JEE Main 2025 Apr 8 Shift 2, Mathematics Q19: Algebra Of Determinants

JEE Main2025Apr 8, Shift 2Mathematics

Let $A = 22 + p 2 + p + q 46 + 2 p 8 + 3 p + 2 q 612 + 3 p 20 + 6 p + 3 q$ .

If $det (adj (adj (3 A))) = 2^{m} \cdot 3^{n}$ , $m, n \in N$ , then $m + n$ is equal to

A
$22$
B
$24$
C
$26$
D
$20$

Solution

Apply column operations to simplify $∣ A ∣$ . First: $C_{3} \to C_{3} - C_{2} - \frac{q}{2} C_{1}$ eliminates $q$ , then $C_{2} \to C_{2} - (1 + \frac{p}{2}) C_{1}$ eliminates $p$ .

The matrix reduces to $200422668$ , giving $∣ A ∣ = 2 (16 - 12) = 8 = 2^{3}$ .

For an $n \times n$ matrix, $∣ adj (adj (M)) ∣ = ∣ M ∣^{(n - 1)^{2}}$ ; here $n = 3$ , so $∣ adj (adj (3 A)) ∣ = ∣3 A ∣^{4}$ .

Compute $∣3 A ∣ = 3^{3} ∣ A ∣ = 27 \cdot 8 = 216$ , so $∣3 A ∣^{4} = 21 6^{4} = (2^{3} \cdot 3^{3})^{4} = 2^{12} \cdot 3^{12}$ .

Therefore $m = 12$ and $n = 12$ , giving $m + n = 24$ , matching option (2).

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JEE Main 2025 Apr 8 Shift 2, Mathematics Q19: Algebra Of Determinants

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