# Given that $A^3=-A$ show that $A$ is not invertible.

Let $A$ be an arbitrary square matrix with real numbers as elements. Given that $A^3=-A$ show that $A$ is not invertible.

This question appeared in my linear algebra book in the chapter on determinants so I assume that I'm supposed to show that $\det (A)=0$. I don't know how to do this, some help would be greatly appreciated.

• Some condition must be missing, consider $$\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}.$$ – Daniel Fischer Oct 27 '17 at 22:30
• If $n$ is odd (where $A$ is an $n\times n$ matrix), then $A$ is not invertible. Indeed, if it was, then $det(A)^2 = det(A^2) = det(-I) = (-1)^n = -1$. – mathworker21 Oct 27 '17 at 22:32
• In general the accompanying polynomial works – Jorge Fernández Hidalgo Oct 27 '17 at 22:32
• en.wikipedia.org/wiki/Companion_matrix – Jorge Fernández Hidalgo Oct 27 '17 at 22:32
• Thank you Daniel, I opened the book again and saw that it was a 3x3 matrix. – David Oct 27 '17 at 22:34

For odd $n$ we have $\det(A)^3=(-1^n)\det(A)$. Which for nonzero $\det(A)$ gives $\det(A)^2=-1$ which is impossible for a real matrix.
• I think you mean $-1$ at the end, bud – mathworker21 Oct 27 '17 at 22:32