TITLE: $A$ 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?
QUESTION [14 upvotes]: $A$ is an invertible matrix with $n$ columns and $n$ rows, where $n$ is an even number. We are given that $A^3+A=0$ and we need to calculate $\det(A^4)$. Here is my solution: 
$$A^3+A=0 \implies A^{-1}(A^3+A)=0 \implies A^2=-I \implies A^4=I \implies \det(A^4)=1.$$ But I did not use the fact that n is even. Am I wrong, or this is not needed? If I'm wrong, please don't tell me the solution yet. Just tell me where I'm wrong. Thanks!

REPLY [18 votes]: 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.