JEE 2023 — Mathematics PYQ

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Question

JEE 2023 — Mathematics PYQ

JEE | Mathematics | 2023

Let for $A = \begin{bmatrix} 1 & 2 & 3 \\ \alpha & 3 & 1 \\ 1 & 1 & 2 \end{bmatrix}, |A| = 2$ . If $|2 \text{ adj } (2 \text{ adj } (2A))| = 32^n$ , then $3n + \alpha$ is equal to:

Choose the correct answer:

Explanation

Step 1: $\alpha$ ki value nikalna

Humein diya gaya hai $|A| = 2$ . Matrix $A$ ka determinant nikalte hain:

$|A| = \begin{vmatrix} 1 & 2 & 3 \\ \alpha & 3 & 1 \\ 1 & 1 & 2 \end{vmatrix} = 2$

$1(6 - 1) - 2(2\alpha - 1) + 3(\alpha - 3) = 2$

$5 - 4\alpha + 2 + 3\alpha - 9 = 2$

$-\alpha - 2 = 2 \implies \alpha = -4$

Step 2: Determinant property ka use karke $n$ nikalna

Property: $|k \text{ adj } M| = k^n |M|^{n-1}$ (jahan $n$ matrix ka order hai, yahan $n=3$ hai).

Humein nikalna hai: $|2 \text{ adj } (2 \text{ adj } (2A))| = 32^n$

Pehle andar wali term simplify karte hain:

$|2A| = 2^3 |A| = 8 \times 2 = 16$
Let $B = 2 \text{ adj } (2A)$ . Iska determinant:

$|B| = |2 \text{ adj } (2A)| = 2^3 |\text{ adj } (2A)| = 8 \times |2A|^{3-1} = 8 \times (16)^2 = 8 \times 256 = 2048 = 2^{11}$
Ab final expression:

$|2 \text{ adj } B| = 2^3 |\text{ adj } B| = 8 \times |B|^{3-1} = 8 \times (2^{11})^2 = 2^3 \times 2^{22} = 2^{25}$

Humein diya hai ki ye $32^n$ ke barabar hai:

$2^{25} = (2^5)^n = 2^{5n}$

$5n = 25 \implies n = 5$

Step 3: Final value nikalna

Humein $3n + \alpha$ find karna hai:

$3(5) + (-4) = 15 - 4 = 11$

Correct Option: (4) 11

Choose the correct answer:

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