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

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

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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. 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, $A^{k}=P*D^{k}*P^{-1}$ $={\begin{bmatrix}1&2\\3&-1\end{bmatrix}}{\begin{bmatrix}5^{k}&0\\0&(-2)^{k}\end{bmatrix}}{\begin{bmatrix}1/7&2/7\\3/7&-1/7\end{bmatrix}}.$