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$A$ 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!

Asaf Karagila
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Omer
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    This is correct. – Aniruddh Agarwal Dec 23 '18 at 13:53
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    You're right for me. Probably the author thought of another way, in which $n$ being even is used (one may even easily prove $\det A=\pm1$). – Bernard Dec 23 '18 at 13:53
  • We have $A^4=-A^2$, so, if $n$ is even, $(\det A)^4=(\det A)^2$, hence $(\det A)^2=1$ and $(\det A)^4=\det(A^4)=1$. Without knowing that $n$ is even, this argument would not work; but yours is definitely better. – egreg Dec 23 '18 at 15:29

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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.

Mark Heavey
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  • In other words, if $n$ is odd and $A$ is invertible, it's impossible to have $A^3 + A = 0$, correct? – Sambo Dec 23 '18 at 14:38
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    @Sambo Sort of; it's impossible for $n$ to be odd, $A$ invertible, and $A^3 + A = 0$ all at once. Remove any one of those conditions and there is no inconsistency. – Mark Heavey Dec 23 '18 at 14:41
  • What is In and what why it's determinant is (-1)^n – user4951 Dec 23 '18 at 16:48
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    @user4951 : $I_n$ is the $n \times n$ identity matrix. No one claims that $\det I_n = (-1)^n$. The claim is about $\det(-I_n)$, which is a matrix with $0$s in every cell except the diagonal, on which there are $n$ copies of $-1$, so the determinant (by minors, say) simplifies to the product of those $n$ copies of $-1$, i.e., $(-1)^n$. – Eric Towers Dec 23 '18 at 17:32