If $$A = \left[\begin{array}{cc} -5 &-6\cr 5 &1 \end{array}\right],$$ then
$A^{-1} =$ $\left[\Rule{0pt}{2.4em}{0pt}\right.$$\left]\Rule{0pt}{2.4em}{0pt}\right.$.

Given $\vec{b} = \left[\begin{array}{c} 5\cr -3 \end{array}\right]$, solve $A\vec{x}=\vec{b}$.

$\vec{x}=$ $\left[\Rule{0pt}{2.4em}{0pt}\right.$$\left]\Rule{0pt}{2.4em}{0pt}\right.$.