WeBWorK Standalone Renderer

Let $A=\left[\begin{array}{ccc} 6 &-8 &0\cr -2 &6 &0\cr -2 &6 &0\cr 6 &-8 &0 \end{array}\right].$
Find a basis for the image of $A$ (or, equivalently, for the linear transformation $T(x)=Ax$ ).

A basis for the image of $A$ is $\Bigg\lbrace$ $\left[\Rule{0pt}{4.8em}{0pt}\right.$ $\left]\Rule{0pt}{4.8em}{0pt}\right.$ , $\left[\Rule{0pt}{4.8em}{0pt}\right.$ $\left]\Rule{0pt}{4.8em}{0pt}\right.$ $\Bigg\rbrace.$