JEE Advanced 2025 — Paper 2 — MATH-II solution

JEE Advanced2025Paper 2MATH-II

Let $I = (1001)$ and $P = (2003)$ . Let $Q = (x y z 4)$ for some non-zero real numbers $x$ , $y$ , and $z$ , for which there is $2 \times 2$ matrix $R$ with all entries being non-zero real numbers, such that $QR = R P$ .

Then which of the following statements is (are) TRUE?

A
The determinant of $Q - 2 I$ is zero
B
The determinant of $Q - 6 I$ is $12$
C
The determinant of $Q - 3 I$ is $15$
D
$y z = 2$

Solution

Step 1: Expand $QR = R P$ column by column.

Write $R = (r_{1} r_{2} r_{3} r_{4})$ . The first column gives $x r_{1} + y r_{3} = 2 r_{1}$ and $z r_{1} + 4 r_{3} = 2 r_{3}$ , leading to $\frac{r _{3}}{r _{1}} = \frac{x - 2}{- y} = \frac{z}{- 2}$ . Hence $2 x - 4 = y z$ .

The second column gives $\frac{r _{4}}{r _{2}} = \frac{x - 3}{- y} = - z$ , hence $x - 3 = y z$ .

Step 2: Solve the two relations.

From $2 x - 4 = y z$ and $x - 3 = y z$ , subtracting yields $x = 1$ and therefore $y z = - 2$ .

Step 3: Characteristic polynomial of $Q$ .

$det (Q - λ I) = (1 - λ) (4 - λ) - y z = λ^{2} - 5 λ + 4 + 2 = λ^{2} - 5 λ + 6$ .

Step 4: Evaluate at the required values.

At $λ = 2$ : $4 - 10 + 6 = 0$ , so $det (Q - 2 I) = 0$ . Option (A) is TRUE.

At $λ = 6$ : $36 - 30 + 6 = 12$ , so $det (Q - 6 I) = 12$ . Option (B) is TRUE.

At $λ = 3$ : $9 - 15 + 6 = 0 \neq = 15$ . Option (C) is FALSE.

Since $y z = - 2 \neq = 2$ , option (D) is FALSE.

$∴$ The correct options are (A) and (B).

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