Chapter 4: Determinants

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

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.

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:

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

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

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

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

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

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

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

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.

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

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

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

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

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

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.

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.