Science:Math Exam Resources/Courses/MATH221/December 2009/Question 08/Solution 1

We can first let $A=\left[\begin{array}{cc}17& 9\\ -30& -16\end{array}\right]$ such that $A$ solves the equation$\left[\begin{array}{c}{x}_{n+1}\\ y_{n+1}\end{array}\right]=A\left[\begin{array}{c}{x}_n\\ y_n\end{array}\right]$ In this problem, we are asked to find $x_{30}$ and $y_{30}$ based on $x_0$ and $y_0$. Since

$\left[\begin{array}{c}{x}_1\\ y_1\end{array}\right]=A\left[\begin{array}{c}{x}_0\\ y_0\end{array}\right]$,

we can write

$\left[\begin{array}{c}{x}_{30}\\ y_{30}\end{array}\right]=A^{30}\left[\begin{array}{c}{x}_0\\ y_0\end{array}\right]$

Now we need to solve for $A^{30}$ This would be easier if we do the diagonalization $A=PD{P}^{-1}$ First we need to find the eigenvalues of $A$ Let

$\text{det}\left[\begin{array}{cc}17-\lambda & 9\\ -30& -16-\lambda \end{array}\right]=(17-\lambda )(-16-\lambda )+270=0$

Solving the equation above and we get ${\lambda }_1=2$ and ${\lambda }_2=-1$ Then we need to find the corresponding eigenvectors When $\lambda =2$, we have

$\left[\begin{array}{cc}17-2& 9\\ -30& -16-2\end{array}\right]v=0$

Solve this to get

$v_1=\left[\begin{array}{c}3\\ -5\end{array}\right]$

When $\lambda =-1$, we have

$\left[\begin{array}{cc}17+1& 9\\ -30& -16+1\end{array}\right]v=0$

Solve this to get

$v_2=\left[\begin{array}{c}1\\ -2\end{array}\right]$

So now we can write the diagonal matrix

$D=\left[\begin{array}{cc}{\lambda }_1& 0\\ 0& {\lambda }_2\end{array}\right]=\left[\begin{array}{cc}2& 0\\ 0& -1\end{array}\right]$

and matrix

$P=\left[\begin{array}{cc}{v}_1& v_2\end{array}\right]=\left[\begin{array}{cc}3& 1\\ -5& -2\end{array}\right]$, $P^{-1}=\left[\begin{array}{cc}2& 1\\ -5& -3\end{array}\right]$

So finally,

$\begin{array}{rl}{A}^{30}& =P{D}^{30}{P}^{-1}\\ & =\left[\begin{array}{cc}3& 1\\ -5& -2\end{array}\right]\left[\begin{array}{cc}{2}^{30}& 0\\ 0& (-1{)}^{30}\end{array}\right]\left[\begin{array}{cc}2& 1\\ -5& -3\end{array}\right]\\ & =\left[\begin{array}{cc}6\cdot 2^{30}-5& 3\cdot 2^{30}-3\\ -10\cdot 2^{30}+10& -5\cdot 2^{30}+6\end{array}\right]\end{array}$

Thus

$\begin{array}{rl}\left[\begin{array}{c}{x}_{30}\\ y_{30}\end{array}\right]& =A^{30}\left[\begin{array}{c}{x}_0\\ y_0\end{array}\right]\\ & =\left[\begin{array}{cc}6\cdot 2^{30}-5& 3\cdot 2^{30}-3\\ -10\cdot 2^{30}+10& -5\cdot 2^{30}+6\end{array}\right]\left[\begin{array}{c}1\\ -1\end{array}\right]\\ & =\left[\begin{array}{c}3\cdot 2^{30}-2\\ -5\cdot 2^{30}+4\end{array}\right]\end{array}$

So $x_{30}=3\cdot 2^{30}-2,y_{30}=-5\cdot 2^{30}+4$