Science:Math Exam Resources/Courses/MATH152/April 2012/Question 04 (c)

MATH152 April 2012

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Question 04 (c)
Consider the matrix ${\begin{aligned}A={\begin{bmatrix}3&2\\2&3\end{bmatrix}}.\end{aligned}}$ Use the information in items (a) and (b) to write A as $MDM^{-1}$, where M is an invertible matrix and D is a diagonal matrix.

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
The columns of M correspond to the eigenvectors and the values of D are the eigenvalues in order of the eigenvectors in the matrix M.

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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. The matrix M is made up of the eigenvectors of A and so $M={\begin{bmatrix}1&1\\-1&1\end{bmatrix}}.$ The matrix D is a diagonal matrix of the eigenvalues listed in the order that we list the eigenvectors in M. Therefore, $D={\begin{bmatrix}1&0\\0&5\end{bmatrix}}.$ We can also compute $M^{-1}$ by recalling that for a 2x2 matrix Q such that $Q={\begin{bmatrix}a&b\\c&d\end{bmatrix}}$ that $Q^{-1}={\frac {1}{\det {Q}}}{\begin{bmatrix}d&-b\\-c&a\end{bmatrix}}.$ Since det(M)=2, we have that $M^{-1}={\frac {1}{2}}{\begin{bmatrix}1&-1\\1&1\end{bmatrix}}={\begin{bmatrix}1/2&-1/2\\1/2&1/2\end{bmatrix}}.$ Multiplying everything together we get $MDM^{-1}={\begin{bmatrix}1&1\\-1&1\end{bmatrix}}{\begin{bmatrix}1&0\\0&5\end{bmatrix}}{\begin{bmatrix}1/2&-1/2\\1/2&1/2\end{bmatrix}}={\begin{bmatrix}3&2\\2&3\end{bmatrix}}$ as we were expecting.

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