Science:Math Exam Resources/Courses/MATH221/December 2007/Question 01

MATH221 December 2007

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Question 01
Find a basis for the null space of the matrix $A={\begin{pmatrix}2&-4&1&t\\1&-2&2&t\\1&-2&1&2t\\1&-2&1&t\end{pmatrix}}.$ Your answer will depend on t.

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. To determine a basis for the null space of A, let $x={\begin{pmatrix}x_{1}\\x_{2}\\x_{3}\\x_{4}\end{pmatrix}}$ and suppose Ax = 0. We want to solve this equation for x. To do so, we append the zero vector to A ${\begin{pmatrix}2&-4&1&t&|&0\\1&-2&2&t&|&0\\1&-2&1&2t&|&0\\1&-2&1&t&|&0\end{pmatrix}}$ and row reduce A in the standard way. Leaving aside these details (exercise), the end result is ${\begin{pmatrix}1&-2&0&0&|&0\\0&0&1&0&|&0\\0&0&0&t&|&0\\0&0&0&0&|&0\end{pmatrix}}$ Therefore we deduce that x belongs to the null space of A if and only if ${\begin{aligned}x_{1}&=2x_{2}\\x_{3}&=0\\tx_{4}&=0\end{aligned}}$. Case 1: $t\neq 0$. Then $x_{4}=0$, so that $x=x_{2}{\begin{pmatrix}2\\1\\0\\0\end{pmatrix}}$ where $x_{2}\in \mathbb {R}$ is a free parameter. In this case we therefore conclude that $\left\{{\begin{pmatrix}2\\1\\0\\0\end{pmatrix}}\right\}$ is a basis for the null space of A. Case 2: $t=0$. Then there is no restriction on $x_{4}$, and we have $x={\begin{pmatrix}2x_{2}\\x_{2}\\0\\x_{4}\end{pmatrix}}=x_{2}{\begin{pmatrix}2\\1\\0\\0\end{pmatrix}}+x_{4}{\begin{pmatrix}0\\0\\0\\1\end{pmatrix}}$ where $x_{2},x_{4}\in \mathbb {R}$. In this case, we conclude that $\left\{{\begin{pmatrix}2\\1\\0\\0\end{pmatrix}},{\begin{pmatrix}0\\0\\0\\1\end{pmatrix}}\right\}$ is a basis for the null space of A.

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