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.

Any element of the set $\left\{\lambda\right\}$ of $A$ eigenvalues satisfy the equations
$$
\lambda^{2} = \left(a - 2\right)\lambda + 2\left(a - 1\right)\,,
\quad
\lambda^{3} = 2\lambda\,,
\quad
\sum_{\lambda}\lambda = a - 2
$$
Then, $\sum_{\lambda}\lambda^{3} = 2\left(a - 2\right)$. Also,
\begin{align}
\sum_{\lambda}\lambda^{3}
&=
\sum_{\lambda}\lambda\left[\left(a - 2\right)\lambda + 2\left(a - 1\right)\right]
=
\left(a - 2\right)\sum_{\lambda}\lambda^{2}
+
2\left(a - 1\right)\left(a - 2\right)
\\[3mm]&=
\left(a - 2\right)^{3} + 6\left(a - 1\right)\left(a - 2\right)
\end{align}
We get
$$
\left(a - 2\right)\left[\left(a - 2\right)^{2} + 6a - 8\right] = 0
$$
$$
\color{#ff0000}{\large%
a = 2\,,\qquad\qquad a = -1 \pm \sqrt{5\,}}
$$
