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

Given: $$A$$ is a $$2 \times 2$$ matrix with all non-zero entries and $$A^2 = I$$.

Let $$a = \text{tr}(A)$$ and $$b = |A| = \det(A)$$.

To begin, Since $$A^2 = I$$, we have $$\det(A^2) = \det(I) = 1$$, so $$(\det A)^2 = 1$$, giving $$b = \pm 1$$.

Next, Taking trace of $$A^2 = I$$: $$\text{tr}(A^2) = \text{tr}(I) = 2$$.

For a $$2 \times 2$$ matrix: $$\text{tr}(A^2) = (\text{tr}A)^2 - 2\det(A)$$.

$$ a^2 - 2b = 2 $$

From here, If $$b = 1$$: $$a^2 = 4$$, so $$a = \pm 2$$. This gives eigenvalues both equal to 1 (or both -1), meaning $$A = \pm I$$. But then off-diagonal entries would be 0, contradicting $$a_{ij} \neq 0$$.

If $$b = -1$$: $$a^2 = 0$$, so $$a = 0$$. The eigenvalues are $$+1$$ and $$-1$$, and the matrix can have all non-zero entries. ✓

Continuing,

$$ 3a^2 + 4b^2 = 3(0) + 4(1) = 4 $$

The correct answer is 4.