Question

The inverse of matrix $\left[\begin{array}{ccc}0& 1& 0\\ 1& 0& 0\\ 0& 0& 1\end{array}\right]$ is

• A. $\left[\begin{array}{ccc}0& 1& 0\\ 1& 0& 0\\ 0& 0& 1\end{array}\right]$
• B. $\left[\begin{array}{ccc}0& -1& 0\\ -1& 0& 0\\ 0& 0& -1\end{array}\right]$
• C. $\left[\begin{array}{ccc}0& 1& 0\\ 0& 0& 1\\ 1& 0& 0\end{array}\right]$
• D. $\left[\begin{array}{ccc}0& -1& 0\\ 0& 0& -1\\ -1& 0& 0\end{array}\right]$

Answer 1

The correct option is A $\left[\begin{array}{ccc}0& 1& 0\\ 1& 0& 0\\ 0& 0& 1\end{array}\right]$

Let I = $\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$

$\stackrel{R_1\leftrightarrow R_2}{\to }$$\left[\begin{array}{ccc}0& 1& 0\\ 1& 0& 0\\ 0& 0& 1\end{array}\right]$ = A

Hence given matrix A is an elementary matrix so $A^{-1}$ = A itself because inverse matrix corresponding to the elementry matrix is the matrix itself .