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
Step-by-Step Solution
Key Concept: For a 2x2 matrix A, A^2 = I implies the characteristic equation is \lambda^2 - tr(A)\lambda + |A| = 0. Since A^2 = I, the eigenvalues are 1 and -1. Thus, tr(A) = 1 + (-1) = 0 or 1 + 1 = 2 or -1 - 1 = -2. Given a_ij != 0, we analyze the trace and determinant.
Let $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$. Given $a, b, c, d \neq 0$ and $A^2 = I$. The characteristic equation is $\lambda^2 - \text{tr}(A)\lambda + |A| = 0$. Since $A^2 = I$, the eigenvalues are $\pm 1$. If eigenvalues are $1, -1$, then $\text{tr}(A) = 0$ and $|A| = -1$. If eigenvalues are $1, 1$, then $\text{tr}(A) = 2$ and $|A| = 1$. If eigenvalues are $-1, -1$, then $\text{tr}(A) = -2$ and $|A| = 1$. For $A^2 = I$, $A^2 = \begin{pmatrix} a^2+bc & b(a+d) \\ c(a+d) & d^2+bc \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$. Since $b, c \neq 0$, $a+d = 0$, so $\text{tr}(A) = a = 0$. But $a_{ij} \neq 0$ is given, so $a$ cannot be 0. Thus, $a+d$ must be 0 is not possible if $a, d \neq 0$. Wait, if $a+d=0$, then $a^2+bc=1$ and $d^2+bc=1$, so $a^2=d^2$, $d = -a$. Then $|A| = ad-bc = -a^2-bc = -1$. So $a^2+bc=1$. This is consistent. Then $a=0$ is not required, only $a+d=0$. So $a$ is the trace, $a = a+d = 0$. Then $b = |A| = -1$. $3(0)^2 + 4(-1)^2 = 4$.
Correct Answer: 4