Recursive formula, eigenvalue problem A sequence $ \{X_i \}_{i \geq 0} $ is defined recursive by $ X_{n+1} = 3 X_{n} - 2X_{n-1}, \qquad n \geq1. $ 
and $ X_0 = \begin{pmatrix}
1 & 0  \newline
1 & 1 
\end{pmatrix},X_1 = \begin{pmatrix}
1 & 1  \newline
1 & 1 
\end{pmatrix}. $ Find an explicit formula for $ X_n $.
My attempt;
We can write $  = \begin{pmatrix}
X_{n+1}   \newline X_{n} \end{pmatrix} = \begin{pmatrix} 3 & -2  \newline 1 & 0 \end{pmatrix} \begin{pmatrix} X_{n} \newline X_{n-1} \end{pmatrix} $ for $ n \geq 1. $ Diagonalizing the matrix yields 
$\begin{pmatrix} X_{n+1}   \newline X_{n} \end{pmatrix} = \begin{pmatrix} 2 & 1  \newline 1 & 1 \end{pmatrix} \begin{pmatrix} 2^n & 0  \newline 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & -1  \newline -1 & 2 \end{pmatrix} \begin{pmatrix} X_{n} \newline X_{n-1} \end{pmatrix}$
This is and old exam; now the answer concludes that $X_n $ can be written as $ X_n = A + 2^n B $, for some matrices A and  B, which can be found by plugging in $ X_1 $ and $ X_0.$ My question is how did he draw this conclusion, is it only from the eigenvalues? How would one approach questions on this form if the matrix couldnt be diagonalized?

Note that $X_{n+1}-2 X_n = X_n - 2 X_{n-1} = \Delta = X_1 - 2 X_0$.
Hence $X_{n+1} = 2 X_n + \Delta$.
The general solution is
$X_n = 2^n X_0 + \sum_{k=0}^{n-1} 2^{n-k-1} \Delta = 2^n X_0+(2^{n}-1)\Delta = 2^n(X_1-X_0)-X_1+2X_0$.
