Science:Math Exam Resources/Courses/MATH307/April 2010/Question 04 (a)

MATH307 April 2010

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Question 04 (a)
Consider the following recursion relation: $x_{n+1}=ax_{n}+bx_{n-1}$ where $n\geq 1$, and $a,b$ $\epsilon$ $\mathbb {R}$ Find the matrix $A_{n}$ so that for all n, we have ${\begin{bmatrix}x_{n+1}\\x_{n}\end{bmatrix}}=A{\begin{bmatrix}x_{1}\\x_{0}\end{bmatrix}}$

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. It’s relatively simple to find the matrix we want. Using the relation given above and the knowledge that $x_{n}=x_{n}$, we can see that the relation can be defined by the matrix equation: ${\begin{bmatrix}x_{n+1}\\x_{n}\end{bmatrix}}={\begin{bmatrix}a&b\\1&0\end{bmatrix}}{\begin{bmatrix}x_{n}\\x_{n-1}\end{bmatrix}}$ but we want to define the $n^{th}$ term by a combination of $x_{1}$ and $x_{0}$ by having a matrix $A_{n}$ We know what $A_{1}$ is, $A_{1}={\begin{bmatrix}a&b\\1&0\end{bmatrix}}$ So let’s use that knowledge and continue: When $n$ = $1$ ${\begin{bmatrix}x_{2}\\x_{1}\end{bmatrix}}={\begin{bmatrix}a&b\\1&0\end{bmatrix}}{\begin{bmatrix}x_{1}\\x_{0}\end{bmatrix}}$ and when $n$ = $2$ ${\begin{bmatrix}x_{3}\\x_{2}\end{bmatrix}}={\begin{bmatrix}a&b\\1&0\end{bmatrix}}{\begin{bmatrix}x_{2}\\x_{1}\end{bmatrix}}$ but we know what ${\begin{bmatrix}x_{2}\\x_{1}\end{bmatrix}}$ is, it’s merely ${\begin{bmatrix}a&b\\1&0\end{bmatrix}}{\begin{bmatrix}x_{1}\\x_{0}\end{bmatrix}}$ so ${\begin{bmatrix}x_{3}\\x_{2}\end{bmatrix}}={\begin{bmatrix}a&b\\1&0\end{bmatrix}}{\begin{bmatrix}a&b\\1&0\end{bmatrix}}{\begin{bmatrix}x_{1}\\x_{0}\end{bmatrix}}$ or ${\begin{bmatrix}x_{3}\\x_{2}\end{bmatrix}}={\begin{bmatrix}a&b\\1&0\end{bmatrix}}^{2}{\begin{bmatrix}x_{1}\\x_{0}\end{bmatrix}}$ continuing along the trend above, we can see that, in general, $A_{n}={\begin{bmatrix}a&b\\1&0\end{bmatrix}}^{n}$