JEE 2023 — Mathematics PYQ

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Question

JEE 2023 — Mathematics PYQ

JEE | Mathematics | 2023

Let $B = \begin{bmatrix} 1 & 3 & \alpha \\ 1 & 2 & 3 \\ \alpha & \alpha & 4 \end{bmatrix}, \alpha > 2$ be the adjoint of a matrix $A$ and $|A| = 2$ , then $\begin{bmatrix} \alpha & -2\alpha & \alpha \end{bmatrix} B \begin{bmatrix} \alpha \\ -2\alpha \\ \alpha \end{bmatrix}$ is equal to:

Choose the correct answer:

Explanation

1. $\alpha$ ki value nikalna

Hume pata hai ki $|adj(A)| = |A|^{n-1}$ hota hai.

Yahan $A$ ek $3 \times 3$ matrix hai ( $n=3$ ), isliye $|B| = |adj(A)| = |A|^{3-1} = |A|^2$ .

Hume $|A| = 2$ diya gaya hai, toh:

|B| = 2^2 = 4

Ab Matrix $B$ ka determinant nikalte hain:

|B| = 1(8 - 3\alpha) - 3(4 - 3\alpha) + \alpha(\alpha - 2\alpha) = 4

8 - 3\alpha - 12 + 9\alpha - \alpha^2 = 4

-\alpha^2 + 6\alpha - 4 = 4

\alpha^2 - 6\alpha + 8 = 0

Quadratic equation ko solve karne par:

(\alpha - 4)(\alpha - 2) = 0 \implies \alpha = 4 \text{ ya } \alpha = 2

Sawal mein diya hai $\alpha > 2$ , isliye hum $\alpha = 4$ lenge.

2. Matrix Expression ko solve karna

Hume nikalna hai: $X^T B X$ , jahan $X = \begin{bmatrix} \alpha \\ -2\alpha \\ \alpha \end{bmatrix}$ .

Pehle $BX$ nikalte hain ( $\alpha = 4$ rakh kar):

B = \begin{bmatrix} 1 & 3 & 4 \\ 1 & 2 & 3 \\ 4 & 4 & 4 \end{bmatrix}, \quad X = \begin{bmatrix} 4 \\ -8 \\ 4 \end{bmatrix}

BX = \begin{bmatrix} 1(4) + 3(-8) + 4(4) \\ 1(4) + 2(-8) + 3(4) \\ 4(4) + 4(-8) + 4(4) \end{bmatrix} = \begin{bmatrix} 4 - 24 + 16 \\ 4 - 16 + 12 \\ 16 - 32 + 16 \end{bmatrix} = \begin{bmatrix} -4 \\ 0 \\ 0 \end{bmatrix}

3. Final Multiplication

Ab $\begin{bmatrix} 4 & -8 & 4 \end{bmatrix} \begin{bmatrix} -4 \\ 0 \\ 0 \end{bmatrix}$ calculate karte hain:

(4 \times -4) + (-8 \times 0) + (4 \times 0)

= -16 + 0 + 0 = -16

Correct Answer

Sahi option (4) $-16$ hai.