Let A = [ a i j ] 2 × 2 , where a i j = 0 for all i , j and A 2 = I . Let a be the sum of all diagonal elements of A and b = | A | . Then 3 a 2 + 4 b 2 is equal to [2023]

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Solution

Q.
Let $A = {[a_{i j}]}_{2 \times 2}$ , where $a_{i j} = 0$ for all $i, j$ and $A^{2} = I$ . Let $a$ be the sum of all diagonal elements of $A$ and $b =$ $| A |$ . Then $3 a^{2} + 4 b^{2}$ is equal to [2023]

1 3

2 7

3 4

4 14

Ans.
(3)

Let
$A = [\begin{matrix} m & n \\ q & p \end{matrix}]$ $∵ A^{2} = I$

$\Rightarrow [\begin{matrix} m^{2} + q n & m n + n p \\ q m + q p & n q + p^{2} \end{matrix}] = [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] \begin{matrix} \end{matrix}$

$\Rightarrow m^{2} + q n = 1$ ...(i); $n (m + p) = 0$ ...(ii)

$q (m + p) = 0$ ...(iii), $n q + p^{2} = 1$ ...(iv)

$\Rightarrow m^{2} - p^{2} = 0 \Rightarrow m = \pm p [Using (i) & (iv)]$

Also, $m + p = 0$

Now, let $a = m + p$ and $b = m p - q n$

$So, 3 a^{2} + 4 b^{2} = 3 \times q^{2} + 4 {(m p - q n)}^{2}$

$= 4 {(- m^{2} - q n)}^{2} = 4 \times 1 = 4$ $[Using (i)]$

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