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

MATH221 December 2009

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Question 08
A discrete dynamical system is described by $\displaystyle x_{n+1}=17x_{n}+9y_{n}$ $\displaystyle y_{n+1}=-30x_{n}-16y_{n}$ Given that $\displaystyle x_{0}=1$, $\displaystyle y_{0}=-1$, find $\displaystyle x_{30}$, $\displaystyle y_{30}$.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. We can first let $A={\begin{bmatrix}17&9\\-30&-16\end{bmatrix}}$ such that $A$ solves the equation${\begin{bmatrix}x_{n+1}\\y_{n+1}\end{bmatrix}}=A{\begin{bmatrix}x_{n}\\y_{n}\end{bmatrix}}$ In this problem, we are asked to find $x_{30}$ and $y_{30}$ based on $x_{0}$ and $y_{0}$. Since ${\begin{bmatrix}x_{1}\\y_{1}\end{bmatrix}}=A{\begin{bmatrix}x_{0}\\y_{0}\end{bmatrix}}$, we can write ${\begin{bmatrix}x_{30}\\y_{30}\end{bmatrix}}=A^{30}{\begin{bmatrix}x_{0}\\y_{0}\end{bmatrix}}$ Now we need to solve for $A^{30}$ This would be easier if we do the diagonalization $A=PDP^{-1}$ First we need to find the eigenvalues of $A$ Let ${\text{det}}{\begin{bmatrix}17-\lambda &9\\-30&-16-\lambda \end{bmatrix}}=(17-\lambda )(-16-\lambda )+270=0$ 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 ${\begin{bmatrix}17-2&9\\-30&-16-2\end{bmatrix}}v=0$ Solve this to get $v_{1}={\begin{bmatrix}3\\-5\end{bmatrix}}$ When $\lambda =-1$, we have ${\begin{bmatrix}17+1&9\\-30&-16+1\end{bmatrix}}v=0$ Solve this to get $v_{2}={\begin{bmatrix}1\\-2\end{bmatrix}}$ So now we can write the diagonal matrix $D={\begin{bmatrix}\lambda _{1}&0\\0&\lambda _{2}\end{bmatrix}}={\begin{bmatrix}2&0\\0&-1\end{bmatrix}}$ and matrix $P={\begin{bmatrix}v_{1}&v_{2}\end{bmatrix}}={\begin{bmatrix}3&1\\-5&-2\end{bmatrix}}$, $P^{-1}={\begin{bmatrix}2&1\\-5&-3\end{bmatrix}}$ So finally, ${\begin{aligned}A^{30}&=PD^{30}P^{-1}\\&={\begin{bmatrix}3&1\\-5&-2\end{bmatrix}}{\begin{bmatrix}2^{30}&0\\0&(-1)^{30}\end{bmatrix}}{\begin{bmatrix}2&1\\-5&-3\end{bmatrix}}\\&={\begin{bmatrix}6\cdot 2^{30}-5&3\cdot 2^{30}-3\\-10\cdot 2^{30}+10&-5\cdot 2^{30}+6\end{bmatrix}}\end{aligned}}$ Thus ${\begin{aligned}{\begin{bmatrix}x_{30}\\y_{30}\end{bmatrix}}&=A^{30}{\begin{bmatrix}x_{0}\\y_{0}\end{bmatrix}}\\&={\begin{bmatrix}6\cdot 2^{30}-5&3\cdot 2^{30}-3\\-10\cdot 2^{30}+10&-5\cdot 2^{30}+6\end{bmatrix}}{\begin{bmatrix}1\\-1\end{bmatrix}}\\&={\begin{bmatrix}3\cdot 2^{30}-2\\-5\cdot 2^{30}+4\end{bmatrix}}\end{aligned}}$ So $x_{30}=3\cdot 2^{30}-2,y_{30}=-5\cdot 2^{30}+4$

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