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