Question

Please write the answer to all 4 uploading problems including this one on the paper and upload the answer in one pdf file after you finish and submit this final, (10 points) Let A - [12 - :) Find A 100 (Hint: when you're picking the linearly independent eigenvectors to form the matrix P. you could pick the eigerwectors that have all integer entries. e.g. instead of you can pick * []). The computation will be largely simplified if you do so) [1]

Answer 1

Solution:

The characteristic equation is

12 - (sum of diagonal elements) + A = 0

$\small \therefore \lambda ^{2}-\lambda =0$

..(2-1) = 0

= 0,1

For  

11 = 0, A+01 X = 0

$\small \begin{bmatrix} 9 & -6\\ 12 & -8 \end{bmatrix}\begin{bmatrix} x\\ y\\ \end{bmatrix}=\begin{bmatrix} 0\\ 0\\ \end{bmatrix}$

$\small 9x-6y=0\Rightarrow x=\frac{2}{3}y$

$\small \therefore v_{1}=\begin{bmatrix} 2\\ 3\\ \end{bmatrix}$

For  $\small \lambda _{2}=1,\: \: \: \: \left [ A-I \right ]X=0$

$\small \begin{bmatrix} 8 & -6\\ 12 & -9 \end{bmatrix}\begin{bmatrix} x\\ y\\ \end{bmatrix}=\begin{bmatrix} 0\\ 0\\ \end{bmatrix}$

$\small 8x-6y=0\Rightarrow x=\frac{3}{4}y$

$\small \therefore v_{2}=\begin{bmatrix} 3\\ 4\\ \end{bmatrix}$

Since the eigenvalues are distinct, $\small A$ is diagonalisable.

$\small \therefore A=PDP^{-1}$ , where

$\small P=\begin{bmatrix} 2 &3 \\ 3 & 4 \end{bmatrix},\: \: \: \: \: \: \: \: \: \: \: \: D=\begin{bmatrix} 0 &0 \\ 0 & 1 \end{bmatrix}$

$\small \therefore A^{100}=PD^{100}P^{-1}$

$\small \therefore A^{100}=\begin{bmatrix} 2 &3 \\ 3 &4 \end{bmatrix}\begin{bmatrix} 0 & 0\\ 0 & 1 \end{bmatrix}\begin{bmatrix} 2 &3 \\ 3 &4 \end{bmatrix}^{-1}$

$\small \therefore A^{100}=\begin{bmatrix} 9& -6\\ 12& -8\end{bmatrix}$