Science:Math Exam Resources/Courses/MATH152/April 2011/Question B 03 (c)

MATH152 April 2011

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Question B 03 (c)
Let A be a $2\times {2}$ matrix which transforms ${\begin{bmatrix}1\\1\end{bmatrix}}\quad {\text{to}}\quad {\begin{bmatrix}2\\2\end{bmatrix}}$ and transforms ${\begin{bmatrix}1\\-1\end{bmatrix}}\quad {\text{to}}\quad {\begin{bmatrix}0\\0\end{bmatrix}}.$ What is the matrix A?

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 matrix A has four unknowns. We are given 2 vector solutions of Ax=b. How can we turn this into 4 equations for our 4 unknowns?

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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. Let's write our matrix A as $A={\begin{bmatrix}a&b\\c&d\end{bmatrix}}$ with yet to be determined entries. In the problem we know that $A{\begin{bmatrix}1\\-1\end{bmatrix}}={\begin{bmatrix}a&b\\c&d\end{bmatrix}}{\begin{bmatrix}1\\-1\end{bmatrix}}={\begin{bmatrix}0\\0\end{bmatrix}}$ and we expand this out to get ${\begin{aligned}a-b&=0,\\c-d&=0.\end{aligned}}$ Therefore, a=b and c=d. We also have that $A{\begin{bmatrix}1\\1\end{bmatrix}}={\begin{bmatrix}a&b\\c&d\end{bmatrix}}{\begin{bmatrix}1\\1\end{bmatrix}}={\begin{bmatrix}2\\2\end{bmatrix}}$ which we can expand out to get ${\begin{aligned}a+b=2a&=2,\\c+d=2c&=2.\end{aligned}}$ Therefore we conclude that a=b=c=d=1. Therefore our matrix A is $A={\begin{bmatrix}1&1\\1&1\end{bmatrix}}.$ Notice that we can indeed confirm that this matrix has eigenvalues 2 and 0 and is not invertible.