JEE Main 2025 Jan 22 Shift 1, Mathematics Q23: Algebra Of Determinants

JEE Main2025Jan 22, Shift 1Mathematics

Let A be a square matrix of order 3 such that $det (A) = - 2$ and $det (3 adj (- 6 adj (3 A))) = 2^{m + n} \cdot 3^{mn}$ , $m > n$ . Then $4 m + 2 n$ is equal to _____.

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Solution

For an $n \times n$ matrix $A$ , $det (adj A) = (det A)^{n - 1}$ and $det (k A) = k^{n} det A$ . Here $n = 3$ .

Step 1: $det (3 A) = 3^{3} \cdot (- 2) = - 54$ .

Step 2: $det (adj (3 A)) = (- 54)^{2} = 2916 = 2^{2} \cdot 3^{6}$ .

Step 3: $- 6 adj (3 A)$ is a $3 \times 3$ matrix. $det (- 6 adj (3 A)) = (- 6)^{3} \cdot 2916 = - 216 \cdot 2916 = - 2^{3} \cdot 3^{3} \cdot 2^{2} \cdot 3^{6} = - 2^{5} \cdot 3^{9}$ .

Step 4: $det (adj (- 6 adj (3 A))) = (- 2^{5} \cdot 3^{9})^{2} = 2^{10} \cdot 3^{18}$ .

Step 5: $det (3 \cdot adj (\dots)) = 3^{3} \cdot 2^{10} \cdot 3^{18} = 2^{10} \cdot 3^{21}$ .

Matching with $2^{m + n} \cdot 3^{mn}$ : $m + n = 10$ and $mn = 21$ . With $m > n$ , $(m, n) = (7, 3)$ .

$4 m + 2 n = 28 + 6 = 34$ .

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JEE Main 2025 Jan 22 Shift 1, Mathematics Q23: Algebra Of Determinants

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