Let $k$ be a natural number $\ge$ 3. Find the value of the constants $x$ and $y$ such that $A^k = xA +yI$ (Both x and y depend on k). So the question states that the matrix
$ A = \begin{bmatrix} 4 & -3 \\ 2 & -1 \end{bmatrix}$
characteristic polynomial is $ p(x) = x^2 - 3x + 2.\\$
I am supposed to find x and y when k is natural $\ge 3$ that $A^k = xA + yI$ 
So I tried to solve for both x and y by using the Cayley-Hamilton Theorem for A which you get: 
$p(A) = A^2 - 3A + 2I => A^2 = 3A - 2I$ 
which then I tried to use this to find a pattern among the first few values of k, which lead me to:
$k= 1 => A = A-0I\\ k=2 => A^2 = 3A - 2I \\ k=3 => A^3 = A^2A = (3A-2I)A = 3A^2-2A = 3(3A-2I)-2A = 7A - 6I \\ k=4 => A^4 = 15A - 14I \\ k=5 => A^5 = 31A -30I \\ k=6 => a^6 = (31A -30I)A = 31(3A-2I)-30A = 63A - 62I \\ k=7 => A^7 = 63(3A-2I)-62A = 127A - 126I$
here is where I got lost because I stopped being able to find a pattern. 
Is there an easier way to be able to find the values of x and y in terms of k?
 A: Write $A^k=x_kA+y_kI$. Then $A^{k+1}=x_kA^2+y_kA=x_k(3A-2I)+y_kA=(3x_k+y_k)A-2x_kI$. In other words:
$$\begin{bmatrix}x_{k+1} \\ y_{k+1} \end{bmatrix}=\begin{bmatrix} 3 & 1 \\ -2 & 0 \end{bmatrix} \begin{bmatrix}x_{k} \\ y_k \end{bmatrix}$$
Diagonalizing the matrix above yields 
$$\begin{bmatrix} 3 & 1 \\ -2 & 0 \end{bmatrix}=\begin{bmatrix} -1 & -1 \\ 2 & 1 \end{bmatrix}\begin{bmatrix} 1 & 0 \\ 0 & 2 \end{bmatrix}\begin{bmatrix} 1 & 1 \\ -2 & -1 \end{bmatrix}$$
which implies that
$$\begin{bmatrix}x_{k} \\ y_k \end{bmatrix}=\begin{bmatrix} -1 & -1 \\ 2 & 1 \end{bmatrix}\begin{bmatrix} 1 & 0 \\ 0 & 2^k \end{bmatrix}\begin{bmatrix} 1 & 1 \\ -2 & -1 \end{bmatrix}\begin{bmatrix}x_0 \\ y_0 \end{bmatrix}$$
A: Use the Cayley-Hamilton theorem in a different way. For any polynomials $f(t)$ and $p(t)$, we have $f(t) = p(t)q(t)+r(t)$ where $r$ has degree less than $p$. If $p$ is the characteristic polynomial of $A$, then $p(A)=0$, which means that any polynomial $f(A)$ can be reduced to a polynomial of degree less than the order of $A$. You probably already knew all of this already. What makes this fact of practical use, however, is that the same is true of $A$’s eigenvalues. That is, if $\lambda$ is an eigenvalue of $A$, then $f(\lambda)=r(\lambda)$. If you know $A$’s eigenvalues, you can therefore construct a system of linear equations in the coefficients of $r$. In the case of repeated eigenvalues, these equations are not independent, but you can generate additional independent equations by differentiating.  
For this problem, the equations are of the form $\lambda x+y=\lambda^k$. The eigenvalues of $A$ are $1$ and $2$, producing the system $$x+y=1 \\ 2x+y=2^k.$$ I expect that you can take it from here.
