Question

If $A=\left[\begin{array}{ccc}-1& 2& 0\\ -1& 1& 1\\ 0& 1& 0\end{array}\right]$, show that $A^2=A^{-1}.$

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have},A=\left[-120 \\ -111 \\ 010\right] \\ \mathrm{Now}, \\ {A}^2=\left[-120 \\ -111 \\ 010\right]\left[-120 \\ -111 \\ 010\right]=\left[1-2+0-2+2+00+2+0 \\ 1-1+0-2+1+10+1+0 \\ 0-1+00+1+00+1+0\right]=\left[-102 \\ 001 \\ -111\right] \\ \mathrm{and}{A}^2\times A=\left[-102 \\ 001 \\ -111\right]\left[-120 \\ -111 \\ 010\right]=\left[1+0+0-2+0+20 \\ 0+0+00+0+10 \\ 0-2+1+11\right]=\left[100 \\ 010 \\ 001\right]=I_3[\mathrm{Identity}\mathrm{matrix}\mathrm{of}\mathrm{order}3] \\ \Rightarrow A^2\times A=I_3 \\ \Rightarrow A^2=A^{-1} \end{gathered}$$