= Solution
For an integer $c$, repeated multiplicativity of the <determinant>, with the inverse identity for negative powers, gives
$$
\boxed{\det(A^c)=(\det A)^c\quad(c\in\mathbb Z).}
$$
For an arbitrary real $c$, an invertible real <matrix> does not by itself define a unique real <matrix power>. For example, the scalar matrix $(-1)$ has no real square root. If a <matrix logarithm> $L$ has been chosen with $e^L=A$ and $A^c$ is defined as $e^{cL}$, the <matrix exponential determinant identity> gives the precise formula
$$
\boxed{\det(A^c)=e^{c\operatorname{tr}L}.}
$$
This cannot always be replaced by an independently chosen scalar branch of $(\det A)^c$. For instance, $A=-I_2$ has the complex square root $iI_2$, whose <determinant> is $-1$, while the principal scalar square root of $\det A=1$ is $+1$. For a positive-definite <symmetric matrix> with its usual positive spectral power, all <eigenvalues> are positive and the familiar formula $(\det A)^c$ holds for every real $c$.
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