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

MATH307 April 2005

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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

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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. 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 $A^{N}=SD^{N}S^{-1}={\begin{pmatrix}1&2\\1&1\end{pmatrix}}{\begin{pmatrix}1^{N}&0\\0&2^{N}\end{pmatrix}}{\begin{pmatrix}-1&2\\1&-1\end{pmatrix}}={\begin{pmatrix}2^{N+1}-1&2-2^{N+1}\\2^{N}-1&2-2^{N}\end{pmatrix}}.$