JEE Main 2025 Jan 24 Shift 1, Mathematics Q24: Algebra Of Matrices

JEE Main2025Jan 24, Shift 1Mathematics

Let $A$ be a $3 \times 3$ matrix such that $X^{T} A X = O$ for all nonzero $3 \times 1$ matrices $X = x y z$ .

If $A 111 = 14 - 5$ , $A 121 = 04 - 8$ , and $det (adj (2 (A + I))) = 2^{α} 3^{β} 5^{γ}$ , $α, β, γ \in N$ , then $α^{2} + β^{2} + γ^{2}$ is

Your answer

Solution

The condition $X^{T} A X = 0$ for every $X$ forces $A$ to be skew-symmetric (the symmetric part of $A$ must be zero). Write

$A = 0 ab - a 0 c - b - c 0$

Apply $A (1, 1, 1)^{T} = (1, 4, - 5)^{T}$ :

$a + b = 1, - a + c = 4, - b - c = - 5 \Rightarrow b + c = 5$

Apply $A (1, 2, 1)^{T} = (0, 4, - 8)^{T}$ :

$2 a + b = 0, - a + c = 4, - 2 c - b = - 8 \Rightarrow b + 2 c = 8$

From $2 a + b = 0$ and $a + b = 1$ : $a = - 1$ , $b = 2$ . Then $c = 3$ .

So $A = 0 - 12 103 - 2 - 30$ .

$2 (A + I) = 2 - 24 226 - 2 - 62$ . Factor 2 from each row: $det (2 (A + I)) = 2^{3} det (A + I)$ .

Direct expansion gives $det (2 (A + I)) = 120$ .

For a $3 \times 3$ matrix $M$ , $det (adj M) = (det M)^{2}$ . Therefore

$det (adj (2 (A + I))) = 12 0^{2} = 14400 = 2^{6} \cdot 3^{2} \cdot 5^{2}$

So $α = 6, β = 2, γ = 2$ and $α^{2} + β^{2} + γ^{2} = 36 + 4 + 4 = 44$ .

$∴$ Answer: $44$ .

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