$A$ is an invertible $n\times n$ matrix, where $n$ is an even number. Given that $A^3+A=0$, calculate $\det(A^4)$. Is there too much information?
Solution 1:
It's not possible for $n$ to be odd, which can be extracted from your solution:
$ A^2 = -I_n \; \Rightarrow \; \det (A^2) = ( \det A )^2 = \det (-I_n) = (-1)^n$. But $A$ is invertible, which means that $( \det A )^2 > 0$ which means that $(-1)^n > 0$ which only occurs when $n$ is even.
So it was not necessary in your solution, but rather the question breaks down if $n$ is odd.