Science:Math Exam Resources/Courses/MATH307/April 2005/Question 05

{{#incat:MER QGQ flag|{{#incat:MER QGH flag|{{#incat:MER QGS flag|}}}}}}

MATH307 April 2005

• Q1 (a) • Q1 (b) • Q2 (a) • Q2 (b) • Q2 (c) • Q2 (d) • Q3 (a) • Q3 (b) • Q3 (c) • Q4 (a) • Q4 (b) • Q4 (c) • Q4 (d) • Q5 • Q6 (a) • Q6 (b) • Q7 • Q8 (a) • Q8 (b) • Q8 (c) • Q8 (d) • Q8 (e) • Q8 (f) • Q9 (a) • Q9 (b) • Q9 (c) • Q9 (d) • Q9 (e) •

Other MATH307 Exams

• December 2012 • April 2013 • April 2012 • April 2006 • December 2008 • December 2010 •

Question 05
Find $A^{N}$ for any positive integer $N$ when $A$ is the matrix $A={\begin{pmatrix}3&-2\\1&0\end{pmatrix}}.$

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
Science:Math Exam Resources/Courses/MATH307/April 2005/Question 05/Hint 1

Checking a solution serves two purposes: helping you if, after having used the hint, you still are stuck on the problem; or if you have solved the problem and would like to check your work.

If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work.
If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

Solution
Recall that if $A$ is diagonalisable, then $A^{N}=SD^{N}S^{-1}$ . Thus, we seek the eigenvalues of $A$ to form the diagonal matrix $D$ , and the eigenvectors of $A$ to make $S$ and $S^{-1}$ . To find the eigenvalues, recall that $A\mathbf {x} =\lambda \mathbf {x} \Rightarrow (A-\lambda I)\mathbf {x} =\mathbf {0} .$ Hence we find the characteristic polynomial and set to 0: $\det(A-\lambda I)=\det \left({\begin{pmatrix}3-\lambda &-2\\1&-\lambda \end{pmatrix}}\right)=\lambda ^{2}-3\lambda +2=(\lambda -2)(\lambda -1)=0.$ This implies that we have two eigenvalues as required: $\lambda _{1}=1,\lambda _{2}=2.$ Now we must find the corresponding eigenvectors ${\begin{aligned}\mathbf {x_{1},x_{2}} :\lambda _{1}=1\Rightarrow A-I={\begin{pmatrix}2&-2\\1&-1\end{pmatrix}}\Rightarrow rref(A-I)={\begin{pmatrix}1&-1\\0&0\end{pmatrix}}\\\Rightarrow x_{1}=x_{2}\Rightarrow \mathbf {x_{1}} ={\begin{pmatrix}1\\1\end{pmatrix}}\end{aligned}}$ and ${\begin{aligned}\lambda _{2}=2\Rightarrow A-2I={\begin{pmatrix}1&-2\\1&-2\end{pmatrix}}\Rightarrow rref(A-2I)={\begin{pmatrix}1&-2\\0&0\end{pmatrix}}\\\Rightarrow x_{1}=2x_{2}\Rightarrow \mathbf {x_{2}} ={\begin{pmatrix}2\\1\end{pmatrix}}.\end{aligned}}$ Now we can construct $S,D$ and find $S^{-1}$ where the columns of $S$ are the eigenvectors and the diagonal entries $D$ are the corresponding eigenvalues: ${\begin{aligned}S={\begin{pmatrix}\mathbf {x_{1}} &\mathbf {x_{2}} \end{pmatrix}}={\begin{pmatrix}1&2\\1&1\end{pmatrix}}\\\Rightarrow S^{-1}={\frac {1}{det(S)}}{\begin{pmatrix}1&-2\\-1&1\end{pmatrix}}={\begin{pmatrix}-1&2\\1&-1\end{pmatrix}},\\D={\begin{pmatrix}\lambda _{1}&0\\0&\lambda _{2}\end{pmatrix}}={\begin{pmatrix}1&0\\0&2\end{pmatrix}}.\end{aligned}}$ It follows that $A^{N}=SD^{N}S^{-1}={\begin{pmatrix}1&2\\1&1\end{pmatrix}}{\begin{pmatrix}1^{N}&0\\0&2^{N}\end{pmatrix}}{\begin{pmatrix}-1&2\\1&-1\end{pmatrix}}={\begin{pmatrix}2^{N+1}-1&2-2^{N+1}\\2^{N}-1&2-2^{N}\end{pmatrix}}.$