Let $\mathrm{A}={\left[{\mathrm{a}}_{\mathrm{ij}}\right]}_{2\times 2}$, where ${\mathrm{a}}_{\mathrm{ij}}\ne 0$ for all i, j and ${\mathrm{A}}^2=\mathrm{I}$.  Let a be the sum of all diagonal elements of A and $\mathrm{b}=\left|\mathrm{A}\right|$. Then $3{\mathrm{a}}^2+4{ \mathrm{b}}^2$is equal to