## Elementary Linear Algebra 7th Edition

a basis for the nullspace of $A$ consists of the vectors \left[\begin{aligned}-1\\ 1\\1 \\0 \end{aligned}\right], \left[\begin{aligned} 2\\ -2\\0\\1 \end{aligned}\right].
Given the matrix $$A= \left[ \begin {array}{cccc} 1&3&-2&4\\ 0&1&-1&2 \\ -2&-6&4&-8\end {array} \right] .$$ The reduced row echelon form is given by $$\left[ \begin {array}{cccc} 1&0&1&-2\\ 0&1&-1&2 \\ 0&0&0&0\end {array} \right] .$$ The corresponding system is \begin{aligned} x_{1} +x_3-2x_4&=0 \\ x_{2} -x_3+2x_4&=0 \end{aligned}. The solution of the above system is $x_1=-s+2t$,$x_2=s-2t$, $x_3=s$ 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} -s+2t\\ s-2t\\s\\t \end{aligned}\right]=s\left[\begin{aligned}-1\\ 1\\1 \\0 \end{aligned}\right]+t\left[\begin{aligned} 2\\ -2\\0\\1 \end{aligned}\right] . So, a basis for the nullspace of $A$ consists of the vectors \left[\begin{aligned}-1\\ 1\\1 \\0 \end{aligned}\right], \left[\begin{aligned} 2\\ -2\\0\\1 \end{aligned}\right].