NCERT Solutions for Class 12 Maths Chapter 4: Determinants - Exercise 4.5

Exercise 4.5

1. Find the adjoint of each of the matrices. 

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

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

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

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

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

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 the 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). \[\left[ \begin{matrix}   \text{-1} & \text{5}  \\   \text{-3} & \text{2}  \\ \end{matrix} \right]\]

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 the 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). \[\left[ \begin{matrix}   \text{1} & \text{2} & \text{3}  \\   \text{0} & \text{2} & \text{4}  \\   \text{0} & \text{0} & \text{5}  \\ \end{matrix} \right]\]

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

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

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).

\[\left[ \begin{matrix}   \text{2} & \text{1} & \text{3}  \\   \text{4} & \text{-1} & \text{0}  \\   \text{-7} & \text{2} & \text{1}  \\ \end{matrix} \right]\]

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

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 the 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,

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

\[\therefore {{\text{A}}^{\text{-1}}}\text{=}-\text{1}\left[ \begin{matrix} 2 & 0 & -1  \\ -9 & -2 & 3  \\ -6 & -1 & 2  \\ \end{matrix} \right]\]

Hence, \[{{\text{A}}^{\text{-1}}}\text{=}\left[ \begin{matrix} -2 & 0 & 1  \\ 9 & 2 & -3  \\ 6 & 1 & -2  \\ \end{matrix} \right]\].

11. Find the inverse of each of the matrices (if it exists). 

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

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

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.

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

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

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 \[\text{A=}\left[ \begin{matrix}   \text{3} & \text{1}  \\   \text{-1} & \text{2}  \\ \end{matrix} \right]\] ,show that \[{{\text{A}}^{\text{2}}}\text{-5A+7I=O}\] . Hence find \[{{\text{A}}^{\text{-1}}}\].

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

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

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

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 \[\text{A}\] be a nonsingular square matrix of order \[\text{3 }\!\!\times\!\!\text{ 3}\] . Then \[\left| \text{adjA} \right|\] is equal to

1. \[\left| \text{A} \right|\]

2. \[{{\left| \text{A} \right|}^{\text{2}}}\]

3. \[{{\left| \text{A} \right|}^{\text{3}}}\]

4. \[\text{3}\left| \text{A} \right|\]

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 \[\text{A}\] is an invertible matrix of order \[\text{2}\] , then \[\text{det}\left( {{\text{A}}^{\text{-1}}} \right)\] is equal to

1. \[\text{det}\left( \text{A} \right)\]

2. \[\dfrac{\text{1}}{\text{det}\left( \text{A} \right)}\]

3. \[\text{1}\]

4. \[\text{0}\]

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.

What are Adjoint and Inverse of a Matrix?

Adjoint of a Matrix

The adjoint of a matrix can be defined as the transpose of the cofactor matrix of the given matrix. 

For a given matrix B, the adjoint will be denoted as adj (B).  

Inverse of a Matrix

The inverse of a matrix is given by another matrix, which when multiplying with the given matrix results in the multiplicative identity. The inverse of a Matrix A will be denoted by A-1.

• The inverse of 2 × 2 matrices can be calculated by using a simple formula. Also, to get the inverse of 3 × 3 matrices, we need to learn about the determinant & adjoint of the matrix.

• The inverse of the matrix can be obtained by using two methods. The inverse of a matrix can be calculated by elementary operations and also by using the adjoint of a matrix. The elementary operations on matrices will be performed through transformations of rows or columns. Also, the inverse of a matrix can be calculated by using the inverse of the matrix formula by using the determinant and the adjoint of the matrix.

Maths NCERT Solutions Class 12 Chapter 4 Exercise 4.5

Class 12 is a very vital year in your academic life. It literally decides which education path you will choose. Class 12 board exams are crucial for everyone. They stay with you wherever you go. The subjects you learn in Class 12 will also help you build a solid conceptual foundation for your graduation and post-graduation years. Thorough preparation for Class 12 exams will also help you succeed in different competitive exams.

Class 12 Maths Chapter 4 Exercise 4.5 deals with Adjoint and Inverse of a Matrix. Exercise 4.5 Class 12 Maths consists of 4 crucial theorems. The conceptual understanding of Exercise 4.5 Maths Class 12 is critical. NCERT Solutions for Class 12 Maths Chapter 4 contains all kinds of different questions from this section of the chapter. Class 12 Exercise 4.5, more or less follows the CBSE exam question pattern. Vedantu’s Class 12 Maths NCERT Solutions Chapter 4 Exercise 4.5 also follows the same pattern. This helps you get accustomed to the different question types from Exercise 4.4 Class 12 Maths that can be asked in your board exams. These are step by solutions that have been devised to simplify all complex concepts in simpler terms. They make understanding tough concepts easier. 

Class 12 Maths Chapter 4 comprise of many important topics and subtopics that are essential from your board exam. Here is a list of all topics and subtopics covered in this chapter.

• Definition of Determinants

• Applications of Determinants

• Types of Determinants

1. First Order Determinant

2. Second-Order Determinant

3. Third Order Determinant

• Properties of Determinants

1. Property 1: When the determinant’s columns and rows are interchanged, then the value of the that determinant remains unaffected 

2. Property 2: The sign of the determinant changes when any two columns or rows of a determinant are interchanged.

3. Property 3: When any two columns or rows of a determinant are identical, and when the corresponding elements are just the same, then the determinant value is zero.

4. Property 4: When each element of a column or row of a determinant is multiplied by a constant K then the value also gets multiplied by constant K

5. Property 5: When some or all elements of a column or row of a determinant are expressed as a sum of two or more terms, then the determinant can be expressed as the sum of two or more determinants.

6. Property 6: When to each element of any column or row of a determinant, the equimultiples of the corresponding elements of other columns or row are added, then the value of the determinant remains the same i.e., the value of the determinant remains the same if the operation is applied.

7. Property 7: If each element of column or row is zero then the determinant’s value is zero

8. Property 8: In a determinant, if all the elements on one side of the principal diagonal are zero, then the value of the determinant is equal to the product of the elements is the principal diagonal.

• Area of a Triangle

• Minors and Cofactors

• Adjoint and Inverse of a Matrix

1. Theorem 1: When A be any given Square Matrix of order n, then A(adj A)=(adj A)A= AI

• Singular and No Singular Matrix

2. Theorem 2: When A and B are non-singular matrices of the same order, then AB and BA are also non-singular matrices of the same order.

3. Theorem 3: The determinant of the product of matrices is equal to the product of their respective determinants, that is, AB= |A| |B|, where A and B are Square Matrices of the same order.

4. Theorem 4: A square matrix A is invertible if and only if A is a non-singular matrix.

• Application of Determinants and Matrices

• The Solution of a System of Linear Equations Using the Inverse of a Matrix
  
  

You should choose Vedantu as your learning hub because it makes learning Class 12 Maths Determinants Exercise 4.5 and all other exercises interesting, stimulating, though-provoking a happy experience. Studying Exercise 4.5 Class 12 Maths with Vedantu’s comprehensive NCERT Solutions, LIVE Master Classes and Free Conceptual Videos, LIVE Courses, Lectures, and LIVE Quizzes Online Tests and Ongoing competition will give you a solid hold over the subject matter. It will give you enough practice to perfect your exam preparation. The online tests will help you get well-acquainted with the examination environment an increase your overall speed and efficiency. Class 12 Maths Ex. 4.5 Solutions have been prepared in a strategic manner to help you learn faster and retain for a long period of time.

NCERT Solutions for Class 12 Maths PDF Download

• Chapter 1 - Relations and Functions

• Chapter 2 - Inverse Trigonometric Functions

• Chapter 3 - Matrices

• Chapter 4 - Determinants

• Chapter 5 - Continuity and Differentiability

• Chapter 6 - Application of Derivatives

• Chapter 7 - Integrals

• Chapter 8 - Application of Integrals

• Chapter 9 - Differential Equations

• Chapter 10 - Vector Algebra

• Chapter 11 - Three Dimensional Geometry

• Chapter 12 - Linear Programming

• Chapter 13 - Probability

NCERT Solution Class 12 Maths of Chapter 4 All Exercises

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