Let $$A = \left[\begin{array}{cc} -6 &0\cr 0 &1 \end{array}\right].$$ Define the linear transformation $T: {\mathbb R}^2 \rightarrow {\mathbb R}^2$ by $T(\vec{x}) = A\vec{x}$. Find the images of $\vec{u} = \left[\begin{array}{c} -3\cr 3 \end{array}\right]$ and $\vec{v} =\left[\begin{array}{c} a\cr b\cr \end{array}\right]$ under $T$.

$T(\vec{u}) =$ $\left[\Rule{0pt}{2.4em}{0pt}\right.$$\left]\Rule{0pt}{2.4em}{0pt}\right.$

$T(\vec{v}) =$ $\left[\Rule{0pt}{2.4em}{0pt}\right.$$\left]\Rule{0pt}{2.4em}{0pt}\right.$