Science:Math Exam Resources/Courses/MATH221/December 2007/Question 03

MATH221 December 2007

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Question 03
Find a 2 x 2 matrix A such that ${\begin{pmatrix}1&2\\0&1\end{pmatrix}}A{\begin{pmatrix}1&3\\0&1\end{pmatrix}}={\begin{pmatrix}1&1\\1&1\end{pmatrix}}.$

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. Letting $B={\begin{pmatrix}1&2\\0&1\end{pmatrix}}$ and $C={\begin{pmatrix}1&3\\0&1\end{pmatrix}}$, the task at hand is to solve $BAC={\begin{pmatrix}1&1\\1&1\end{pmatrix}}$ for A. Now B and C are invertible matrices (as they have have non-zero determinant), so we can solve for A by multiplying the above equation on the left by $B^{-1}$, and on the right by $C^{-1}$, giving $A=B^{-1}{\begin{pmatrix}1&1\\1&1\end{pmatrix}}C^{-1}.$ It is straightforward to compute $B^{-1}={\begin{pmatrix}1&-2\\0&1\end{pmatrix}}$ and $C^{-1}={\begin{pmatrix}1&-3\\0&1\end{pmatrix}}$. Therefore $A={\begin{pmatrix}1&-2\\0&1\end{pmatrix}}{\begin{pmatrix}1&1\\1&1\end{pmatrix}}{\begin{pmatrix}1&-3\\0&1\end{pmatrix}}={\begin{pmatrix}-1&2\\1&-2\end{pmatrix}}.$

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