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Find $A$ ∈ $M_2$ $(\Bbb{R})$ which satisfies $A^4 = I$ but $A^2 ≠ I$.

I had a similar problem to this in my homework that was to find a 2x2 matrix such that $A^2 = -I$ without using any zero entries. That one was relatively easy and I came up with the answer $A = \begin{bmatrix}-1 & -2\\1& 1\end{bmatrix}$. However, I am having some trouble with this problem. Any help is appreciated.

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  • $\begingroup$ @Batman How'd that be possible? $det(A)\ne0$, $A$ is invertible $\endgroup$ – Ottavio Bartenor Nov 14 '15 at 22:49
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Your $A = \begin{bmatrix}-1 & -2\\1& 1\end{bmatrix}$ matrix is good as well. If $A^2=-I\ne I$, $A^4=I$

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Try with : $$A=\begin{pmatrix}0&-1\\1&0\end{pmatrix},$$ the rotation of angle $\frac{\pi}{2}.$

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