text{X=}\left[ \begin{matrix} \text{x}  \\ \text{y}  \\ \end{matrix} \right]\] and \[\text{B=}\left[ \begin{matrix} \text{-2}  \\ \text{3}  \\ \end{matrix} \right]\] such that, this system of equations can be written in the form of \[\text{AX=B}\].

Determining the value of $A$, we have:

\[\Rightarrow \left| \text{A} \right|\text{=8+3}\]

\[\therefore \left| \text{A} \right|=11\ne 0\]

Thus, \[\text{A}\] is non-singular.

 $\therefore $ It’s inverse exists.

We know that the adjoint of a square matrix is the transpose of its cofactor matrix.

$\therefore adjA=\left[ \begin{matrix} 4 & 1  \\ -3 & 2  \\ \end{matrix} \right]$

Now,

\[\Rightarrow {{\text{A}}^{\text{-1}}}\text{=}\dfrac{\text{1}}{\left| \text{A} \right|}\left( \text{adjA} \right)\]

\[\Rightarrow {{\text{A}}^{\text{-1}}}\text{=}\dfrac{\text{1}}{\text{11}}\left[ \begin{matrix} \text{4} & \text{1}  \\ \text{-3} & \text{2}  \\ \end{matrix} \right]\]

\[\therefore \text{X=}{{\text{A}}^{\text{-1}}}\text{B=}\dfrac{\text{1}}{\text{11}}\left[ \begin{matrix} \text{4} & \text{1}  \\ \text{-3} & \text{2}  \\ \end{matrix} \right]\left[ \begin{matrix} \text{-2}  \\ \text{3}  \\ \end{matrix} \right]\]

\[\Rightarrow \left[ \begin{matrix} \text{x}  \\ \text{y}  \\ \end{matrix} \right]\text{=}\dfrac{\text{1}}{\text{11}}\left[ \begin{matrix} \text{-8+3}  \\ \text{6+6}  \\ \end{matrix} \right]\]

\[\Rightarrow \left[ \begin{matrix} \text{x}  \\ \text{y}  \\ \end{matrix} \right]\text{=}\dfrac{\text{1}}{\text{11}}\left[ \begin{matrix} \text{-5}  \\ \text{12}  \\ \end{matrix} \right]\]

\[\Rightarrow \left[ \begin{matrix} \text{x}  \\ \text{y}  \\ \end{matrix} \right]\text{=}\left[ \begin{matrix} \text{-}\dfrac{\text{5}}{\text{11}}  \\ \dfrac{\text{12}}{\text{11}}  \\ \end{matrix} \right]\]

Thus, \[x=-\dfrac{5}{11}\] and \[y=\dfrac{12}{11}\].

9. Solve system of linear equations, using matrix method.

\[\mathbf{\text{4x-3y=3}}\]

\[\mathbf{\text{3x-5y=7}}\]

Ans: Given equations,

\[\text{4x-3y=3}\]

\[\text{3x-5y=7}\].

Let \[\text{A=}\left[ \begin{matrix} \text{4} & \text{-3}  \\ \text{3} & \text{-5}  \\ \end{matrix} \right]\] , \[\text{X=}\left[ \begin{matrix} \text{x}  \\ \text{y}  \\ \end{matrix} \right]\] and \[\text{B=}\left[ \begin{matrix} \text{3}  \\ \text{7}  \\ \end{matrix} \right]\] such that, this system of equations can be written in the form of \[\text{AX=B}\].

Determining the value of $A$, we have:

\[\Rightarrow \left| \text{A} \right|\text{=-20+9}\]

\[\therefore \left| \text{A} \right|\text{=-11}\ne \text{0}\]

Thus, \[\text{A}\] is a non-singular matrix.

Hence, its inverse exists.

Formula for inverse is ${{A}^{-1}}=\dfrac{adjA}{\left| A \right|}$.

Finding cofactors,

${{A}_{11}}={{(-1)}^{1+1}}(-5)=-5$

${{A}_{12}}={{(-1)}^{1+2}}(3)=-3$

${{A}_{21}}={{(-1)}^{2+1}}(-3)=3$

${{A}_{22}}={{(-1)}^{2+2}}(4)=4$

Cofactor matrix is $\left[ \begin{matrix} -5 & -3  \\ 3 & 4  \\ \end{matrix} \right]$.

Taking its transpose to get adjoint matrix as $\left[ \begin{matrix} -5 & 3  \\ -3 & 4  \\ \end{matrix} \right]$.

Therefore inverse is

\[\Rightarrow {{\text{A}}^{\text{-1}}}\text{=-}\dfrac{\text{1}}{\text{11}}\left[ \begin{matrix} \text{-5} & \text{3}  \\ \text{-3} & \text{4}  \\ \end{matrix} \right]\]

\[\therefore \text{X=}{{\text{A}}^{\text{-1}}}\text{B=-}\dfrac{\text{1}}{\text{11}}\left[ \begin{matrix} \text{-5} & \text{3}  \\ \text{-3} & \text{4}  \\ \end{matrix} \right]\left[ \begin{matrix} \text{3}  \\ \text{7}  \\ \end{matrix} \right]\]

\[\Rightarrow \left[ \begin{matrix} \text{x}  \\ \text{y}  \\ \end{matrix} \right]\text{=}\dfrac{\text{1}}{\text{11}}\left[ \begin{matrix} \text{5} & \text{-3}  \\ \text{3} & \text{-4}  \\ \end{matrix} \right]\left[ \begin{matrix} \text{3}  \\ \text{7}  \\ \end{matrix} \right]\]

\[\Rightarrow \left[ \begin{matrix} \text{x}  \\ \text{y}  \\ \end{matrix} \right]\text{=}\dfrac{\text{1}}{\text{11}}\left[ \begin{matrix} \text{15-21}  \\ \text{9-28}  \\ \end{matrix} \right]\]

\[\Rightarrow \left[ \begin{matrix} \text{x}  \\ \text{y}  \\ \end{matrix} \right]\text{=}\left[ \begin{matrix} \text{-}\dfrac{\text{6}}{\text{11}}  \\ \text{-}\dfrac{\text{19}}{\text{11}}  \\ \end{matrix} \right]\]

Thus, \[x=-\dfrac{6}{11}\] and \[y=-\dfrac{19}{11}\].

10. Solve system of linear equations, using matrix method.

\[\mathbf{\text{5x+2y=3}}\]

\[\mathbf{\text{3x+2y=5}}\]

Ans: Given equations,

\[\text{5x+2y=3}\]

\[\text{5x+2y=3}\]

Let \[\text{A=}\left[ \begin{matrix} \text{5} & \text{2}  \\ \text{3} & \text{2}  \\ \end{matrix} \right]\] , \[\text{X=}\left[ \begin{matrix} \text{x}  \\ \text{y}  \\ \end{matrix} \right]\] and \[\text{B=}\left[ \begin{matrix} \text{3}  \\ \text{5}  \\ \end{matrix} \right]\] such that, this system of equations can be written in the form of \[\text{AX=B}\].

Determining the value of $A$, we have:

\[\Rightarrow \left| \text{A} \right|\text{=10}-6\]

\[\therefore \left| \text{A} \right|\text{=4}\ne \text{0}\]

Thus \[\text{A}\] is non-singular,

Therefore, its inverse exists.

Formula for inverse is ${{A}^{-1}}=\dfrac{adjA}{\left| A \right|}$.

Finding cofactors,

${{A}_{11}}={{(-1)}^{1+1}}(2)=2$

${{A}_{12}}={{(-1)}^{1+2}}(3)=-3$

${{A}_{21}}={{(-1)}^{2+1}}(2)=-2$

${{A}_{22}}={{(-1)}^{2+2}}(5)=5$

Cofactor matrix is $\left[ \begin{matrix} 2 & -3  \\ -2 & 5  \\ \end{matrix} \right]$.

Taking its transpose to get adjoint matrix as $\left[ \begin{matrix} 2 & -2  \\ -3 & 5  \\ \end{matrix} \right]$.

Therefore inverse is

\[\Rightarrow {{\text{A}}^{\text{-1}}}\text{=}\dfrac{\text{1}}{4}\left[ \begin{matrix} 2 & -2  \\ -3 & 5  \\ \end{matrix} \right]\]

\[\therefore \text{X=}{{\text{A}}^{\text{-1}}}\text{B=}\dfrac{\text{1}}{4}\left[ \begin{matrix} 2 & -2  \\ -3 & 5  \\ \end{matrix} \right]\left[ \begin{matrix} \text{3}  \\ 5  \\ \end{matrix} \right]\]

\[\Rightarrow \left[ \begin{matrix} \text{x}  \\ \text{y}  \\ \end{matrix} \right]\text{=}\dfrac{\text{1}}{4}\left[ \begin{matrix} 2 & -2  \\ -3 & 5  \\ \end{matrix} \right]\left[ \begin{matrix} \text{3}  \\ 5  \\ \end{matrix} \right]\]

\[\Rightarrow \left[ \begin{matrix} \text{x}  \\ \text{y}  \\ \end{matrix} \right]\text{=}\dfrac{\text{1}}{4}\left[ \begin{matrix} 6-10  \\ -9+25  \\ \end{matrix} \right]\]

\[\Rightarrow \left[ \begin{matrix} \text{x}  \\ \text{y}  \\ \end{matrix} \right]\text{=}\left[ \begin{matrix} -\dfrac{4}{4}  \\ \dfrac{\text{16}}{4}  \\ \end{matrix} \right]\]

Thus, \[x=-1\] and \[y=4\].

11. Solve system of linear equations, using matrix method.

\[\mathbf{\text{2x+y+z=1}}\]

\[\mathbf{\text{x-2y-z=}\dfrac{\text{3}}{\text{2}}}\]

\[\mathbf{\text{3y-5z=9}}\]

Ans: Given equations,

\[\text{2x+y+z=1}\]

\[\text{x-2y-z=}\dfrac{\text{3}}{\text{2}}\]

\[\text{3y-5z=9}\]

Let \[\text{A=}\left[ \begin{matrix} \text{2} & \text{1} & \text{1}  \\ \text{1} & \text{-2} & \text{-1}  \\ \text{0} & \text{3} & \text{-5}  \\ \end{matrix} \right]\] , \[\text{X=}\left[ \begin{matrix} \text{x}  \\ \text{y}  \\ \text{z}  \\ \end{matrix} \right]\] and \[\text{B=}\left[ \begin{matrix} \text{1}  \\ \dfrac{\text{3}}{\text{2}}  \\ \text{9}  \\ \end{matrix} \right]\] such that, this system of equations can be written in the form of \[\text{AX=B}\].

The determinant of $A$ is found by expanding along the first column:

\[\Rightarrow \left| \text{A} \right|\text{=2}\left( \text{10+3} \right)\text{-1}\left( \text{-5-3} \right)\text{+0}\]

\[\Rightarrow \left| \text{A} \right|\text{=2}\left( \text{13} \right)\text{-1}\left( \text{-8} \right)\]

\[\Rightarrow \left| \text{A} \right|=\text{34}\ne \text{0}\]

Thus, \[\text{A}\] is a non-singular matrix.

$\therefore $ Its inverse exists.

Hence,

\[\Rightarrow {{\text{A}}_{11}}\text{=}{{\left( \text{-1} \right)}^{1+1}}{{\text{M}}_{11}}={{\left( \text{-1} \right)}^{2}}{{\text{M}}_{11}}\]

\[\Rightarrow {{A}_{11}}\text{=13}\]

\[\Rightarrow {{\text{A}}_{12}}\text{=}{{\left( \text{-1} \right)}^{1+2}}{{\text{M}}_{12}}={{\left( \text{-1} \right)}^{3}}{{\text{M}}_{12}}\]

\[\Rightarrow {{A}_{12}}=5\]

\[\Rightarrow {{\text{A}}_{13}}\text{=}{{\left( \text{-1} \right)}^{1+3}}{{\text{M}}_{13}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{13}}\]

\[\Rightarrow {{A}_{13}}=3\]

Similarly,

\[\Rightarrow {{\text{A}}_{21}}\text{=}{{\left( \text{-1} \right)}^{2+1}}{{\text{M}}_{21}}={{\left( \text{-1} \right)}^{3}}{{\text{M}}_{21}}\]

\[\Rightarrow {{\text{A}}_{21}}\text{=8}\]

\[\Rightarrow {{\text{A}}_{22}}\text{=}{{\left( \text{-1} \right)}^{2+2}}{{\text{M}}_{22}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{22}}\]

\[\Rightarrow {{\text{A}}_{22}}=-10\]

\[\Rightarrow {{\text{A}}_{23}}\text{=}{{\left( \text{-1} \right)}^{2+3}}{{\text{M}}_{23}}={{\left( \text{-1} \right)}^{5}}{{\text{M}}_{23}}\]

\[\Rightarrow {{\text{A}}_{23}}\text{=}-6\]

and

\[\Rightarrow {{\text{A}}_{31}}\text{=}{{\left( \text{-1} \right)}^{3+1}}{{\text{M}}_{31}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{31}}\]

\[\Rightarrow {{\text{A}}_{31}}=1\]

\[\Rightarrow {{\text{A}}_{32}}\text{=}{{\left( \text{-1} \right)}^{3+2}}{{\text{M}}_{32}}={{\left( \text{-1} \right)}^{5}}{{\text{M}}_{32}}\]

\[\Rightarrow {{\text{A}}_{32}}=3\]

\[\Rightarrow {{\text{A}}_{33}}\text{=}{{\left( \text{-1} \right)}^{3+3}}{{\text{M}}_{33}}={{\left( \text{-1} \right)}^{6}}{{\text{M}}_{33}}\]

\[\Rightarrow {{\text{A}}_{33}}=-5\].

Cofactor matrix is \[\left[ \begin{matrix} 13 & 5 & 3  \\ 8 & -10 & -6  \\ 1 & 3 & -5  \\ \end{matrix} \right]\].

We know that adjoint of a matrix is the transpose of its cofactor matrix.

\[\Rightarrow adjA={{\left[ \begin{matrix} 13 & 5 & 3  \\ 8 & -10 & -6  \\ 1 & 3 & -5  \\ \end{matrix} \right]}^{T}}\]

\[\Rightarrow adjA=\left[ \begin{matrix} \text{13} & \text{8} & \text{1}  \\ \text{5} & \text{-10} & \text{3}  \\ \text{3} & \text{-16} & \text{-5}  \\ \end{matrix} \right]\]

The inverse of a matrix is given by:

\[\Rightarrow {{\text{A}}^{\text{-1}}}\text{=}\dfrac{\text{1}}{\left| \text{A} \right|}\left( \text{adjA} \right)\]

\[\therefore {{\text{A}}^{\text{-1}}}\text{=}\dfrac{\text{1}}{\text{34}}\left[ \begin{matrix} \text{13} & \text{8} & \text{1}  \\ \text{5} & \text{-10} & \text{3}  \\ \text{3} & \text{-16} & \text{-5}  \\ \end{matrix} \right]\]

\[\therefore \text{X=}{{\text{A}}^{\text{-1}}}\text{B=}\dfrac{\text{1}}{\text{34}}\left[ \begin{matrix} \text{13} & \text{8} & \text{1}  \\ \text{5} & \text{-10} & \text{3}  \\ \text{3} & \text{-16} & \text{-5}  \\ \end{matrix} \right]\left[ \begin{matrix} \text{1}  \\ \dfrac{\text{3}}{\text{2}}  \\ \text{9}  \\ \end{matrix} \right]\]

\[\Rightarrow \left[ \begin{matrix} \text{x}  \\ \text{y}  \\ \text{z}  \\ \end{matrix} \right]\text{=}\dfrac{\text{1}}{\text{34}}\left[ \begin{matrix} \text{13+12+9}  \\ \text{5-15+27}  \\ \text{3-9-45}  \\ \end{matrix} \right]\]

\[\Rightarrow \left[ \begin{matrix} \text{x}  \\ \text{y}  \\ \text{z}  \\ \end{matrix} \right]\text{=}\dfrac{\text{1}}{\text{34}}\left[ \begin{matrix} \text{34}  \\ \text{17}  \\ \text{-51}  \\ \end{matrix} \right]\]

\[\Rightarrow \left[ \begin{matrix} \text{x}  \\ \text{y}  \\ \text{z}  \\ \end{matrix} \right]\text{=}\left[ \begin{matrix} \text{1}  \\ \dfrac{\text{1}}{\text{2}}  \\ \text{-}\dfrac{\text{3}}{\text{2}}  \\ \end{matrix} \right]\]

Thus, \[x=1\] and \[y=\dfrac{1}{2}\] and \[z=-\dfrac{3}{2}\].

12. Solve system of linear equations, using matrix method.

\[\mathbf{\text{x-y+z=4}}\]

\[\mathbf{\text{2x+y-3z=0}}\]

\[\mathbf{\text{x+y+z=2}}\]

Ans: Given equations,

\[\text{x-y+z=4}\]

\[\text{2x+y-3z=0}\]

\[\text{x+y+z=2}\]

Let \[\text{A=}\left[ \begin{matrix} \text{1} & \text{1} & \text{1}  \\ \text{2} & \text{1} & \text{-3}  \\ \text{1} & \text{1} & \text{1}  \\ \end{matrix} \right]\] , \[\text{X=}\left[ \begin{matrix} \text{x}  \\ \text{y}  \\ \text{z}  \\ \end{matrix} \right]\] and \[\text{B=}\left[ \begin{matrix} \text{4}  \\ \text{0}  \\ \text{2}  \\ \end{matrix} \right]\] such that, this system of equations can be written in the form of \[\text{AX=B}\].

Determining the value of $A$, we have:

\[\Rightarrow \left| \text{A} \right|\text{=1}\left( \text{1+3} \right)\text{+1}\left( \text{2+3} \right)\text{+1}\left( \text{2-1} \right)\]

\[\Rightarrow \left| \text{A} \right|\text{=4+5+1}\]

\[\therefore \left| \text{A} \right|\text{=10}\ne \text{0}\]

Thus \[\text{A}\] is non-singular.

$\therefore $ Its inverse exists.

Hence,

\[\Rightarrow {{\text{A}}_{11}}\text{=}{{\left( \text{-1} \right)}^{1+1}}{{\text{M}}_{11}}={{\left( \text{-1} \right)}^{2}}{{\text{M}}_{11}}\]

\[\Rightarrow {{A}_{11}}\text{=4}\]

\[\Rightarrow {{\text{A}}_{12}}\text{=}{{\left( \text{-1} \right)}^{1+2}}{{\text{M}}_{12}}={{\left( \text{-1} \right)}^{3}}{{\text{M}}_{12}}\]

\[\Rightarrow {{A}_{12}}=-5\]

\[\Rightarrow {{\text{A}}_{13}}\text{=}{{\left( \text{-1} \right)}^{1+3}}{{\text{M}}_{13}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{13}}\]

\[\Rightarrow {{A}_{13}}=1\]

Similarly,

\[\Rightarrow {{\text{A}}_{21}}\text{=}{{\left( \text{-1} \right)}^{2+1}}{{\text{M}}_{21}}={{\left( \text{-1} \right)}^{3}}{{\text{M}}_{21}}\]

\[\Rightarrow {{\text{A}}_{21}}\text{=2}\]

\[\Rightarrow {{\text{A}}_{22}}\text{=}{{\left( \text{-1} \right)}^{2+2}}{{\text{M}}_{22}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{22}}\]

\[\Rightarrow {{\text{A}}_{22}}=0\]

\[\Rightarrow {{\text{A}}_{23}}\text{=}{{\left( \text{-1} \right)}^{2+3}}{{\text{M}}_{23}}={{\left( \text{-1} \right)}^{5}}{{\text{M}}_{23}}\]

\[\Rightarrow {{\text{A}}_{23}}\text{=}-\text{2}\]

and

\[\Rightarrow {{\text{A}}_{31}}\text{=}{{\left( \text{-1} \right)}^{3+1}}{{\text{M}}_{31}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{31}}\]

\[\Rightarrow {{\text{A}}_{31}}=2\]

\[\Rightarrow {{\text{A}}_{32}}\text{=}{{\left( \text{-1} \right)}^{3+2}}{{\text{M}}_{32}}={{\left( \text{-1} \right)}^{5}}{{\text{M}}_{32}}\]

\[\Rightarrow {{\text{A}}_{32}}=5\]

\[\Rightarrow {{\text{A}}_{33}}\text{=}{{\left( \text{-1} \right)}^{3+3}}{{\text{M}}_{33}}={{\left( \text{-1} \right)}^{6}}{{\text{M}}_{33}}\]

\[\Rightarrow {{\text{A}}_{33}}=3\].

Cofactor matrix is \[\left[ \begin{matrix} 4 & -5 & 1  \\ 2 & 0 & -2  \\ 2 & 5 & 3  \\ \end{matrix} \right]\].

We know that adjoint of a matrix is the transpose of its cofactor matrix.

\[\Rightarrow adjA={{\left[ \begin{matrix} 4 & -5 & 1  \\ 2 & 0 & -2  \\ 2 & 5 & 3  \\ \end{matrix} \right]}^{T}}\]

The inverse of a matrix is given by:

\[\Rightarrow {{\text{A}}^{\text{-1}}}\text{=}\dfrac{\text{1}}{\left| \text{A} \right|}\left( \text{adjA} \right)\]

\[\therefore {{\text{A}}^{\text{-1}}}\text{=}\dfrac{\text{1}}{\text{10}}\left[ \begin{matrix} \text{4} & \text{2} & \text{2}  \\ \text{-5} & \text{0} & \text{5}  \\ \text{1} & \text{-2} & \text{3}  \\ \end{matrix} \right]\]

\[\therefore \text{X=}{{\text{A}}^{\text{-1}}}\text{B}\]

\[\Rightarrow \left[ \begin{matrix} \text{x}  \\ \text{y}  \\ \text{z}  \\ \end{matrix} \right]\text{=}\dfrac{\text{1}}{\text{10}}\left[ \begin{matrix} \text{4} & \text{2} & \text{2}  \\ \text{-5} & \text{0} & \text{5}  \\ \text{1} & \text{-2} & \text{3}  \\ \end{matrix} \right]\left[ \begin{matrix} \text{4}  \\ \text{0}  \\ \text{2}  \\ \end{matrix} \right]\]

\[\Rightarrow \left[ \begin{matrix} \text{x}  \\ \text{y}  \\ \text{z}  \\ \end{matrix} \right]\text{=}\dfrac{\text{1}}{\text{10}}\left[ \begin{matrix} \text{16+0+4}  \\ \text{-20+0+10}  \\ \text{4+0+6}  \\ \end{matrix} \right]\]

\[\Rightarrow \left[ \begin{matrix} \text{x}  \\ \text{y}  \\ \text{z}  \\ \end{matrix} \right]\text{=}\dfrac{\text{1}}{\text{10}}\left[ \begin{matrix} \text{20}  \\ \text{-10}  \\ \text{10}  \\ \end{matrix} \right]\]

\[\Rightarrow \left[ \begin{matrix} \text{x}  \\ \text{y}  \\ \text{z}  \\ \end{matrix} \right]\text{=}\left[ \begin{matrix} \text{2}  \\ \text{-1}  \\ \text{1}  \\ \end{matrix} \right]\]

Thus, \[x=2\] , \[y=-1\] and \[z=1\].

13. Solve system of linear equations, using matrix method.

\[\mathbf{\text{2x+3y+3z=5}}\]

\[\mathbf{\text{x-2y+z=-4}}\]

\[\mathbf{\text{3x-y-2z=3}}\]

Ans: Given equations,

\[\text{2x+3y+3z=5}\]

\[\text{x-2y+z=-4}\]

\[\text{3x-y-2z=3}\]

Let \[\text{A=}\left[ \begin{matrix} \text{2} & \text{3} & \text{3}  \\ \text{1} & \text{-2} & \text{1}  \\ \text{3} & \text{-1} & \text{-2}  \\ \end{matrix} \right]\] , \[\text{X=}\left[ \begin{matrix} \text{x}  \\ \text{y}  \\ \text{z}  \\ \end{matrix} \right]\] and \[\text{B=}\left[ \begin{matrix} \text{5}  \\ \text{-4}  \\ \text{3}  \\ \end{matrix} \right]\] such that, this system of equations can be written in the form of \[\text{AX=B}\].

Determining the value of $A$, we have:

\[\Rightarrow \left| \text{A} \right|\text{=2}\left( \text{4+1} \right)\text{-3}\left( \text{2-3} \right)\text{+3}\left( \text{-1+6} \right)\]

\[\Rightarrow \left| \text{A} \right|\text{=2}\left( \text{5} \right)\text{-3}\left( \text{-5} \right)\text{+3}\left( \text{5} \right)\]

\[\Rightarrow \left| \text{A} \right|\text{=10+15+15}\]

\[\Rightarrow \left| \text{A} \right|\text{=40}\ne \text{0}\]

Thus, \[\text{A}\] is a non-singular matrix.

$\therefore $ It’s inverse exists.

Hence,

\[\Rightarrow {{\text{A}}_{11}}\text{=}{{\left( \text{-1} \right)}^{1+1}}{{\text{M}}_{11}}={{\left( \text{-1} \right)}^{2}}{{\text{M}}_{11}}\]

\[\Rightarrow {{A}_{11}}\text{=5}\]

\[\Rightarrow {{\text{A}}_{12}}\text{=}{{\left( \text{-1} \right)}^{1+2}}{{\text{M}}_{12}}={{\left( \text{-1} \right)}^{3}}{{\text{M}}_{12}}\]

\[\Rightarrow {{A}_{12}}=5\]

\[\Rightarrow {{\text{A}}_{13}}\text{=}{{\left( \text{-1} \right)}^{1+3}}{{\text{M}}_{13}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{13}}\]

\[\Rightarrow {{A}_{13}}=5\]

Similarly,

\[\Rightarrow {{\text{A}}_{21}}\text{=}{{\left( \text{-1} \right)}^{2+1}}{{\text{M}}_{21}}={{\left( \text{-1} \right)}^{3}}{{\text{M}}_{21}}\]

\[\Rightarrow {{\text{A}}_{21}}\text{=3}\]

\[\Rightarrow {{\text{A}}_{22}}\text{=}{{\left( \text{-1} \right)}^{2+2}}{{\text{M}}_{22}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{22}}\]

\[\Rightarrow {{\text{A}}_{22}}=-13\]

\[\Rightarrow {{\text{A}}_{23}}\text{=}{{\left( \text{-1} \right)}^{2+3}}{{\text{M}}_{23}}={{\left( \text{-1} \right)}^{5}}{{\text{M}}_{23}}\]

\[\Rightarrow {{\text{A}}_{23}}\text{=11}\]

and

\[\Rightarrow {{\text{A}}_{31}}\text{=}{{\left( \text{-1} \right)}^{3+1}}{{\text{M}}_{31}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{31}}\]

\[\Rightarrow {{\text{A}}_{31}}=9\]

\[\Rightarrow {{\text{A}}_{32}}\text{=}{{\left( \text{-1} \right)}^{3+2}}{{\text{M}}_{32}}={{\left( \text{-1} \right)}^{5}}{{\text{M}}_{32}}\]

\[\Rightarrow {{\text{A}}_{32}}=1\]

\[\Rightarrow {{\text{A}}_{33}}\text{=}{{\left( \text{-1} \right)}^{3+3}}{{\text{M}}_{33}}={{\left( \text{-1} \right)}^{6}}{{\text{M}}_{33}}\]

\[\Rightarrow {{\text{A}}_{33}}=-7\].

Cofactor matrix is \[\left[ \begin{matrix} 5 & 5 & 5  \\ 3 & -13 & 11  \\ 9 & 1 & -7  \\ \end{matrix} \right]\].

We know that adjoint of a matrix is the transpose of its cofactor matrix.

\[\Rightarrow adjA={{\left[ \begin{matrix} 5 & 5 & 5  \\ 3 & -13 & 11  \\ 9 & 1 & -7  \\ \end{matrix} \right]}^{T}}\]

The inverse of a matrix is given by:

\[\Rightarrow {{\text{A}}^{\text{-1}}}\text{=}\dfrac{\text{1}}{\left| \text{A} \right|}\left( \text{adjA} \right)\]

\[\therefore {{\text{A}}^{\text{-1}}}\text{=}\dfrac{\text{1}}{\text{40}}\left[ \begin{matrix} \text{5} & \text{3} & \text{9}  \\ \text{5} & \text{-13} & \text{1}  \\ \text{5} & \text{11} & \text{-7}  \\ \end{matrix} \right]\]

\[\therefore \text{X=}{{\text{A}}^{\text{-1}}}\text{B}\]

\[\Rightarrow \left[ \begin{matrix} \text{x}  \\ \text{y}  \\ \text{z}  \\ \end{matrix} \right]\text{=}\dfrac{\text{1}}{\text{40}}\left[ \begin{matrix} \text{5} & \text{3} & \text{9}  \\ \text{5} & \text{-13} & \text{1}  \\ \text{5} & \text{11} & \text{-7}  \\ \end{matrix} \right]\left[ \begin{matrix} \text{5}  \\ \text{-4}  \\ \text{3}  \\ \end{matrix} \right]\]

\[\Rightarrow \left[ \begin{matrix} \text{x}  \\ \text{y}  \\ \text{z}  \\ \end{matrix} \right]\text{=}\dfrac{\text{1}}{\text{40}}\left[ \begin{matrix} \text{25-12+27}  \\ \text{25+52+3}  \\ \text{25-44-21}  \\ \end{matrix} \right]\]

\[\Rightarrow \left[ \begin{matrix} \text{x}  \\ \text{y}  \\ \text{z}  \\ \end{matrix} \right]\text{=}\dfrac{\text{1}}{\text{40}}\left[ \begin{matrix} \text{40}  \\ \text{80}  \\ \text{-40}  \\ \end{matrix} \right]\]

\[\Rightarrow \left[ \begin{matrix} \text{x}  \\ \text{y}  \\ \text{z}  \\ \end{matrix} \right]\text{=}\left[ \begin{matrix} \text{1}  \\ \text{2}  \\ \text{-1}  \\ \end{matrix} \right]\]

Thus, \[x=1\], \[y=2\] and \[z=-1\].

14. Solve system of linear equations, using matrix method.

\[\mathbf{\text{x-y+2z=7}}\]

\[\mathbf{\text{3x+4y-5z=-5}}\]

\[\mathbf{\text{2x-y+3z=12}}\]

Ans: Given equations,

\[\text{x-y+2z=7}\]

\[\text{3x+4y-5z=-5}\]

\[\text{2x-y+3z=12}\]

Let \[\text{A=}\left[ \begin{matrix} \text{1} & \text{-1} & \text{2}  \\ \text{3} & \text{4} & \text{-5}  \\ \text{2} & \text{-1} & \text{3}  \\ \end{matrix} \right]\] , \[\text{X=}\left[ \begin{matrix} \text{x}  \\ \text{y}  \\ \text{z}  \\ \end{matrix} \right]\] and \[\text{B=}\left[ \begin{matrix} \text{7}  \\ \text{-5}  \\ \text{12}  \\ \end{matrix} \right]\] such that, this system of equations can be written in the form of \[\text{AX=B}\].

Determining the value of $A$, we have:

\[\Rightarrow \left| \text{A} \right|\text{=1}\left( \text{12-5} \right)\text{+1}\left( \text{9+10} \right)\text{+2}\left( \text{-3-8} \right)\]

\[\Rightarrow \left| \text{A} \right|\text{=7+19-22}\]

\[\therefore \left| \text{A} \right|\text{=4}\ne \text{0}\]

Thus, \[\text{A}\] is a non-singular matrix.

$\therefore $ It’s inverse exists.

Now,

\[\Rightarrow {{\text{A}}_{11}}\text{=}{{\left( \text{-1} \right)}^{1+1}}{{\text{M}}_{11}}={{\left( \text{-1} \right)}^{2}}{{\text{M}}_{11}}\]

\[\Rightarrow {{A}_{11}}\text{=7}\]

\[\Rightarrow {{\text{A}}_{12}}\text{=}{{\left( \text{-1} \right)}^{1+2}}{{\text{M}}_{12}}={{\left( \text{-1} \right)}^{3}}{{\text{M}}_{12}}\]

\[\Rightarrow {{A}_{12}}=-19\]

\[\Rightarrow {{\text{A}}_{13}}\text{=}{{\left( \text{-1} \right)}^{1+3}}{{\text{M}}_{13}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{13}}\]

\[\Rightarrow {{A}_{13}}=-11\]

Similarly,

\[\Rightarrow {{\text{A}}_{21}}\text{=}{{\left( \text{-1} \right)}^{2+1}}{{\text{M}}_{21}}={{\left( \text{-1} \right)}^{3}}{{\text{M}}_{21}}\]

\[\Rightarrow {{\text{A}}_{21}}\text{=1}\]

\[\Rightarrow {{\text{A}}_{22}}\text{=}{{\left( \text{-1} \right)}^{2+2}}{{\text{M}}_{22}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{22}}\]

\[\Rightarrow {{\text{A}}_{22}}=-1\]

\[\Rightarrow {{\text{A}}_{23}}\text{=}{{\left( \text{-1} \right)}^{2+3}}{{\text{M}}_{23}}={{\left( \text{-1} \right)}^{5}}{{\text{M}}_{23}}\]

\[\Rightarrow {{\text{A}}_{23}}\text{=}-1\]

and

\[\Rightarrow {{\text{A}}_{31}}\text{=}{{\left( \text{-1} \right)}^{3+1}}{{\text{M}}_{31}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{31}}\]

\[\Rightarrow {{\text{A}}_{31}}=-3\]

\[\Rightarrow {{\text{A}}_{32}}\text{=}{{\left( \text{-1} \right)}^{3+2}}{{\text{M}}_{32}}={{\left( \text{-1} \right)}^{5}}{{\text{M}}_{32}}\]

\[\Rightarrow {{\text{A}}_{32}}=11\]

\[\Rightarrow {{\text{A}}_{33}}\text{=}{{\left( \text{-1} \right)}^{3+3}}{{\text{M}}_{33}}={{\left( \text{-1} \right)}^{6}}{{\text{M}}_{33}}\]

\[\Rightarrow {{\text{A}}_{33}}=7\].

Cofactor matrix is \[\left[ \begin{matrix} 7 & -19 & -11  \\ 1 & -1 & -1  \\ -3 & 11 & 7  \\ \end{matrix} \right]\].

We know that the adjoint of a matrix is the transpose of its cofactor matrix.

\[\Rightarrow adjA={{\left[ \begin{matrix} 7 & -19 & -11  \\ 1 & -1 & -1  \\ -3 & 11 & 7  \\ \end{matrix} \right]}^{T}}\]

\[\Rightarrow adjA=\left[ \begin{matrix} 7 & 1 & -3  \\ -19 & -1 & 11  \\ -11 & -1 & 7  \\ \end{matrix} \right]\]

The inverse of a matrix is given by:

\[\Rightarrow {{\text{A}}^{\text{-1}}}\text{=}\dfrac{\text{1}}{\left| \text{A} \right|}\left( \text{adjA} \right)\]

\[\Rightarrow {{\text{A}}^{\text{-1}}}\text{=}\dfrac{\text{1}}{\text{4}}\left[ \begin{matrix} \text{7} & \text{1} & \text{-3}  \\ \text{-19} & \text{-1} & \text{11}  \\ \text{-11} & \text{-1} & \text{7}  \\ \end{matrix} \right]\]

\[\therefore \text{X=}{{\text{A}}^{\text{-1}}}\text{B}\]

\[\Rightarrow \left[ \begin{matrix} \text{x}  \\ \text{y}  \\ \text{z}  \\ \end{matrix} \right]\text{=}\dfrac{\text{1}}{\text{4}}\left[ \begin{matrix} \text{7} & \text{1} & \text{-3}  \\ \text{-19} & \text{-1} & \text{11}  \\ \text{-11} & \text{-1} & \text{7}  \\ \end{matrix} \right]\left[ \begin{matrix} \text{7}  \\ \text{-5}  \\ \text{12}  \\ \end{matrix} \right]\]

\[\Rightarrow \left[ \begin{matrix} \text{x}  \\ \text{y}  \\ \text{z}  \\ \end{matrix} \right]\text{=}\dfrac{\text{1}}{\text{4}}\left[ \begin{matrix} \text{49-5-36}  \\ \text{-133+5+132}  \\ \text{-77+5+84}  \\ \end{matrix} \right]\]

\[\Rightarrow \left[ \begin{matrix} \text{x}  \\ \text{y}  \\ \text{z}  \\ \end{matrix} \right]\text{=}\dfrac{\text{1}}{\text{4}}\left[ \begin{matrix} \text{8}  \\ \text{4}  \\ \text{12}  \\ \end{matrix} \right]\]

\[\Rightarrow \left[ \begin{matrix} \text{x}  \\ \text{y}  \\ \text{z}  \\ \end{matrix} \right]\text{=}\left[ \begin{matrix} \text{2}  \\ \text{1}  \\ \text{3}  \\ \end{matrix} \right]\]

Thus \[\text{x=2}\] , \[\text{y=1}\] and \[\text{z=3}\]

15. If $\mathbf{A=\left[ \begin{matrix} 2 & -3 & 5  \\ 3 & 2 & -4  \\ 1 & 1 & -2  \\ \end{matrix} \right] }$, find $\mathbf{{{A}^{-1}}}$. Using $\mathbf{{{A}^{-1}}}$ solve the system of equations

$\mathbf{2x-3y+5z=11}$

$\mathbf{3x+2y-4z=-5}$

$\mathbf{x+y-2z=-3}$

Ans: Given equations,

\[2x-3y+5z=11\]

\[3x+2y-4z=-5\]

\[x+y-2z=-3\]

Let \[\text{A=}\left[ \begin{matrix} 2 & -3 & 5  \\ \text{3} & 2 & -4  \\ 1 & \text{1} & -2  \\ \end{matrix} \right]\] , \[\text{X=}\left[ \begin{matrix} \text{x}  \\ \text{y}  \\ \text{z}  \\ \end{matrix} \right]\] and \[\text{B=}\left[ \begin{matrix} 11  \\ -5  \\ -3  \\ \end{matrix} \right]\] such that, this system of equations can be written in the form of \[\text{AX=B}\].

Determining the value of $A$, we have:

\[\Rightarrow \left| \text{A} \right|\text{=2}\left( \text{-4+4} \right)\text{+3}\left( \text{-6+4} \right)\text{+5}\left( \text{3-2} \right)\]

\[\Rightarrow \left| \text{A} \right|\text{=0-6+5}\]

\[\therefore \left| \text{A} \right|\text{=-1}\ne \text{0}\]

Thus, \[\text{A}\] is a non-singular matrix.

$\therefore $ It’s inverse exists.

Now,

\[\Rightarrow {{\text{A}}_{11}}\text{=}{{\left( \text{-1} \right)}^{1+1}}{{\text{M}}_{11}}={{\left( \text{-1} \right)}^{2}}{{\text{M}}_{11}}\]

\[\Rightarrow {{A}_{11}}\text{=0}\]

\[\Rightarrow {{\text{A}}_{12}}\text{=}{{\left( \text{-1} \right)}^{1+2}}{{\text{M}}_{12}}={{\left( \text{-1} \right)}^{3}}{{\text{M}}_{12}}\]

\[\Rightarrow {{A}_{12}}=2\]

\[\Rightarrow {{\text{A}}_{13}}\text{=}{{\left( \text{-1} \right)}^{1+3}}{{\text{M}}_{13}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{13}}\]

\[\Rightarrow {{A}_{13}}=1\]

Similarly,

\[\Rightarrow {{\text{A}}_{21}}\text{=}{{\left( \text{-1} \right)}^{2+1}}{{\text{M}}_{21}}={{\left( \text{-1} \right)}^{3}}{{\text{M}}_{21}}\]

\[\Rightarrow {{\text{A}}_{21}}\text{=-1}\]

\[\Rightarrow {{\text{A}}_{22}}\text{=}{{\left( \text{-1} \right)}^{2+2}}{{\text{M}}_{22}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{22}}\]

\[\Rightarrow {{\text{A}}_{22}}=-9\]

\[\Rightarrow {{\text{A}}_{23}}\text{=}{{\left( \text{-1} \right)}^{2+3}}{{\text{M}}_{23}}={{\left( \text{-1} \right)}^{5}}{{\text{M}}_{23}}\]

\[\Rightarrow {{\text{A}}_{23}}\text{=}-5\]

and

\[\Rightarrow {{\text{A}}_{31}}\text{=}{{\left( \text{-1} \right)}^{3+1}}{{\text{M}}_{31}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{31}}\]

\[\Rightarrow {{\text{A}}_{31}}=2\]

\[\Rightarrow {{\text{A}}_{32}}\text{=}{{\left( \text{-1} \right)}^{3+2}}{{\text{M}}_{32}}={{\left( \text{-1} \right)}^{5}}{{\text{M}}_{32}}\]

\[\Rightarrow {{\text{A}}_{32}}=23\]

\[\Rightarrow {{\text{A}}_{33}}\text{=}{{\left( \text{-1} \right)}^{3+3}}{{\text{M}}_{33}}={{\left( \text{-1} \right)}^{6}}{{\text{M}}_{33}}\]

\[\Rightarrow {{\text{A}}_{33}}=13\].

Cofactor matrix is \[\left[ \begin{matrix} 0 & 2 & 1  \\ -1 & -9 & -5  \\ 2 & 23 & 13  \\ \end{matrix} \right]\].

We know that adjoint of a matrix is the transpose of its cofactor matrix.

\[\Rightarrow adjA={{\left[ \begin{matrix} 0 & 2 & 1  \\ -1 & -9 & -5  \\ 2 & 23 & 13  \\ \end{matrix} \right]}^{T}}\]

\[\Rightarrow adjA=\left[ \begin{matrix} 0 & -1 & 2  \\ 2 & -9 & 23  \\ 1 & -5 & 13  \\ \end{matrix} \right]\]

The inverse of a matrix is given by:

\[\Rightarrow {{\text{A}}^{\text{-1}}}\text{=}\dfrac{\text{1}}{\left| \text{A} \right|}\left( \text{adjA} \right)\]

\[\Rightarrow {{\text{A}}^{\text{-1}}}\text{=}\dfrac{\text{1}}{-1}\left[ \begin{matrix} 0 & -1 & 2  \\ 2 & -9 & 23  \\ 1 & -5 & 13  \\ \end{matrix} \right]\]

\[\therefore \text{X=}{{\text{A}}^{\text{-1}}}\text{B}\]

\[\Rightarrow \left[ \begin{matrix} \text{x}  \\ \text{y}  \\ \text{z}  \\ \end{matrix} \right]\text{=-1}\left[ \begin{matrix} 0 & -1 & 2  \\ 2 & -9 & 23  \\ 1 & -5 & 13  \\ \end{matrix} \right]\left[ \begin{matrix} 11  \\ -5  \\ -3  \\ \end{matrix} \right]\]

\[\Rightarrow \left[ \begin{matrix} \text{x}  \\ \text{y}  \\ \text{z}  \\ \end{matrix} \right]\text{=-1}\left[ \begin{matrix} \text{5-6}  \\ \text{22+45-69}  \\ \text{11+25-39}  \\ \end{matrix} \right]\] \[\Rightarrow \left[ \begin{matrix} \text{x}  \\ \text{y}  \\ \text{z}  \\ \end{matrix} \right]\text{=-1}\left[ \begin{matrix} -1  \\ -2  \\ -3  \\ \end{matrix} \right]\]

\[\Rightarrow \left[ \begin{matrix} \text{x}  \\ \text{y}  \\ \text{z}  \\ \end{matrix} \right]\text{=}\left[ \begin{matrix} 1  \\ 2  \\ \text{3}  \\ \end{matrix} \right]\]         

Thus \[\text{x=1}\] , \[\text{y=2}\] and \[\text{z=3}\].

16. The cost of \[\mathbf{\text{4kg}}\] onion, \[\mathbf{\text{3kg}}\] wheat and \[\mathbf{\text{2kg}}\] rice is Rs\[\mathbf{\text{60}}\]. The cost of  \[\mathbf{\text{2kg}}\] onion, \[\mathbf{\text{4kg}}\]wheat and \[\mathbf{\text{6kg}}\] rice is Rs\[\mathbf{\text{90}}\]. The cost of \[\mathbf{\text{6kg}}\] onion \[\mathbf{\text{2kg}}\] wheat and \[\mathbf{\text{3kg}}\] rice is Rs\[\mathbf{\text{70}}\].

Find cost of each item per kg by matrix method.

Ans: Let us suppose that the cost of onions, wheat and rice per kg be Rs \[\text{x}\] , Rs \[\text{y}\] and Rs \[\text{z}\] respectively.

Then, the given situation can be represented by a system of equations as:

\[\Rightarrow \text{4x+3y+2z=60}\]

\[\Rightarrow \text{2x+4y+6z=90}\]

\[\Rightarrow \text{6x+2y+3z=70}\]

Let \[\text{A=}\left[ \begin{matrix} \text{4} & \text{3} & \text{2}  \\ \text{2} & \text{4} & \text{6}  \\ \text{6} & \text{2} & \text{3}  \\ \end{matrix} \right]\] , \[\text{X=}\left[ \begin{matrix} \text{x}  \\ \text{y}  \\ \text{z}  \\ \end{matrix} \right]\] and \[\text{B=}\left[ \begin{matrix} \text{60}  \\ \text{90}  \\ \text{70}  \\ \end{matrix} \right]\] such that, this system of equations can be written in the form of \[\text{AX=B}\].

\[\Rightarrow \left| \text{A} \right|\text{=4}\left( \text{12-12} \right)\text{-3}\left( \text{6-36} \right)\text{+2}\left( \text{4-24} \right)\]

\[\Rightarrow \left| \text{A} \right|\text{=0+90-40}\]

\[\Rightarrow \left| \text{A} \right|\text{=50}\ne \text{0}\]

Thus,

\[\Rightarrow {{\text{A}}_{11}}\text{=}{{\left( \text{-1} \right)}^{1+1}}{{\text{M}}_{11}}={{\left( \text{-1} \right)}^{2}}{{\text{M}}_{11}}\]

\[\Rightarrow {{A}_{11}}\text{=0}\]

\[\Rightarrow {{\text{A}}_{12}}\text{=}{{\left( \text{-1} \right)}^{1+2}}{{\text{M}}_{12}}={{\left( \text{-1} \right)}^{3}}{{\text{M}}_{12}}\]

\[\Rightarrow {{A}_{12}}=30\]

\[\Rightarrow {{\text{A}}_{13}}\text{=}{{\left( \text{-1} \right)}^{1+3}}{{\text{M}}_{13}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{13}}\]

\[\Rightarrow {{A}_{13}}=-20\]

Similarly,

\[\Rightarrow {{\text{A}}_{21}}\text{=}{{\left( \text{-1} \right)}^{2+1}}{{\text{M}}_{21}}={{\left( \text{-1} \right)}^{3}}{{\text{M}}_{21}}\]

\[\Rightarrow {{\text{A}}_{21}}\text{=}-5\]

\[\Rightarrow {{\text{A}}_{22}}\text{=}{{\left( \text{-1} \right)}^{2+2}}{{\text{M}}_{22}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{22}}\]

\[\Rightarrow {{\text{A}}_{22}}=0\]

\[\Rightarrow {{\text{A}}_{23}}\text{=}{{\left( \text{-1} \right)}^{2+3}}{{\text{M}}_{23}}={{\left( \text{-1} \right)}^{5}}{{\text{M}}_{23}}\]

\[\Rightarrow {{\text{A}}_{23}}\text{=10}\]

And

\[\Rightarrow {{\text{A}}_{31}}\text{=}{{\left( \text{-1} \right)}^{3+1}}{{\text{M}}_{31}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{31}}\]

\[\Rightarrow {{\text{A}}_{31}}=10\]

\[\Rightarrow {{\text{A}}_{32}}\text{=}{{\left( \text{-1} \right)}^{3+2}}{{\text{M}}_{32}}={{\left( \text{-1} \right)}^{5}}{{\text{M}}_{32}}\]

\[\Rightarrow {{\text{A}}_{32}}=-20\]

\[\Rightarrow {{\text{A}}_{33}}\text{=}{{\left( \text{-1} \right)}^{3+3}}{{\text{M}}_{33}}={{\left( \text{-1} \right)}^{6}}{{\text{M}}_{33}}\]

\[\Rightarrow {{\text{A}}_{33}}=10\].

Cofactor matrix is \[\left[ \begin{matrix} 0 & 30 & -20  \\ -5 & 0 & 10  \\ 10 & -20 & 10  \\ \end{matrix} \right]\].

We know that adjoint of a matrix is the transpose of its cofactor matrix.

\[\Rightarrow adjA={{\left[ \begin{matrix} 0 & 30 & -20  \\ -5 & 0 & 10  \\ 10 & -20 & 10  \\ \end{matrix} \right]}^{T}}\]

\[\therefore \left( \text{adjA} \right)\text{=}\left[ \begin{matrix} \text{0} & \text{-5} & \text{10}  \\ \text{30} & \text{0} & \text{-20}  \\ \text{-20} & \text{10} & \text{10}  \\ \end{matrix} \right]\]

\[\Rightarrow {{\text{A}}^{\text{-1}}}\text{=}\dfrac{\text{1}}{\left| \text{A} \right|}\left( \text{adjA} \right)\]

\[\therefore {{\text{A}}^{\text{-1}}}\text{=}\dfrac{\text{1}}{\text{50}}\left[ \begin{matrix} \text{0} & \text{-5} & \text{10}  \\ \text{30} & \text{0} & \text{-20}  \\ \text{-20} & \text{10} & \text{10}  \\ \end{matrix} \right]\]

Since, \[\text{X=}{{\text{A}}^{\text{-1}}}\text{B}\]

\[\Rightarrow \text{X=}\dfrac{\text{1}}{\text{50}}\left[ \begin{matrix} \text{0} & \text{-5} & \text{10}  \\ \text{30} & \text{0} & \text{-20}  \\ \text{-20} & \text{10} & \text{10}  \\ \end{matrix} \right]\left[ \begin{matrix} \text{60}  \\ \text{90}  \\ \text{70}  \\ \end{matrix} \right]\]

\[\Rightarrow \left[ \begin{matrix} \text{x}  \\ \text{y}  \\ \text{z}  \\ \end{matrix} \right]\text{=}\dfrac{\text{1}}{\text{50}}\left[ \begin{matrix} \text{0+450+700}  \\ \text{1800+0-1400}  \\ \text{-1200+900+700}  \\ \end{matrix} \right]\]

\[\Rightarrow \left[ \begin{matrix} \text{x}  \\ \text{y}  \\ \text{z}  \\ \end{matrix} \right]\text{=}\dfrac{\text{1}}{\text{50}}\left[ \begin{matrix} \text{250}  \\ \text{400}  \\ \text{400}  \\ \end{matrix} \right]\]

\[\Rightarrow \left[ \begin{matrix} \text{x}  \\ \text{y}  \\ \text{z}  \\ \end{matrix} \right]\text{=}\left[ \begin{matrix} \text{5}  \\ \text{8}  \\ \text{8}  \\ \end{matrix} \right]\]

Thus \[\text{x=5}\] , \[\text{y=8}\] , and \[\text{z=8}\]

Hence, the cost of onions is Rs \[\text{5}\] per kg, the cost of wheat is Rs \[\text{8}\] per kg, and the cost of rice is Rs \[\text{8}\] per kg.

Miscellaneous Exercise Solutions

1. Prove that the determinant \[\mathbf{\left| \begin{matrix} x & \sin \theta  & \cos \theta   \\ -\sin \theta  & -x & 1  \\ \cos \theta  & 1 & x  \\ \end{matrix} \right|}\] is independent of \[\mathbf{\theta} \].

Ans: Let \[\Delta =\left| \begin{matrix} x & \sin \theta  & \cos \theta   \\ -\sin \theta  & -x & 1  \\ \cos \theta  & 1 & x  \\ \end{matrix} \right|\]

Solving it, we have:

\[\Rightarrow \Delta =x\left( {{x}^{2}}-1 \right)-\sin \theta \left( -x\sin \theta -\cos \theta  \right)+\cos \theta \left( -\sin \theta +x\cos \theta  \right)\]

\[\Rightarrow \Delta ={{x}^{3}}-x+x{{\sin }^{2}}\theta +\sin \theta \cos \theta -\sin \theta \cos \theta +x{{\cos }^{2}}\theta \]

\[\Rightarrow \Delta ={{x}^{3}}-x+x\left( {{\sin }^{2}}\theta +{{\cos }^{2}}\theta  \right)\]

\[\Rightarrow \Delta \text{=}{{\text{x}}^{\text{3}}}\text{-x+x}\]

\[\therefore \Delta \text{=}{{\text{x}}^{\text{3}}}\]

Hence, \[\text{ }\!\!\Delta\!\!\text{ }\] is independent of \[\theta \] .

2. Evaluate \[\mathbf{\left| \begin{matrix} \cos \alpha \cos \beta  & \cos \alpha \sin \beta  & -\sin \alpha   \\ -\sin \beta  & \cos \beta  & 0  \\ \sin \alpha \cos \beta  & \sin \alpha \sin \beta  & \cos \alpha   \\ \end{matrix} \right|}\]

Ans: Let \[\Delta =\left| \begin{matrix} \cos \alpha \cos \beta  & \cos \alpha \sin \beta  & -\sin \alpha   \\ -\sin \beta  & \cos \beta  & 0  \\ \sin \alpha \cos \beta  & \sin \alpha \sin \beta  & \cos \alpha   \\ \end{matrix} \right|\]

Expanding along column \[{{\text{C}}_{3}}\]

\[\Rightarrow \Delta =-\sin \alpha \left( -\sin \alpha {{\sin }^{2}}\beta +{{\cos }^{2}}\beta \sin \alpha  \right)+\cos \alpha \left( \cos \alpha {{\cos }^{2}}\beta +\cos \alpha {{\sin }^{2}}\beta  \right)\]

\[\Rightarrow \Delta ={{\sin }^{2}}\alpha \left( s{{\sin }^{2}}\beta +{{\cos }^{2}}\beta  \right)+{{\cos }^{2}}\alpha \left( {{\cos }^{2}}\beta +{{\sin }^{2}}\beta  \right)\]

\[\Rightarrow \Delta ={{\sin }^{2}}\alpha \left( 1 \right)+{{\cos }^{2}}\alpha \left( 1 \right)\]

\[\therefore \Delta \text{=1}\]

3. If $\mathbf{{{\mathbf{A}}^{-1}}=\left[ \begin{matrix} 3 & -1 & 1  \\ -15 & 6 & -5  \\ 5 & -2 & 2  \\ \end{matrix} \right]$ and $B=\left[ \begin{matrix} 1 & 2 & -2  \\ -1 & 3 & 0  \\ 0 & -2 & 1  \\ \end{matrix} \right]$, find ${{\left( AB \right)}^{-1}}}$.

Ans: The below result will be used for simplification,

${{\left( AB \right)}^{-1}}={{B}^{-1}}{{A}^{-1}}$

Finding inverse of matrix B. so, the determinant is

$\left| B \right|=1\left( 3-0 \right)-2\left( -1-0 \right)-2\left( 2-0 \right)$

$\Rightarrow \left| B \right|=3+2-4$

$\therefore \left| B \right|=1$

Now finding the cofactors,

${{B}_{11}}={{(-1)}^{1+1}}\left| \begin{matrix} 3 & 0  \\ -2 & 1  \\ \end{matrix} \right|\Rightarrow {{B}_{11}}=3$

${{B}_{12}}={{(-1)}^{1+2}}\left| \begin{matrix} -1 & 0  \\ 0 & 1  \\ \end{matrix} \right|\Rightarrow {{B}_{12}}=1$

${{B}_{13}}={{(-1)}^{1+3}}\left| \begin{matrix} -1 & 3  \\ 0 & -2  \\ \end{matrix} \right|\Rightarrow {{B}_{13}}=2$

${{B}_{21}}={{(-1)}^{2+1}}\left| \begin{matrix} 2 & -2  \\ -2 & 1  \\ \end{matrix} \right|\Rightarrow {{B}_{21}}=-\left( 2-4 \right)=2$

${{B}_{22}}={{(-1)}^{2+2}}\left| \begin{matrix} 1 & -2  \\ 0 & 1  \\ \end{matrix} \right|\Rightarrow {{B}_{22}}=1$

${{B}_{23}}={{(-1)}^{2+3}}\left| \begin{matrix} 1 & 2  \\ 0 & -2  \\ \end{matrix} \right|\Rightarrow {{B}_{23}}=2$

${{B}_{31}}={{(-1)}^{3+1}}\left| \begin{matrix} 2 & -2  \\ 3 & 0  \\ \end{matrix} \right|\Rightarrow {{B}_{31}}=6$

${{B}_{32}}={{(-1)}^{3+2}}\left| \begin{matrix} 1 & -2  \\ -1 & 0  \\ \end{matrix} \right|\Rightarrow {{B}_{32}}=2$

${{B}_{33}}={{(-1)}^{3+3}}\left| \begin{matrix} 1 & 2  \\ -1 & 3  \\ \end{matrix} \right|\Rightarrow {{B}_{33}}=\left( 3+2 \right)=5$

The cofactor matrix is $\left[ \begin{matrix} 3 & 1 & 2  \\ 2 & 1 & 2  \\ 6 & 2 & 5  \\ \end{matrix} \right]$. The adjoint will be the transpose of cofactor matrix.

$adj\left( B \right)=\left[ \begin{matrix} 3 & 2 & 6  \\ 1 & 1 & 2  \\ 2 & 2 & 5  \\ \end{matrix} \right]$

The inverse is given by ${{B}^{-1}}=\frac{adj\left( B \right)}{\left| B \right|}$. So,

${{B}^{-1}}=\left[ \begin{matrix} 3 & 2 & 6  \\ 1 & 1 & 2  \\ 2 & 2 & 5  \\ \end{matrix} \right]$

Now, it is already given that ${{A}^{-1}}=\left[ \begin{matrix} 3 & -1 & 1  \\ -15 & 6 & -5  \\ 5 & -2 & 2  \\ \end{matrix} \right]$.

So, we can compute ${{\left( AB \right)}^{-1}}={{B}^{-1}}{{A}^{-1}}$ as below,

${{\left( AB \right)}^{-1}}=\left[ \begin{matrix} 3 & 2 & 6  \\ 1 & 1 & 2  \\ 2 & 2 & 5  \\ \end{matrix} \right]\left[ \begin{matrix} 3 & -1 & 1  \\ -15 & 6 & -5  \\ 5 & -2 & 2  \\ \end{matrix} \right]$

$\Rightarrow {{\left( AB \right)}^{-1}}=\left[ \begin{matrix} 9-30+30 & -3+12-12 & 3-10+12  \\ 3-15+10 & -1+6-4 & 1-5+4  \\ 6-30+25 & -2+12-10 & 2-10+10  \\ \end{matrix} \right]$

$\therefore {{\left( AB \right)}^{-1}}=\left[ \begin{matrix} 9 & -3 & 5  \\ -2 & 1 & 0  \\ 1 & 0 & 2  \\ \end{matrix} \right]$

4. Let \[\mathbf{\text{A=}\left[ \begin{matrix} \text{1} & \text{2} & \text{1}  \\ \text{2} & \text{3} & \text{1}  \\ \text{1} & \text{1} & \text{5}  \\ \end{matrix} \right]}\] verify that

1. \[\mathbf{{{\left[ \text{adjA} \right]}^{\text{-1}}}\text{=adj}\left( {{\text{A}}^{\text{-1}}} \right)}\]

Ans: Given, \[\text{A=}\left[ \begin{matrix} \text{1} & \text{2} & \text{1}  \\ \text{2} & \text{3} & \text{1}  \\ \text{1} & \text{1} & \text{5}  \\ \end{matrix} \right]\]

\[\Rightarrow \left| \text{A} \right|\text{=1}\left( \text{15-1} \right)\text{-2}\left( \text{10-1} \right)\text{+1}\left( \text{2-3} \right)\]

\[\Rightarrow \left| A \right|\text{=14-18-1}\]

\[\therefore \left| A \right|\text{=}-5\]

Thus,

\[\Rightarrow {{\text{A}}_{11}}\text{=}{{\left( \text{-1} \right)}^{1+1}}{{\text{M}}_{11}}={{\left( \text{-1} \right)}^{2}}{{\text{M}}_{11}}\]

\[\Rightarrow {{A}_{11}}\text{=14}\]

\[\Rightarrow {{\text{A}}_{12}}\text{=}{{\left( \text{-1} \right)}^{1+2}}{{\text{M}}_{12}}={{\left( \text{-1} \right)}^{3}}{{\text{M}}_{12}}\]

\[\Rightarrow {{A}_{12}}=-9\]

\[\Rightarrow {{\text{A}}_{13}}\text{=}{{\left( \text{-1} \right)}^{1+3}}{{\text{M}}_{13}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{13}}\]

\[\Rightarrow {{A}_{13}}=-1\]

Similarly,

\[\Rightarrow {{\text{A}}_{21}}\text{=}{{\left( \text{-1} \right)}^{2+1}}{{\text{M}}_{21}}={{\left( \text{-1} \right)}^{3}}{{\text{M}}_{21}}\]

\[\Rightarrow {{\text{A}}_{21}}\text{=-9}\]

\[\Rightarrow {{\text{A}}_{22}}\text{=}{{\left( \text{-1} \right)}^{2+2}}{{\text{M}}_{22}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{22}}\]

\[\Rightarrow {{\text{A}}_{22}}=4\]

\[\Rightarrow {{\text{A}}_{23}}\text{=}{{\left( \text{-1} \right)}^{2+3}}{{\text{M}}_{23}}={{\left( \text{-1} \right)}^{5}}{{\text{M}}_{23}}\]

\[\Rightarrow {{\text{A}}_{23}}\text{=1}\]

And

\[\Rightarrow {{\text{A}}_{31}}\text{=}{{\left( \text{-1} \right)}^{3+1}}{{\text{M}}_{31}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{31}}\]

\[\Rightarrow {{\text{A}}_{31}}=-1\]

\[\Rightarrow {{\text{A}}_{32}}\text{=}{{\left( \text{-1} \right)}^{3+2}}{{\text{M}}_{32}}={{\left( \text{-1} \right)}^{5}}{{\text{M}}_{32}}\]

\[\Rightarrow {{\text{A}}_{32}}=1\]

\[\Rightarrow {{\text{A}}_{33}}\text{=}{{\left( \text{-1} \right)}^{3+3}}{{\text{M}}_{33}}={{\left( \text{-1} \right)}^{6}}{{\text{M}}_{33}}\]

\[\Rightarrow {{\text{A}}_{33}}=-1\].

Cofactor matrix is \[\left[ \begin{matrix} 14 & -9 & -1  \\ -9 & 4 & 1  \\ -1 & 1 & -1  \\ \end{matrix} \right]\].

We know that adjoint of a matrix is the transpose of its cofactor matrix.

\[\Rightarrow adjA={{\left[ \begin{matrix} 14 & -9 & -1  \\ -9 & 4 & 1  \\ -1 & 1 & -1  \\ \end{matrix} \right]}^{T}}\] \[\therefore \text{adjA=}\left[ \begin{matrix} 14 & -9 & -1  \\ -9 & 4 & 1  \\ -1 & 1 & -1  \\ \end{matrix} \right]\]

Let us denote the adjoint of A as B. So, \[B\text{=}\left[ \begin{matrix} 14 & -9 & -1  \\ -9 & 4 & 1  \\ -1 & 1 & -1  \\ \end{matrix} \right]\].

The inverse of A is given by

\[{{\text{A}}^{\text{-1}}}\text{=}\dfrac{\text{1}}{\left| \text{A} \right|}\left( \text{adjA} \right)\]

\[\Rightarrow {{\text{A}}^{\text{-1}}}\text{=-}\dfrac{\text{1}}{5}\left[ \begin{matrix} 14 & -9 & -1  \\ -9 & 4 & 1  \\ -1 & 1 & -1  \\ \end{matrix} \right]\]

\[\therefore {{A}^{\text{-1}}}\text{=}\dfrac{\text{1}}{5}\left[ \begin{matrix} -14 & 9 & 1  \\ 9 & -4 & -1  \\ 1 & -1 & 1  \\ \end{matrix} \right]\]

Now, we have to verify \[{{\left[ \text{adjA} \right]}^{\text{-1}}}\text{=adj}\left( {{\text{A}}^{\text{-1}}} \right)\].

Let us compute the RHS first, i.e. the adjoint of \[{{A}^{\text{-1}}}\text{=}\dfrac{\text{1}}{5}\left[ \begin{matrix} -14 & 9 & 1  \\ 9 & -4 & -1  \\ 1 & -1 & 1  \\ \end{matrix} \right]\] or \[{{A}^{\text{-1}}}\text{=}\left[ \begin{matrix} -\dfrac{14}{5} & \dfrac{9}{5} & \dfrac{\text{1}}{5}  \\ \dfrac{9}{5} & -\dfrac{4}{5} & -\dfrac{\text{1}}{5}  \\ \dfrac{\text{1}}{5} & -\dfrac{\text{1}}{5} & \dfrac{\text{1}}{5}  \\ \end{matrix} \right]\].

Thus,

\[\Rightarrow {{\text{A}}_{11}}\text{=}{{\left( \text{-1} \right)}^{1+1}}{{\text{M}}_{11}}={{\left( \text{-1} \right)}^{2}}{{\text{M}}_{11}}\]

\[\Rightarrow {{A}_{11}}\text{=-}\dfrac{\text{1}}{5}\]

\[\Rightarrow {{\text{A}}_{12}}\text{=}{{\left( \text{-1} \right)}^{1+2}}{{\text{M}}_{12}}={{\left( \text{-1} \right)}^{3}}{{\text{M}}_{12}}\]

\[\Rightarrow {{A}_{12}}=-\dfrac{2}{5}\]

\[\Rightarrow {{\text{A}}_{13}}\text{=}{{\left( \text{-1} \right)}^{1+3}}{{\text{M}}_{13}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{13}}\]

\[\Rightarrow {{A}_{13}}=-\dfrac{\text{1}}{5}\]

Similarly,

\[\Rightarrow {{\text{A}}_{21}}\text{=}{{\left( \text{-1} \right)}^{2+1}}{{\text{M}}_{21}}={{\left( \text{-1} \right)}^{3}}{{\text{M}}_{21}}\]

\[\Rightarrow {{\text{A}}_{21}}\text{=-}\dfrac{2}{5}\]

\[\Rightarrow {{\text{A}}_{22}}\text{=}{{\left( \text{-1} \right)}^{2+2}}{{\text{M}}_{22}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{22}}\]

\[\Rightarrow {{\text{A}}_{22}}=-\dfrac{3}{5}\]

\[\Rightarrow {{\text{A}}_{23}}\text{=}{{\left( \text{-1} \right)}^{2+3}}{{\text{M}}_{23}}={{\left( \text{-1} \right)}^{5}}{{\text{M}}_{23}}\]

\[\Rightarrow {{\text{A}}_{23}}\text{=-}\dfrac{\text{1}}{5}\]

And

\[\Rightarrow {{\text{A}}_{31}}\text{=}{{\left( \text{-1} \right)}^{3+1}}{{\text{M}}_{31}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{31}}\]

\[\Rightarrow {{\text{A}}_{31}}=-\dfrac{\text{1}}{5}\]

\[\Rightarrow {{\text{A}}_{32}}\text{=}{{\left( \text{-1} \right)}^{3+2}}{{\text{M}}_{32}}={{\left( \text{-1} \right)}^{5}}{{\text{M}}_{32}}\]

\[\Rightarrow {{\text{A}}_{32}}=-\dfrac{\text{1}}{5}\]

\[\Rightarrow {{\text{A}}_{33}}\text{=}{{\left( \text{-1} \right)}^{3+3}}{{\text{M}}_{33}}={{\left( \text{-1} \right)}^{6}}{{\text{M}}_{33}}\]

\[\Rightarrow {{\text{A}}_{33}}=-1\].

Cofactor matrix is \[\left[ \begin{matrix} -\dfrac{\text{1}}{5} & -\dfrac{2}{5} & -\dfrac{\text{1}}{5}  \\ -\dfrac{2}{5} & -\dfrac{3}{5} & -\dfrac{\text{1}}{5}  \\ -\dfrac{\text{1}}{5} & -\dfrac{\text{1}}{5} & -1  \\ \end{matrix} \right]\].  We know that adjoint of a matrix is the transpose of its cofactor matrix.

\[\Rightarrow adj{{A}^{-1}}={{\left[ \begin{matrix} -\dfrac{\text{1}}{5} & -\dfrac{2}{5} & -\dfrac{\text{1}}{5}  \\ -\dfrac{2}{5} & -\dfrac{3}{5} & -\dfrac{\text{1}}{5}  \\ -\dfrac{\text{1}}{5} & -\dfrac{\text{1}}{5} & -1  \\ \end{matrix} \right]}^{T}}\]

\[\therefore \text{adj}{{A}^{-1}}\text{=}\left[ \begin{matrix} -\dfrac{\text{1}}{5} & -\dfrac{2}{5} & -\dfrac{\text{1}}{5}  \\ -\dfrac{2}{5} & -\dfrac{3}{5} & -\dfrac{\text{1}}{5}  \\ -\dfrac{\text{1}}{5} & -\dfrac{\text{1}}{5} & -1  \\ \end{matrix} \right]\]

Next, moving to the LHS, i.e. the inverse of adjoint of A. We have adjoint of A as matrix B,

\[B\text{=}\left[ \begin{matrix} 14 & -9 & -1  \\ -9 & 4 & 1  \\ -1 & 1 & -1  \\ \end{matrix} \right]\]

Determinant of B is

$\left| B \right|=\left[ 14\left( -4-1 \right)+9\left( 9+1 \right)-1\left( -9+4 \right) \right]$

$\Rightarrow \left| B \right|=\left[ -70+90+5 \right]$

$\therefore \left| B \right|=25$

Now the cofactors,

Thus,

\[\Rightarrow {{B}_{11}}\text{=}{{\left( \text{-1} \right)}^{1+1}}{{\text{M}}_{11}}={{\left( \text{-1} \right)}^{2}}{{\text{M}}_{11}}\]

\[\Rightarrow {{B}_{11}}\text{=-5}\]

\[\Rightarrow {{B}_{12}}\text{=}{{\left( \text{-1} \right)}^{1+2}}{{\text{M}}_{12}}={{\left( \text{-1} \right)}^{3}}{{\text{M}}_{12}}\]

\[\Rightarrow {{B}_{12}}=-10\]

\[\Rightarrow {{B}_{13}}\text{=}{{\left( \text{-1} \right)}^{1+3}}{{\text{M}}_{13}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{13}}\]

\[\Rightarrow {{B}_{13}}=-5\]

Similarly,

\[\Rightarrow {{B}_{21}}\text{=}{{\left( \text{-1} \right)}^{2+1}}{{\text{M}}_{21}}={{\left( \text{-1} \right)}^{3}}{{\text{M}}_{21}}\]

\[\Rightarrow {{B}_{21}}\text{=-10}\]

\[\Rightarrow {{B}_{22}}\text{=}{{\left( \text{-1} \right)}^{2+2}}{{\text{M}}_{22}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{22}}\]

\[\Rightarrow {{B}_{22}}=-15\]

\[\Rightarrow {{B}_{23}}\text{=}{{\left( \text{-1} \right)}^{2+3}}{{\text{M}}_{23}}={{\left( \text{-1} \right)}^{5}}{{\text{M}}_{23}}\]

\[\Rightarrow {{B}_{23}}\text{=-5}\]

And

\[\Rightarrow {{B}_{31}}\text{=}{{\left( \text{-1} \right)}^{3+1}}{{\text{M}}_{31}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{31}}\]

\[\Rightarrow {{B}_{31}}=-5\]

\[\Rightarrow {{B}_{32}}\text{=}{{\left( \text{-1} \right)}^{3+2}}{{\text{M}}_{32}}={{\left( \text{-1} \right)}^{5}}{{\text{M}}_{32}}\]

\[\Rightarrow {{B}_{32}}=-5\]

\[\Rightarrow {{B}_{33}}\text{=}{{\left( \text{-1} \right)}^{3+3}}{{\text{M}}_{33}}={{\left( \text{-1} \right)}^{6}}{{\text{M}}_{33}}\]

\[\Rightarrow {{B}_{33}}=-25\].

Cofactor matrix is \[\left[ \begin{matrix} -5 & -10 & -5  \\ -10 & -15 & -5  \\ -5 & -5 & -25  \\ \end{matrix} \right]\].

We know that adjoint of a matrix is the transpose of its cofactor matrix.

\[\Rightarrow adjB={{\left[ \begin{matrix} -5 & -10 & -5  \\ -10 & -15 & -5  \\ -5 & -5 & -25  \\ \end{matrix} \right]}^{T}}\]

\[\therefore \text{adjB=}\left[ \begin{matrix} -5 & -10 & -5  \\ -10 & -15 & -5  \\ -5 & -5 & -25  \\ \end{matrix} \right]\]

Now the inverse is found as

\[{{B}^{\text{-1}}}\text{=}\dfrac{\text{1}}{\left| B \right|}\left( \text{adjB} \right)\]

\[\Rightarrow {{B}^{\text{-1}}}\text{=}\dfrac{\text{1}}{25}\left[ \begin{matrix} -5 & -10 & -5  \\ -10 & -15 & -5  \\ -5 & -5 & -25  \\ \end{matrix} \right]\]

\[\therefore {{B}^{\text{-1}}}\text{=}\left[ \begin{matrix} -\dfrac{\text{1}}{5} & -\dfrac{2}{5} & -\dfrac{\text{1}}{5}  \\ -\dfrac{2}{5} & -\dfrac{3}{5} & -\dfrac{\text{1}}{5}  \\ -\dfrac{\text{1}}{5} & -\dfrac{\text{1}}{5} & -1  \\ \end{matrix} \right]\]

So, LHS is \[{{\left[ \text{adjA} \right]}^{\text{-1}}}\text{=}\left[ \begin{matrix} -\dfrac{\text{1}}{5} & -\dfrac{2}{5} & -\dfrac{\text{1}}{5}  \\ -\dfrac{2}{5} & -\dfrac{3}{5} & -\dfrac{\text{1}}{5}  \\ -\dfrac{\text{1}}{5} & -\dfrac{\text{1}}{5} & -1  \\ \end{matrix} \right]\]. 

Since LHS=RHS, it is verified that \[{{\left[ \text{adjA} \right]}^{\text{-1}}}\text{=adj}\left( {{\text{A}}^{\text{-1}}} \right)\].

2. \[\mathbf{{{\left( {{\text{A}}^{\text{-1}}} \right)}^{\text{-1}}}\text{=A}}\]

Ans: Since \[{{A}^{\text{-1}}}\text{=}\left[ \begin{matrix} -\dfrac{14}{5} & \dfrac{9}{5} & \dfrac{\text{1}}{5}  \\ \dfrac{9}{5} & -\dfrac{4}{5} & -\dfrac{\text{1}}{5}  \\ \dfrac{\text{1}}{5} & -\dfrac{\text{1}}{5} & \dfrac{\text{1}}{5}  \\ \end{matrix} \right]\]  and \[\text{adj}{{A}^{-1}}\text{=}\left[ \begin{matrix} -\dfrac{\text{1}}{5} & -\dfrac{2}{5} & -\dfrac{\text{1}}{5}  \\ -\dfrac{2}{5} & -\dfrac{3}{5} & -\dfrac{\text{1}}{5}  \\ -\dfrac{\text{1}}{5} & -\dfrac{\text{1}}{5} & -1  \\ \end{matrix} \right]\], thus,

\[\Rightarrow \left| {{\text{A}}^{\text{-1}}} \right|\text{=}\left[ -\dfrac{14}{5}\left( -\dfrac{4}{25}-\dfrac{1}{25} \right)-\dfrac{9}{5}\left( \dfrac{9}{25}+\dfrac{1}{25} \right)+\dfrac{1}{5}\left( -\dfrac{9}{25}+\dfrac{4}{25} \right) \right]\]

\[\Rightarrow \left| {{\text{A}}^{\text{-1}}} \right|\text{=}\left[ \dfrac{70}{125}-\dfrac{90}{125}-\dfrac{5}{125} \right]\]

\[\therefore \left| {{\text{A}}^{\text{-1}}} \right|\text{=-}\dfrac{1}{5}\]

Also, we know that an inverse of a matrix is given by:

\[\Rightarrow {{\left( {{\text{A}}^{\text{-1}}} \right)}^{-1}}\text{=}\dfrac{\text{adj}{{\text{A}}^{\text{-1}}}}{\left| \text{A} \right|}\]

\[\Rightarrow {{\left( {{\text{A}}^{\text{-1}}} \right)}^{-1}}\text{=-5}\left[ \begin{matrix} -\dfrac{\text{1}}{5} & -\dfrac{2}{5} & -\dfrac{\text{1}}{5}  \\ -\dfrac{2}{5} & -\dfrac{3}{5} & -\dfrac{\text{1}}{5}  \\ -\dfrac{\text{1}}{5} & -\dfrac{\text{1}}{5} & -1  \\ \end{matrix} \right]\]

\[\Rightarrow {{\left( {{\text{A}}^{\text{-1}}} \right)}^{-1}}\text{=}\left[ \begin{matrix} 1 & 2 & 1  \\ 2 & 3 & 1  \\ 1 & 1 & 1  \\ \end{matrix} \right]\]

\[\Rightarrow {{\left( {{\text{A}}^{\text{-1}}} \right)}^{-1}}\text{=A}\]

Hence verified that \[{{\left( {{\text{A}}^{\text{-1}}} \right)}^{\text{-1}}}\text{=A}\].

5. Evaluate \[\mathbf{\left| \begin{matrix} \text{x} & \text{y} & \text{x+y}  \\ \text{y} & \text{x+y} & \text{x}  \\ \text{x+y} & \text{x} & \text{y}  \\ \end{matrix} \right|}\]

Ans: Given, \[\text{ }\!\!\Delta\!\!\text{ =}\left| \begin{matrix} \text{x} & \text{y} & \text{x+y}  \\ \text{y} & \text{x+y} & \text{x}  \\ \text{x+y} & \text{x} & \text{y}  \\ \end{matrix} \right|\]

Applying \[{{R}_{1}}\to {{R}_{1}}\text{+}{{\text{R}}_{2}}\text{+}{{\text{R}}_{3}}\]

\[\text{ }\!\!\Delta\!\!\text{ =}\left| \begin{matrix} \text{2}\left( \text{x+y} \right) & \text{2}\left( \text{x+y} \right) & \text{2}\left( \text{x+y} \right)  \\ \text{y} & \text{x+y} & \text{x}  \\ \text{x+y} & \text{x} & \text{y}  \\ \end{matrix} \right|\]

Taking \[\text{2}\left( \text{x+y} \right)\] common from \[{{\text{R}}_{1}}\]

\[\Rightarrow \Delta \text{=2}\left( \text{x+y} \right)\left| \begin{matrix} \text{1} & \text{1} & \text{1}  \\ \text{y} & \text{x+y} & \text{x}  \\ \text{x+y} & \text{x} & \text{y}  \\ \end{matrix} \right|\]

Applying the row operations \[{{\text{C}}_{2}}\to {{C}_{2}}-{{C}_{1}}\] and \[{{C}_{3}}\to {{C}_{3}}-{{C}_{1}}\]

\[\text{ }\!\!\Delta\!\!\text{ =2}\left( \text{x+y} \right)\left| \begin{matrix} \text{1} & \text{0} & 0  \\ \text{y} & \text{x} & \text{x-y}  \\ \text{x+y} & \text{-y} & \text{-x}  \\ \end{matrix} \right|\]

Expanding along \[{{\text{R}}_{1}}\]

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =2}\left( \text{x+y} \right)\left[ 1\left( \left( x\times \left( \text{-x} \right) \right)-\left( -y\left( x-y \right) \right) \right)\text{+0+}0 \right]\]

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =2}\left( \text{x+y} \right)\left[ 1\left( \left( \text{-}{{\text{x}}^{2}} \right)-\left( -yx+{{y}^{2}} \right) \right) \right]\]

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =2}\left( \text{x+y} \right)\left[ -{{\text{x}}^{\text{2}}}\text{+xy}-{{\text{y}}^{\text{2}}} \right]\]

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =-2}\left( \text{x+y} \right)\left[ {{\text{x}}^{\text{2}}}-xy+{{\text{y}}^{\text{2}}} \right]\]

Applying the identity,

\[\therefore \text{ }\!\!\Delta\!\!\text{ =-2}\left[ {{\text{x}}^{3}}\text{+}{{\text{y}}^{3}} \right]\]

6. Evaluate \[\mathbf{\left| \begin{matrix} \text{1} & \text{x} & \text{y}  \\ \text{1} & \text{x+y} & \text{y}  \\ \text{1} & \text{x} & \text{x+y}  \\ \end{matrix} \right|}\]

Ans: Given, \[\text{ }\!\!\Delta\!\!\text{ =}\left| \begin{matrix} \text{1} & \text{x} & \text{y}  \\ \text{1} & \text{x+y} & \text{y}  \\ \text{1} & \text{x} & \text{x+y}  \\ \end{matrix} \right|\]

Applying the row operations \[{{\text{R}}_{2}}\to {{\text{R}}_{2}}-{{\text{R}}_{1}}\] and \[{{\text{R}}_{3}}\to {{\text{R}}_{3}}-{{\text{R}}_{1}}\]

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}\left| \begin{matrix} \text{1} & \text{x} & \text{y}  \\ \text{0} & \text{y} & \text{0}  \\ \text{0} & \text{0} & \text{x}  \\ \end{matrix} \right|\]

Expanding along \[{{\text{C}}_{1}}\]

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =1}\left( \text{xy-0} \right)\]

\[\therefore \text{ }\!\!\Delta\!\!\text{ }=\text{xy}\]

7. Solve the system of the following equations

\[\mathbf{\dfrac{\text{2}}{\text{x}}\text{+}\dfrac{\text{3}}{\text{y}}\text{+}\dfrac{\text{10}}{\text{z}}\text{=4}}\]

\[\mathbf{\dfrac{\text{4}}{\text{x}}\text{+}\dfrac{\text{6}}{\text{y}}\text{+}\dfrac{\text{5}}{\text{z}}\text{=1}}\]

\[\mathbf{\dfrac{\text{6}}{\text{x}}\text{+}\dfrac{\text{9}}{\text{y}}\text{+}\dfrac{\text{20}}{\text{z}}\text{=2}}\]

Ans: Given equations,

\[\dfrac{\text{2}}{\text{x}}\text{+}\dfrac{\text{3}}{\text{y}}\text{+}\dfrac{\text{10}}{\text{z}}\text{=4}\]

\[\dfrac{\text{4}}{\text{x}}\text{+}\dfrac{\text{6}}{\text{y}}\text{+}\dfrac{\text{5}}{\text{z}}\text{=1}\]

\[\dfrac{\text{6}}{\text{x}}\text{+}\dfrac{\text{9}}{\text{y}}\text{+}\dfrac{\text{20}}{\text{z}}\text{=2}\]

Let \[\dfrac{\text{1}}{\text{x}}\text{=p}\] , \[\dfrac{\text{1}}{\text{y}}\text{=q}\] and \[\dfrac{\text{1}}{\text{z}}\text{=r}\]

Then the given system of equations becomes:

\[\text{2p+3q+10r=4}\]

\[\text{4p-6q+5r=1}\]

\[\text{6p+9q+20r=2}\]

Let \[\text{A=}\left[ \begin{matrix} \text{2} & \text{3} & \text{10}  \\ \text{4} & \text{-6} & \text{5}  \\ \text{6} & \text{9} & \text{-20}  \\ \end{matrix} \right]\] , \[\text{X=}\left[ \begin{matrix} \text{p}  \\ \text{q}  \\ \text{r}  \\ \end{matrix} \right]\] and \[\text{B=}\left[ \begin{matrix} \text{4}  \\ \text{1}  \\ \text{2}  \\ \end{matrix} \right]\] such that, this system can be written in the form of \[\text{AX=B}\].

Now,

\[\Rightarrow \left| A \right|\text{=2}\left( \text{120-45} \right)\text{-3}\left( \text{-80-30} \right)\text{+10}\left( \text{36+36} \right)\]

\[\Rightarrow \left| A \right|\text{=150+330+720}\]

\[\Rightarrow \left| A \right|\text{=1200}\]

Thus, \[\text{A}\] is a non-singular matrix.

$\therefore $ Its inverse exists.

So,

\[\Rightarrow {{\text{A}}_{11}}\text{=}{{\left( \text{-1} \right)}^{1+1}}{{\text{M}}_{11}}={{\left( \text{-1} \right)}^{2}}\left| \begin{matrix} -6 & 5  \\ 9 & -20  \\ \end{matrix} \right|\]

\[\Rightarrow {{A}_{11}}\text{=75}\]

\[\Rightarrow {{\text{A}}_{12}}\text{=}{{\left( \text{-1} \right)}^{1+2}}{{\text{M}}_{12}}={{\left( \text{-1} \right)}^{3}}\left| \begin{matrix} 4 & 5  \\ 6 & -20  \\ \end{matrix} \right|\]

\[\Rightarrow {{A}_{12}}=110\]

\[\Rightarrow {{\text{A}}_{13}}\text{=}{{\left( \text{-1} \right)}^{1+3}}{{\text{M}}_{13}}={{\left( \text{-1} \right)}^{4}}\left| \begin{matrix} 4 & -6  \\ 6 & 9  \\ \end{matrix} \right|\]

\[\Rightarrow {{A}_{13}}=72\]

Similarly,

\[\Rightarrow {{\text{A}}_{21}}\text{=}{{\left( \text{-1} \right)}^{2+1}}{{\text{M}}_{21}}={{\left( \text{-1} \right)}^{3}}\left| \begin{matrix} 3 & 10  \\ 9 & -20  \\ \end{matrix} \right|\]

\[\Rightarrow {{\text{A}}_{21}}\text{=150}\]

\[\Rightarrow {{\text{A}}_{22}}\text{=}{{\left( \text{-1} \right)}^{2+2}}{{\text{M}}_{22}}={{\left( \text{-1} \right)}^{4}}\left| \begin{matrix} 2 & 10  \\ 6 & -20  \\ \end{matrix} \right|\]

\[\Rightarrow {{\text{A}}_{22}}=100\]

\[\Rightarrow {{\text{A}}_{23}}\text{=}{{\left( \text{-1} \right)}^{2+3}}{{\text{M}}_{23}}={{\left( \text{-1} \right)}^{5}}\left| \begin{matrix} 2 & 3  \\ 6 & 9  \\ \end{matrix} \right|\]

\[\Rightarrow {{\text{A}}_{23}}\text{=0}\]

and

\[\Rightarrow {{\text{A}}_{31}}\text{=}{{\left( \text{-1} \right)}^{3+1}}{{\text{M}}_{31}}={{\left( \text{-1} \right)}^{4}}\left| \begin{matrix} 3 & 10  \\ -6 & 5  \\ \end{matrix} \right|\]

\[\Rightarrow {{\text{A}}_{31}}=75\]

\[\Rightarrow {{\text{A}}_{32}}\text{=}{{\left( \text{-1} \right)}^{3+2}}{{\text{M}}_{32}}={{\left( \text{-1} \right)}^{5}}\left| \begin{matrix} 2 & 10  \\ 4 & 5  \\ \end{matrix} \right|\]

\[\Rightarrow {{\text{A}}_{32}}=30\]

\[\Rightarrow {{\text{A}}_{33}}\text{=}{{\left( \text{-1} \right)}^{3+3}}{{\text{M}}_{33}}={{\left( \text{-1} \right)}^{6}}\left| \begin{matrix} 2 & 3  \\ 4 & -6  \\ \end{matrix} \right|\]

\[\Rightarrow {{\text{A}}_{33}}=-24\].

The inverse of a matrix is given by:

\[\Rightarrow {{\text{A}}^{\text{-1}}}\text{=}\dfrac{\text{1}}{\left| \text{A} \right|}\left( \text{adjA} \right)\]

\[\therefore {{\text{A}}^{\text{-1}}}\text{=}\dfrac{\text{1}}{\text{1200}}\left[ \begin{matrix} \text{75} & \text{150} & \text{75}  \\ \text{110} & \text{-100} & \text{30}  \\ \text{72} & \text{0} & \text{-24}  \\ \end{matrix} \right]\]

Now, \[\text{X=}{{\text{A}}^{\text{-1}}}\text{B}\]

\[\Rightarrow \left[ \begin{matrix} \text{p}  \\ \text{q}  \\ \text{r}  \\ \end{matrix} \right]\text{=}\dfrac{\text{1}}{\text{1200}}\left[ \begin{matrix} \text{75} & \text{150} & \text{75}  \\ \text{110} & \text{-100} & \text{30}  \\ \text{72} & \text{0} & \text{-24}  \\ \end{matrix} \right]\left[ \begin{matrix} \text{4}  \\ \text{1}  \\ \text{2}  \\ \end{matrix} \right]\]

\[\Rightarrow \left[ \begin{matrix} \text{p}  \\ \text{q}  \\ \text{r}  \\ \end{matrix} \right]\text{=}\dfrac{\text{1}}{\text{1200}}\left[ \begin{matrix} \text{300+150+150}  \\ \text{440-100+60}  \\ \text{288+0-48}  \\ \end{matrix} \right]\]

\[\Rightarrow \left[ \begin{matrix} \text{p}  \\ \text{q}  \\ \text{r}  \\ \end{matrix} \right]\text{=}\dfrac{\text{1}}{\text{1200}}\left[ \begin{matrix} \text{600}  \\ \text{400}  \\ \text{240}  \\ \end{matrix} \right]\]

\[\Rightarrow \left[ \begin{matrix} \text{p}  \\ \text{q}  \\ \text{r}  \\ \end{matrix} \right]\text{=}\left[ \begin{matrix} \dfrac{\text{1}}{\text{2}}  \\ \dfrac{\text{1}}{\text{3}}  \\ \dfrac{\text{1}}{\text{5}}  \\ \end{matrix} \right]\]

\[\therefore \text{P=}\dfrac{\text{1}}{\text{2}}\], \[\text{q=}\dfrac{\text{1}}{\text{3}}\] and \[\text{r=}\dfrac{\text{1}}{\text{5}}\]

Thus \[\text{x=2}\] , \[\text{y=3}\] and \[\text{z=5}\].

8. Choose the correct answer.

If \[\mathbf{\text{a,b,c}}\] are in \[\mathbf{\text{A}\text{.P}}\] , then the determinant

\[\mathbf{\left| \begin{matrix} \text{x+2} & \text{x+3} & \text{x+2a}  \\ \text{x+3} & \text{x+4} & \text{x+2b}  \\ \text{x+4} & \text{x+5} & \text{x+2c}  \\ \end{matrix} \right|}\]

1. \[\mathbf{\text{0}}\]

2. \[\mathbf{\text{1}}\]

3. \[\mathbf{\text{X}}\]

4. \[\mathbf{\text{2X}}\]

Ans: Given, \[\text{ }\!\!\Delta\!\!\text{ =}\left| \begin{matrix} \text{x+2} & \text{x+3} & \text{x+2a}  \\ \text{x+3} & \text{x+4} & \text{x+2b}  \\ \text{x+4} & \text{x+5} & \text{x+2c}  \\ \end{matrix} \right|\]

Since \[\text{a}\] , \[\text{b}\], and \[\text{c}\] are in A.P.  \[\text{2b=a+c}\],

\[\Rightarrow \Delta \text{=}\left| \begin{matrix} \text{x+2} & \text{x+3} & \text{x+2a}  \\ \text{x+3} & \text{x+4} & \text{x+}\left( \text{a+c} \right)  \\ \text{x+4} & \text{x+5} & \text{x+2c}  \\ \end{matrix} \right|\]          

Applying $R_1 \rightarrow R_1 - R_2  \, \,  \text{and} R_3 \rightarrow R_3 - R_2$

\[\Rightarrow \Delta \text{=}\left| \begin{matrix} \text{-1} & \text{-1} & \text{a-c}  \\ \text{x+3} & \text{x+4} & \text{x+}\left( \text{a+c} \right)  \\ \text{1} & \text{1} & \text{c-a}  \\ \end{matrix} \right|\]

Applying \[{{\text{R}}_{\text{1}}}\to {{\text{R}}_{\text{1}}}\text{-}{{\text{R}}_{\text{3}}}\]

\[\Rightarrow \text{ }\!\!\Delta\!\!\text{ =}\left| \begin{matrix} \text{0} & \text{0} & \text{0}  \\ \text{x+3} & \text{x+4} & \text{x+a+c}  \\ \text{1} & \text{1} & \text{c-a}  \\ \end{matrix} \right|\]

\[\therefore \Delta =0\]

9. Choose the correct answer.If \[\mathbf{\text{X,Y,Z}}\] are nonzero real numbers, then the inverse of matrix \[\mathbf{\text{A=}\left[ \begin{matrix} \text{x} & \text{0} & \text{0}  \\ \text{0} & \text{y} & \text{0}  \\ \text{0} & \text{0} & \text{z}  \\ \end{matrix} \right]}\]

1. \[\mathbf{\left[ \begin{matrix} {{\text{x}}^{\text{-1}}} & \text{0} & \text{0}  \\ \text{0} & {{\text{y}}^{\text{-1}}} & \text{0}  \\ \text{0} & \text{0} & {{\text{z}}^{\text{-1}}}  \\ \end{matrix} \right]}\]

2. \[\mathbf{\text{xyz}\left[ \begin{matrix} {{\text{x}}^{\text{-1}}} & \text{0} & \text{0}  \\ \text{0} & {{\text{y}}^{\text{-1}}} & \text{0}  \\ \text{0} & \text{0} & {{\text{z}}^{\text{-1}}}  \\ \end{matrix} \right]}\]

3. \[\mathbf{\dfrac{\text{1}}{\text{xyz}}\left[ \begin{matrix} \text{x} & \text{0} & \text{0}  \\ \text{0} & \text{y} & \text{0}  \\ \text{0} & \text{0} & \text{z}  \\ \end{matrix} \right]}\]

4. \[\mathbf{\dfrac{\text{1}}{\text{xyz}}\left[ \begin{matrix} \text{1} & \text{0} & \text{0}  \\ \text{0} & \text{1} & \text{0}  \\ \text{0} & \text{0} & \text{1}  \\ \end{matrix} \right]}\]

Ans: Given, \[\text{A=}\left[ \begin{matrix} \text{x} & \text{0} & \text{0}  \\ \text{0} & \text{y} & \text{0}  \\ \text{0} & \text{0} & \text{z}  \\ \end{matrix} \right]\]

\[\Rightarrow \left| \text{A} \right|\text{=x}\left( \text{yz-0} \right)\]

\[\therefore \left| \text{A} \right|\text{=xyz}\ne \text{0}\]

Thus,

\[\Rightarrow {{\text{A}}_{11}}\text{=}{{\left( \text{-1} \right)}^{1+1}}{{\text{M}}_{11}}={{\left( \text{-1} \right)}^{2}}{{\text{M}}_{11}}\]

\[\Rightarrow {{A}_{11}}\text{=yz}\]

\[\Rightarrow {{\text{A}}_{12}}\text{=}{{\left( \text{-1} \right)}^{1+2}}{{\text{M}}_{12}}={{\left( \text{-1} \right)}^{3}}{{\text{M}}_{12}}\]

\[\Rightarrow {{A}_{12}}=0\]

\[\Rightarrow {{\text{A}}_{13}}\text{=}{{\left( \text{-1} \right)}^{1+3}}{{\text{M}}_{13}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{13}}\]

\[\Rightarrow {{A}_{13}}=0\]

Similarly,

\[\Rightarrow {{\text{A}}_{21}}\text{=}{{\left( \text{-1} \right)}^{2+1}}{{\text{M}}_{21}}={{\left( \text{-1} \right)}^{3}}{{\text{M}}_{21}}\]

\[\Rightarrow {{\text{A}}_{21}}\text{=0}\]

\[\Rightarrow {{\text{A}}_{22}}\text{=}{{\left( \text{-1} \right)}^{2+2}}{{\text{M}}_{22}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{22}}\]

\[\Rightarrow {{\text{A}}_{22}}=xz\]

\[\Rightarrow {{\text{A}}_{23}}\text{=}{{\left( \text{-1} \right)}^{2+3}}{{\text{M}}_{23}}={{\left( \text{-1} \right)}^{5}}{{\text{M}}_{23}}\]

\[\Rightarrow {{\text{A}}_{23}}\text{=0}\]

and

\[\Rightarrow {{\text{A}}_{31}}\text{=}{{\left( \text{-1} \right)}^{3+1}}{{\text{M}}_{31}}={{\left( \text{-1} \right)}^{4}}{{\text{M}}_{31}}\]

\[\Rightarrow {{\text{A}}_{31}}=0\]

\[\Rightarrow {{\text{A}}_{32}}\text{=}{{\left( \text{-1} \right)}^{3+2}}{{\text{M}}_{32}}={{\left( \text{-1} \right)}^{5}}{{\text{M}}_{32}}\]

\[\Rightarrow {{\text{A}}_{32}}=0\]

\[\Rightarrow {{\text{A}}_{33}}\text{=}{{\left( \text{-1} \right)}^{3+3}}{{\text{M}}_{33}}={{\left( \text{-1} \right)}^{6}}{{\text{M}}_{33}}\]

\[\Rightarrow {{\text{A}}_{33}}=xy\].

Cofactor matrix is \[\left[ \begin{matrix} yz & 0 & 0  \\ 0 & xz & 0  \\ 0 & 0 & xy  \\ \end{matrix} \right]\].

We know that adjoint of a matrix is the transpose of its cofactor matrix.

\[\Rightarrow adjA={{\left[ \begin{matrix} yz & 0 & 0  \\ 0 & xz & 0  \\ 0 & 0 & xy  \\ \end{matrix} \right]}^{T}}\]

\[\therefore \text{adjA=}\left[ \begin{matrix} \text{yz} & \text{0} & \text{0}  \\ \text{0} & \text{xz} & \text{0}  \\ \text{0} & \text{0} & \text{xy}  \\ \end{matrix} \right]\]

The inverse of a matrix is given by:

\[\Rightarrow {{\text{A}}^{\text{-1}}}=\dfrac{\text{1}}{\left| \text{A} \right|}\left( \text{adjA} \right)\]

\[\Rightarrow {{\text{A}}^{\text{-1}}}\text{=}\dfrac{\text{1}}{\text{xyz}}\left[ \begin{matrix} \text{yz} & \text{0} & \text{0}  \\ \text{0} & \text{xz} & \text{0}  \\ \text{0} & \text{0} & \text{xy}  \\ \end{matrix} \right]\]

\[\Rightarrow {{\text{A}}^{\text{-1}}}\text{=}\left[ \begin{matrix} \dfrac{\text{yz}}{\text{xyz}} & \text{0} & \text{0}  \\ \text{0} & \dfrac{\text{xz}}{\text{xyz}} & \text{0}  \\ \text{0} & \text{0} & \dfrac{\text{xy}}{\text{xyz}}  \\ \end{matrix} \right]\]

\[\Rightarrow {{\text{A}}^{\text{-1}}}\text{=}\left[ \begin{matrix} \dfrac{\text{1}}{\text{x}} & \text{0} & \text{0}  \\ \text{0} & \dfrac{\text{1}}{\text{y}} & \text{0}  \\ \text{0} & \text{0} & \dfrac{\text{1}}{\text{z}}  \\ \end{matrix} \right]\]

\[\therefore {{\text{A}}^{\text{-1}}}\text{=}\left[ \begin{matrix} {{\text{x}}^{\text{-1}}} & \text{0} & \text{0}  \\ \text{0} & {{\text{y}}^{\text{-1}}} & \text{0}  \\ \text{0} & \text{0} & {{\text{z}}^{\text{-1}}}  \\ \end{matrix} \right]\]

Thus, A. \[\left[ \begin{matrix} {{\text{x}}^{\text{-1}}} & \text{0} & \text{0}  \\ \text{0} & {{\text{y}}^{\text{-1}}} & \text{0}  \\ \text{0} & \text{0} & {{\text{z}}^{\text{-1}}}  \\ \end{matrix} \right]\] is the correct answer.

10. Choose the correct answer.

Let \[\mathbf{\text{A}=\left[ \begin{matrix} 1 & \sin \theta  & 1  \\ -\sin \theta  & 1 & \sin \theta   \\ -1 & -\sin \theta  & 1  \\ \end{matrix} \right]}\], 

where \[\mathbf{\text{0}\le \text{ }\!\!\theta\!\!\text{ }\le \text{2n}}\] , then

1. \[\mathbf{\text{Det}\left( \text{A} \right)\text{=0}}\]

2. \[\mathbf{text{Det}\left( \text{A} \right)\in \left( \text{2,}\infty  \right)}\]

3. \[\mathbf{\text{Det}\left( \text{A} \right)\in \left( \text{2,4} \right)}\]

4. \[\mathbf{\text{Det}\left( \text{A} \right)\left[ \text{2,4} \right]}\]

Ans: Given, \[A=\left[ \begin{matrix} 1 & \sin \theta  & 1  \\ -\sin \theta  & 1 & \sin \theta   \\ -1 & -\sin \theta  & 1  \\ \end{matrix} \right]\]

\[\therefore \left| A \right|=1\left( 1+{{\sin }^{2}}\theta  \right)-\sin \theta \left( -\sin \theta +\sin \theta  \right)+1\left( {{\sin }^{2}}\theta +1 \right)\]

\[\Rightarrow \left| A \right|=1+{{\sin }^{2}}\theta +{{\sin }^{2}}\theta +1\]

\[\Rightarrow \left| A \right|=2+2{{\sin }^{2}}\theta \]

\[\Rightarrow \left| A \right|=2\left( 1+{{\sin }^{2}}\theta  \right)\]

We know, \[0\le \sin \theta \le 1\]

\[\Rightarrow 1\le 1+{{\sin }^{2}}\theta \le 2\]

\[\Rightarrow 2\le 2\left( 1+{{\sin }^{2}}\theta  \right)\le 4\]

Thus, D. \[\text{Det}\left( \text{A} \right)\left[ \text{2,4} \right]\] is the correct answer.

NCERT Solutions for Class 12 Maths – Free PDF Download

Chapter 4 – Determinants

4.1 Introduction

In Chapter 4 of Class 12 Maths NCERT Solutions, you will learn about determinants and their basic concepts and theorems. The chapter covers determinants up to order three with real entries, as well as various properties of determinants such as minors, cofactors, and applications of determinants. Additionally, the chapter includes topics like the inconsistency of the system of linear equations and solutions of linear equations in two or three variables using the inverse of a matrix.

4.2 Determinant

In Chapter 4 of Class 12 Maths, students can learn how to find the determinant of a matrix. This includes understanding the determinant of a matrix of order one, which is relatively easy to comprehend. The section also covers the determination of the matrix of order two and order three. It explains how to expand the determinant along the row or column containing the maximum number of zeros. Additionally, it helps in understanding positive multiplication rather than negative multiplication.

4.3 Area of Triangle

The Determinants Class 12 PDF covers how to calculate the area of a triangle using vertices and expressions. It explains the concept of the area as a positive quantity and the absolute value of the determinant. It also discusses the use of positive and negative values of the determinant in calculations.

4.4 Minors and Cofactors

This section teaches students how to simplify the expansion of a determinant using minors and cofactors. It also explains the definitions of minors and cofactors. The text covers five methods to calculate the area of a triangle, including using the sum of the product of elements of any row or column with their corresponding cofactor.

4.5 The Adjoint and Inverse of a Matrix

This section is about matrices, focusing on their inverse and existence. It explains how to find the adjoint of a matrix and provides five definitions, including the adjoint of a square matrix and the theorem to verify it. Students will learn how to discover and apply these theorems to verify the definitions of various types of problems.

4.6 Applications of Determinants and Matrices

In Chapter 4 maths class 12, students learn about using determinants and matrices to solve systems of linear equations with two or three variables. The chapter covers checking if the system of equations is consistent, explaining both consistent and inconsistent systems in detail. Students also learn how to solve these equations using the inverse of a matrix, express linear equations as matrix equations, and solve them using the inverse of the coefficient matrix. The chapter also discusses the uniqueness of matrix equations in providing unique solutions for systems of equations, known as the matrix method.

The NCERT Solutions for Class 12 Maths Chapter 4 Determinants' Key features

The following concepts and equations are covered in the NCERT Solutions for chapter 4 Maths class 12 at Vedantu.

• The determinant has zero value if any two rows or any two columns are proportionate or identical.

• If and only if A is not singular, a square matrix A has an inverse.

• When A ≠ 0, the unique solution to the equation AX = B is X = A–1 B.

• A system of equations has a solution if it is consistent.

• There cannot be a solution to an inconsistent set of equations.

• In the matrix equation AX = B, with a square matrix A:

There is a unique solution if | A| ≠ 0.

If (adj A) B ≠ 0 and |A| = 0, then no solution exists.

In case (adj A) B = 0 and | A| = 0, the consistency of the system is uncertain.

Overview of Deleted Syllabus for CBSE Class 12 Determinants 2024-25

Class 12 Maths Chapter 4: Exercises Breakdown

Conclusion

NCERT Solutions for Class 12 Maths Chapter 4 - Determinants, provided by Vedantu, offers comprehensive guidance for understanding this crucial topic. It is important to focus on grasping the fundamental concepts of determinants, such as properties, operations, and applications in solving equations and matrices. In previous year question papers, around 6-8 questions were typically asked from this chapter. Paying attention to solved examples and practice exercises can enhance your problem-solving skills. Moreover, understanding the theoretical aspects alongside practical applications is essential for a deeper comprehension. By practicing with these solutions, students can build a strong foundation in determinants, which is crucial for higher-level math and various competitive exams.

Other Related Study Material for CBSE Class 12 Maths Chapter 4

Chapter-Specific NCERT Solutions for Class 12 Maths

Given below are the chapter-wise NCERT Solutions for Class 12 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.