Finding A solution to a matrix equation I have the following problem in a past exam that involves matrix equations:
Suppose that $X$is a $2 \times2$ matrix satisfying:
$X^{2} = 6X +I$
a) Find the values of $\alpha$ and $\beta$ such that $X^{3} = \alpha X + \beta I$
b) Hence find a solution of the matrix equation 
$X^{3} = \begin{bmatrix} 36 &-1\\ 2&35\end{bmatrix} X + \begin{bmatrix} 9 &8\\ 2&10\end{bmatrix}$
I need a hint for part (b) as I am unable to use the result for part (a). 
For part a the values for $\alpha $ and $\beta$ are 37 and 6. In particular, subbing in the values I have found in part (a), gives me the relation
$X^{3} = 37X + 6I$, however I fail to see how I can apply this to part b, as there is a matrix instead of a real number

Suppose $X=\begin{bmatrix} a & b \\ c &d \end{bmatrix}$. 
It is given that $X^{3} = \begin{bmatrix} 36 &-1\\ 2&35\end{bmatrix} X + \begin{bmatrix} 9 &8\\ 2&10\end{bmatrix}$.
Therefore,$\begin{bmatrix} 36 &-1\\ 2&35\end{bmatrix} \times \begin{bmatrix} a & b \\ c &d \end{bmatrix}+ \begin{bmatrix} 9 &8\\ 2&10\end{bmatrix} = \begin{bmatrix} 36a-c+9 & 36b-d+8 \\ 2a+35c+2 & 2b+35d+10 \end{bmatrix}$. 
Since, $X^3=37X+6I$, we have $X^3= \begin{bmatrix} 37a+6 &37b\\ 37c&37d+6\end{bmatrix}$
Comparing both $X^3$'s. [4 variables and 4 distinct eqations]
We get, $37a+6-36a+c-9=0\Rightarrow \boxed{c+a=3}$,
$37b-36b+d-8=0\Rightarrow \boxed{d+b=8}$,
$37c-2a-35c-2=0\Rightarrow \boxed{c-a=1}$, and
$37d+6-2b-35d-10=0\Rightarrow \boxed{d-b=2}$.
On solving these simultaneous equations we get $\bf{\boxed{a=1,b=3,c=2,d=5}}$.
Therefore, our matrix $X$ is $\begin{bmatrix} 1 & 3 \\ 2&5\end{bmatrix}$.
