<s>[INST] Matrix Algebra Question (Linear Algebra) Find all values of $a$ such that $A^3 = 2A$, where
$$A = \begin{bmatrix} -2 & 2 \\ -1 & a \end{bmatrix}.$$
The matrix I got for $A^3$ at the end didn't match up, but I probably made a multiplication mistake somewhere.
 [/INST] Here's a different approach that avoids matrix multiplication in favor of determinants:
We want to find $a$ such that $A^3=2A$. This is equivalent to $A^3-2A=0$, and to $(A^2-2I)A=0$.
Thus either $A=0$, which is impossible, or $A^2-2I$ is a zero divisor.
Thus \begin{align}0&=\det (A^2-2I) \\ &= \det(A+\sqrt2I)\det(A-\sqrt2 I),\end{align} so one of these determinants must be $0$. That is, $(-2+\sqrt 2)(a+\sqrt 2)+2=0$ or $(-2-\sqrt 2)(a-\sqrt 2)+2=0$. These have the same sole solution: $a=2$.
Note that we get the value of $A^2$ too: plugging in $a=2$ gives $\det A = -2\ne 0$, so $A$ is not a zero divisor, so $A^2-2I=0$, so $A^2=2I$.
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