Let $$A=\left[\begin{array}{cccc} -6 &3 &-6 &6\cr 4 &-2 &4 &-4 \end{array}\right].$$
Find a basis for the kernel of $A$ (or, equivalently, for the linear transformation $T(x)=Ax$).

A basis for the kernel of $A$ is

$\Bigg\lbrace$ $\left[\Rule{0pt}{4.8em}{0pt}\right.$$\left]\Rule{0pt}{4.8em}{0pt}\right.$, $\left[\Rule{0pt}{4.8em}{0pt}\right.$$\left]\Rule{0pt}{4.8em}{0pt}\right.$, $\left[\Rule{0pt}{4.8em}{0pt}\right.$$\left]\Rule{0pt}{4.8em}{0pt}\right.$ $\Bigg\rbrace.$