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

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MATH221 April 2009

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Question 08
Find a formula for $A^{k}$, where $A={\begin{bmatrix}-1&2\\3&4\end{bmatrix}}.$ You may leave your final answer as a product of three matrices.

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

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

To find a formula for $A^{k}$, we first need to find matrices P and D such that $A=P*D*P^{-1}$, where D is a diagonal matrix, whose diagonal entries are eigenvalues of A, and their corresponding eigenvectors are the columns of the matrix P. Finding Eigenvalues: $(A-I\lambda )={\begin{bmatrix}-1-\lambda &2\\3&4-\lambda \end{bmatrix}}\Rightarrow (-1-\lambda )(4-\lambda )-(3*2)=\lambda ^{2}-3\lambda -10$ $=(\lambda -5)(\lambda +2))\Rightarrow \lambda _{1}=5,\lambda _{2}=-2$ Finding Eigenvectors: $A{\vec {v}}=\lambda {\vec {v}}\Rightarrow (A-I\lambda ){\vec {v}}={\vec {0}}$ $(A-I\lambda _{1}){\vec {v_{1}}}={\begin{bmatrix}-6&2\\3&-1\end{bmatrix}}{\vec {v_{1}}}={\vec {0}}\Rightarrow 3v_{11}=v_{21}\Rightarrow {\vec {v_{1}}}={\begin{bmatrix}1\\3\end{bmatrix}}$ $(A-I\lambda _{2}){\vec {v_{2}}}={\begin{bmatrix}1&2\\3&6\end{bmatrix}}{\vec {v_{2}}}={\vec {0}}\Rightarrow -v_{12}=2v_{22}\Rightarrow {\vec {v_{2}}}={\begin{bmatrix}2\\-1\end{bmatrix}}$ So we have $D={\begin{bmatrix}\lambda _{1}&0\\0&\lambda _{2}\end{bmatrix}}={\begin{bmatrix}5&0\\0&-2\end{bmatrix}}$, $P={\begin{bmatrix}{\vec {v_{1}}}&{\vec {v_{2}}}\end{bmatrix}}={\begin{bmatrix}1&2\\3&-1\end{bmatrix}}$, and $P^{-1}={\frac {\begin{bmatrix}-1&-2\\-3&1\end{bmatrix}}{(-1)*1-2*3}}={\begin{bmatrix}1/7&2/7\\3/7&-1/7\end{bmatrix}}$ Since $P*P^{-1}=I,A^{k}=(P*D*P^{-1})^{k}$ $=(P*D*P^{-1})*(P*D*P^{-1})*...*(P*D*P^{-1})*(P*D*P^{-1})$ $=P*D*(P^{-1}*P)*D*(P^{-1}*P)*D*...*(P^{-1}P)*D*P^{-1}$ $=P*D^{k}*P^{-1}.$ Hence, $A^{k}=P*D^{k}*P^{-1}$ $={\begin{bmatrix}1&2\\3&-1\end{bmatrix}}{\begin{bmatrix}5^{k}&0\\0&(-2)^{k}\end{bmatrix}}{\begin{bmatrix}1/7&2/7\\3/7&-1/7\end{bmatrix}}.$

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