## Elementary Linear Algebra 7th Edition

a basis for the nullspace of $A$ consists of the vectors \left[\begin{aligned}2\\ -1\\1 \\0 \end{aligned}\right], \left[\begin{aligned} -5\\ 1\\0\\1 \end{aligned}\right].
Given the matrix $$\left[ \begin {array}{cccc} 1&4&2&1\\ 0&1&1&-1 \\ -2&-8&-4&-2\end {array} \right] .$$ The reduced row echelon form is given by $$\left[ \begin {array}{cccc} 1&0&-2&5\\ 0&1&1&-1 \\ 0&0&0&0\end {array} \right] .$$ The corresponding system is \begin{aligned} x_{1} -2x_3+5x_4&=0 \\ x_{2} +x_3-x_4&=0 \end{aligned}. The solution of the above system is $x_1=2s-5t$,$x_2=-s+t$, $x_3=s$ and $x_4=t$. This means that the solution space of $Ax = 0$ consists of the vectors on the following form x= \left[\begin{aligned} x_{1}\\ x_{2} \\x_3 \\x_4 \end{aligned}\right]= \left[\begin{aligned} 2s-5t\\-s+t\\s\\t \end{aligned}\right]=s\left[\begin{aligned}2\\ -1\\1 \\0 \end{aligned}\right]+t\left[\begin{aligned} -5\\ 1\\0\\1 \end{aligned}\right] . So, a basis for the nullspace of $A$ consists of the vectors \left[\begin{aligned}2\\ -1\\1 \\0 \end{aligned}\right], \left[\begin{aligned} -5\\ 1\\0\\1 \end{aligned}\right].