Matrix Exponentiation $A^{15}$ A is a 2x2 matrix. Let it satisfy $A^2 = A - I$, where $I = \begin{bmatrix} 1 & 0\\ 0 & 1\end{bmatrix}$. I want to find $A^{15}$:
My Approach:
I isolated $A$:
$A^2-A=-I$
$A(A-I)=-I$
From that, I just tried to solve the system that would generate:
$\begin{bmatrix}a & b \\ c & d\end{bmatrix}\begin{bmatrix}a-1 & b \\ c & d-1\end{bmatrix} = \begin{bmatrix}-1 & 0 \\ 0 & -1\end{bmatrix}$
Turns out this doesn't help me get closer to the answer as it leads to:
$\begin{bmatrix}a(a-1)+bc & b(d-1)+ab\\c\left(a-1\right)+dc &d(d-1)+bc\end{bmatrix} = \begin{bmatrix}-1 & 0 \\ 0 & -1\end{bmatrix}$
Is that the correct approach, or else, which property am I missing in order to solve this problem?

We can show immediately that $A^3=-I$, so $A^{15}=(A^3)^5=(-I)^5=-I$, for instance:
$$
\begin{aligned}
A^3+I
&=(A+I)\underbrace{(A^2-A+I)}_{=0}\\
&=0\ .
\\&\qquad\qquad\text{Alternatively:}
\\
A^{15}+I
&=(A^3+I)(A^{12}-A^9+A^6-A^3+I)\\
&=(A+I)\underbrace{(A^2-A+I)}_{=0}(A^{12}-A^9+A^6-A^3+I)\\
&=0\ .
\end{aligned}
$$
Or one can compute the eigenvalues $\lambda1,\lambda_2$, of $A$, see that $\lambda_1^3=\lambda_2^3=-1$, are the eigenvalues of $A^3$, so $A^3=-I$, etc.
