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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.

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

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  • $\begingroup$ I think you mean $-1$ at the end, bud $\endgroup$ – mathworker21 Oct 27 '17 at 22:32
  • $\begingroup$ Also I think you mean "noninvertible A", bud $\endgroup$ – mathworker21 Oct 27 '17 at 22:33

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