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

MATH221 December 2007

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[hide] 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.

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[show] Hint

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[show] Solution

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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