$A^{5}=B^{5}$ and $A^{3} B^{2}=A^{2} B^{3}$
Now, $A^{5}-A^{3} B^{2}=B^{5}-A^{2} B^{3}$
$\Rightarrow A^{3}\left(A^{2}-B^{2}\right)+B^{3}\left(A^{2}-B^{2}\right)=0$
$\Rightarrow\left(A^{3}+B^{3}\right)\left(A^{2}-B^{2}\right)=0$
Post multiplying inverse of $A^{2}-B^{2}$ : $A^{3}+B^{3}=0$
$A\left[ {\begin{array}{*{20}{c}}
1&2&3 \\
0&2&3 \\
0&1&1
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
0&0&1 \\
1&0&0 \\
0&1&0
\end{array}} \right]$
તો $A^{-1}$ મેળવો.