If $A = \left[\begin{array}{cc} -3 &7\cr -8 &3 \end{array}\right]$, then $\det(A)=$ is .

Thus, $A^{-1} =$ $\left[\Rule{0pt}{2.4em}{0pt}\right.$ $\left]\Rule{0pt}{2.4em}{0pt}\right.$ and $\det(A^{-1})=$ .