Question

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

Answer 1

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

`[{\left( {{A^{ - 1}}} \right)^2} = {\left( {{A^2}} \right)^{ - 1}}]`