NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.4 - Determinants

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.

Conclusion

The NCERT Solutions for Exercise 4.4 Class 12 Maths by Vedantu provide a comprehensive understanding of important concepts such as the condition for the existence of the inverse of a matrix, the adjoint and inverse of a matrix, and nonsingular matrices. These exercise 4.4 class 12 maths solutions are crucial for mastering the topic of determinants and their applications. Students should focus on understanding the conditions under which a matrix is invertible, how to calculate the adjoint, and the properties of nonsingular matrices. Practicing 4.4 maths class 12 solutions will enhance your problem-solving skills and help you perform better in exams. For a better understanding, Vedantu’s detailed explanations and step-by-step solutions and resources are available for free.

Class 12 Maths Chapter 4: Exercises Breakdown

CBSE Class 12 Maths Chapter 4 Other Study Materials

NCERT Solutions for Class 12 Maths | Chapter-wise List

Given below are the chapter-wise NCERT 12 Maths solutions PDF. Using these chapter-wise class 12th maths ncert solutions, you can get clear understanding of the concepts from all chapters.

Related Links for NCERT Class 12 Maths in Hindi

Explore these essential links for NCERT Class 12 Maths in Hindi, providing detailed solutions, explanations, and study resources to help students excel in their mathematics exams.

Important Related Links for NCERT Class 12 Maths