Pls long solution hard

We're going to consider the matrix

$\begin{pmatrix} -4 & -15 \\ 2 & 7 \end{pmatrix}$

a)
Let ${P} = \begin{pmatrix} 3 & -5 \\ -1 & 2 \end{pmatrix}$.
Find the 2x2 matrix D such that

${P} {D} {P}^{-1} = \begin{pmatrix} -4 & -15 \\ 2 & 7 \end{pmatrix}$

$\text{Let $M = \begin{pmatrix} -4 & -15 \\ 2 & 7 \end{pmatrix} $ }$

Formula:

$\begin{array}{|rcll|} \hline P^{-1} &=& \dfrac{1}{\begin{vmatrix} 3 & -5 \\ -1 & 2 \end{vmatrix}}\begin{pmatrix} 2 & 5 \\ 1 & 3 \end{pmatrix} \\\\ &=& \dfrac{1}{6-5}\begin{pmatrix} 2 & 5 \\ 1 & 3 \end{pmatrix} \\\\ &=& \dfrac{1}{1}\begin{pmatrix} 2 & 5 \\ 1 & 3 \end{pmatrix} \\\\ \mathbf{P^{-1}} & \mathbf{=} & \mathbf{\begin{pmatrix} 2 & 5 \\ 1 & 3 \end{pmatrix}} \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline PDP^{-1} &=& M \quad | \quad \cdot P \\ PDP^{-1}P &=& MP \quad | \quad P^{-1}P = I(\text{Identity matrix}) \\ PDI &=& MP \quad | \quad \cdot P^{-1} \\ P^{-1}PDI &=& P^{-1}MP \quad | \quad P^{-1}P = I(\text{Identity matrix}) \\ IDI &=& P^{-1}MP \\ \mathbf{D} & \mathbf{=} & \mathbf{P^{-1}MP} \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline \mathbf{D} & \mathbf{=} & \mathbf{P^{-1}MP} \\ &=& \begin{pmatrix} 2 & 5 \\ 1 & 3 \end{pmatrix}\begin{pmatrix} -4 & -15 \\ 2 & 7 \end{pmatrix}\begin{pmatrix} 3 & -5 \\ -1 & 2 \end{pmatrix} \\ &=& \begin{pmatrix} 2 & 5 \\ 2 & 6 \end{pmatrix}\begin{pmatrix} 3 & -5 \\ -1 & 2 \end{pmatrix} \\ \mathbf{D} & \mathbf{=} & \mathbf{\begin{pmatrix} 1 & 0 \\ 0 & 2 \end{pmatrix}} \quad | \quad (\text{Diagonal matrix}) \\ \hline \end{array}$

(b)

Find a formula for  $\mathbf{D^n}$ where D is the matrix you found in part (a).
(You don't need to prove your answer, but explain how you found it.)

$\begin{array}{|rcll|} \hline D &=& \begin{pmatrix} 1 & 0 \\ 0 & 2 \end{pmatrix} \\\\ D^2 &=& \begin{pmatrix} 1 & 0 \\ 0 & 2 \end{pmatrix}\begin{pmatrix} 1 & 0 \\ 0 & 2 \end{pmatrix} \\ &=& \begin{pmatrix} 1^2 & 0 \\ 0 & 2^2 \end{pmatrix} \\\\ D^3 &=& \begin{pmatrix} 1^2 & 0 \\ 0 & 2^2 \end{pmatrix}\begin{pmatrix} 1 & 0 \\ 0 & 2 \end{pmatrix} \\ D^3 &=& \begin{pmatrix} 1^3 & 0 \\ 0 & 2^3 \end{pmatrix} \\ \ldots \\ \mathbf{D^n} & \mathbf{=} & \mathbf{\begin{pmatrix} 1^n & 0 \\ 0 & 2^n \end{pmatrix}} \\ \hline \end{array}$

c)
Using parts (a) and (b), find a formula for

$\begin{pmatrix} -4 & -15 \\ 2 & 7 \end{pmatrix}^n$

$\begin{array}{|rcll|} \hline \mathbf{\begin{pmatrix} -4 & -15 \\ 2 & 7 \end{pmatrix}^n} &=& PD^nP^{-1} \\ &=& \begin{pmatrix} 3 & -5 \\ -1 & 2 \end{pmatrix}\begin{pmatrix} 1^n & 0 \\ 0 & 2^n \end{pmatrix}\begin{pmatrix} 2 & 5 \\ 1 & 3 \end{pmatrix} \\\\ &=& \begin{pmatrix} \mathbf{6-5\cdot 2^n} & \mathbf{15(1-2^n)} \\ \mathbf{2(2^n-1)} & \mathbf{6\cdot 2^n-5} \end{pmatrix} \\ \hline \end{array}$