## Elementary Linear Algebra 7th Edition

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