Chapter 4: Determinants

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

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

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

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.

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.

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,

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

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

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,

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

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

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,

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

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

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,

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

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

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,

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

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

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,

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

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

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