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

MATH152 April 2011

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[hide] 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?

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[show] 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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[show] Solution

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.