# If matrix A is invertible, is (A^2)^-1 = (A^-1)^2?

No take for example the matrix

$A = \left( {\begin{array}{*{20}{c}} 1&0\\ 1&1 \end{array}} \right)$

we can easily find that

${\left( {{A^2}} \right)^{ - 1}} = \left( {\begin{array}{*{20}{c}} 1&0\\ { - 2}&1 \end{array}} \right)$

but

${\left( {{A^{ - 1}}} \right)^2} = \left( {\begin{array}{*{20}{c}} 1&0\\ 1&1 \end{array}} \right)$

hence it is not always true that

$\setminus \left[\left\{\setminus \le f t {\left(\left\{\left\{{A}^{- 1}\right\}\right\} \setminus r i g h t\right)}^{2}\right\} = \left\{\setminus \le f t {\left(\left\{\left\{{A}^{2}\right\}\right\} \setminus r i g h t\right)}^{- 1}\right\} \setminus\right]$