Recurrence relation involving matrices 
I've done part $(i)-(iv) [(i) (C), (ii) (D),(iii) (F), (iv) (E)].$
I would appreciate if someone could show me how to solve this part (v). 
 A: If $M = P D P^{-1}$, then $
X_n = M^n X_0 = P D^n P^{-1} X_0
$.
A: Find $P$ so that $P^{-1}MP$ is a diagonal matrix $D$. Then $D^n=P^{-1}M^nP$, so 
$$X_n=M^nX_0=PD^nP^{-1}X_0\;,$$
which is easy to calculate.
A: By multiplying the matrices, we get the following recurrence relation:
$$\tag1x_{n+1}=2x_n+y_n$$
$$\tag2y_{n+1}=x_n+2y_n$$
Multiplying $(2)$ by $2$ and subtracting $(1)$,
$$2y_{n+1}-x_{n+1}=3y_n$$
Replacing $n$ by $n-1$ (which means we are now assuming $n\gt1$,
$$2y_n-x_n=3y_{n-1}$$
$$\tag3x_n=2y_n-3y_{n-1}$$
Substituting $(3)$ in $(2)$,
$$y_{n+1}=2y_n-3y_{n-1}+2y_n$$
$$\tag4y_{n+1}=4y_n-3y_{n-1}$$
For $n=1$,
$$x_1=2x_0+y_0=4$$
$$y_1=x_0+2y_0=2$$
Thus, $(4)$ is a linear homogeneous recurrence relation, and knowing $y_0=2,y_1=2$, it can be easily solved using the characteristic equation.
Similarly, multiplying $(1)$ by $2$ and subtracting $(2)$ will lead to a linear homogeneous recurrence relation for $x_n$.
A: The recurrence relation can be solved by
$$\mathbf{X}_n = \mathbf{M}^n \mathbf{X}_0$$
So, you have to compute the diagonlization of $\mathbf{M}$
$$\mathbf{M} = \mathbf{P}^{-1} \left( \begin{matrix} 1 & 0 \\ 0 & 3\end{matrix} \right)\mathbf{P} $$
where $$\mathbf{P} = \left( \begin{matrix} 1 & 1 \\ -1 & 1\end{matrix} \right) , \mathbf{P}^{-1} = \frac 12 \left( \begin{matrix} 1 & -1 \\ 1 & 1\end{matrix} \right)$$
So that
$$\mathbf{X}_n = \mathbf{P}^{-1} \left( \begin{matrix} 1 & 0 \\ 0 & 3^n\end{matrix} \right)\mathbf{P} \left( \begin{matrix} 2 \\ 0 \end{matrix} \right)$$
