Question

The inverse of matrix A= $\left [ \begin{matrix}
   {0} & {1} & {0} \\
   {1} & {0} &{0} \\
   {0} & {0} & {1} \\
\end{matrix} \right]$ is
A. A
B.${{A}^{T}} $
C. $\left [ \begin{matrix}
   {1} & {0} & {0} \\
   {0} & {1} &{0} \\
   {0} & {0} & {1} \\
\end{matrix} \right]$
D. $\left [ \begin{matrix}
    {1} & {0} & {0} \\
   {1} & {0} &{0} \\
   {0} & {1} & {0} \\
\end{matrix} \right]$

Answer 1

Hint:
We have to find ${{A}^{-1}}$ of this 3×3 matrix $\left [ \begin{matrix}
   {0} & {1} & {0} \\
   {1} & {0} &{0} \\
   {0} & {0} & {1} \\
\end{matrix} \right]$ , so for that we will solve for $adjA$ and $|A|$. And then dividing both of them will get our required solution. As for finding ${{A}^{-1}}$ the formula suggests ${{A}^{-1}}=\dfrac{adjA}{|A|}$.



Formula Used:
${{A}^{-1}}=\dfrac{adjA}{|A|}$
$|A|$ = $[{{a}_{11}}({{a}_{22}}{{a}_{33}}-{{a}_{23}}{{a}_{32}})-{{a}_{12}}({{a}_{21}}{{a}_{33}}-{{a}_{23}}{{a}_{31}})+{{a}_{13}}({{a}_{21}}{{a}_{32}}-{{a}_{22}}{{a}_{31}})]$


Complete step-by-step Solution:

We are given matrix $A$ = $\left [ \begin{matrix}
   {0} & {1} & {0} \\
   {1} & {0} &{0} \\
   {0} & {0} & {1} \\
\end{matrix} \right]$ and we have to find ${{A}^ {-1}} $
So first of all as the formula suggests we have to find $adjA$,
$adjA=\left[ \begin{matrix}
   {{a}_{22}}{{a}_{33}}-{{a}_{23}}{{a}_{32}} & {{a}_{21}}{{a}_{33}}-{{a}_{23}}{{a}_{31}} & {{a}_{21}}{{a}_{32}}-{{a}_{22}}{{a}_{31}} \\
   {{a}_{12}}{{a}_{33}}-{{a}_{13}}{{a}_{32}} & {{a}_{11}}{{a}_{33}}-{{a}_{13}}{{a}_{31}} & {{a}_{11}}{{a}_{32}}-{{a}_{12}}{{a}_{31}} \\
   {{a}_{12}}{{a}_{23}}-{{a}_{13}}{{a}_{22}} & {{a}_{11}}{{a}_{23}}-{{a}_{13}}{{a}_{21}} & {{a}_{11}}{{a}_{22}}-{{a}_{12}}{{a}_{21}} \\
\end{matrix} \right]$
$adjA=\left[ \begin{matrix}
   \begin{matrix}
   0(1)-0(0) \\
   1(1)-0(0) \\
   1(0)-0(0) \\
\end{matrix} & \begin{matrix}
   1(1)-0(0) \\
   0(1)-0(0) \\
   0(0)-0(1) \\
\end{matrix} & \begin{matrix}
   1(0)-0(0) \\
   0(0)-1(0) \\
   0(0)-1(1) \\
\end{matrix} \\
\end{matrix} \right]$
$adjA=\left[ \begin{matrix}
   \begin{matrix}
   0 \\
   1 \\
   0 \\
\end{matrix} & \begin{matrix}
   1 \\
   0 \\
   0 \\
\end{matrix} & \begin{matrix}
   0 \\
   0 \\
   -1 \\
\end{matrix} \\
\end{matrix} \right]$
Now as we have done above that the signs of the adj will change as per the formula
$adjA=\left[ \begin{matrix}
   \begin{matrix}
   0 \\
   -1 \\
   0 \\
\end{matrix} & \begin{matrix}
   -1 \\
   0 \\
   0 \\
\end{matrix} & \begin{matrix}
   0 \\
   0 \\
   -1 \\
\end{matrix} \\
\end{matrix} \right]$
Now |A| of matrix$\left[ \begin{matrix}
   \begin{matrix}
   0 \\
   1 \\
   0 \\
\end{matrix} & \begin{matrix}
   1 \\
   0 \\
   0 \\
\end{matrix} & \begin{matrix}
   0 \\
   0 \\
   1 \\
\end{matrix} \\
\end{matrix} \right]$ is
$|A|=0(0-0)-1(1-0) +0(0-0)$
$\Rightarrow 0(0)-1+0 $
$\Rightarrow |A|=-1 $
Now as ${{A}^ {-1}} =\dfrac{adjA}{|A|} $
$\Rightarrow {{A}^{-1}}=\dfrac{\left[ \begin{matrix}
   \begin{matrix}
   0 \\
   -1 \\
   0 \\
\end{matrix} & \begin{matrix}
   -1 \\
   0 \\
   0 \\
\end{matrix} & \begin{matrix}
   0 \\
   0 \\
   -1 \\
\end{matrix} \\
\end{matrix} \right]}{-1}$=$\left[ \begin{matrix}
   0 & 1 & 0 \\
   1 & 0 & 0 \\
   0 & 0 & 1 \\
\end{matrix} \right]$= A itself only .

Hence option A is correct.

Note:
${{A}^{-1}}$ exists only when $|A|\ne 0$. We have to remember the formula for ${{A}^{-1}}$. Sometimes students make mistakes while solving adjA and |A| for a 3×3 matrix. If the inverse of a matrix exists, we can find the adjoint of the given matrix and divide it by the determinant of the matrix.