JEE Advanced 2025 Paper 2 Q05 Mathematics Matrices & Determinants Determinants Hard

JEE Advanced 2025 Paper 2 · Q05 · Determinants

Let $I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$ and $P = \begin{pmatrix} 2 & 0 \\ 0 & 3 \end{pmatrix}$. Let $Q = \begin{pmatrix} x & y \\ z & 4 \end{pmatrix}$ for some non-zero real numbers $x, y, z$, for which there exists a $2\times 2$ matrix $R$ with all entries being non-zero real numbers, such that $QR = RP$.

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

  1. A. The determinant of $Q - 2I$ is zero
  2. B. The determinant of $Q - 6I$ is $12$
  3. C. The determinant of $Q - 3I$ is $15$
  4. D. $yz = 2$
JEE Advanced multi-correct — pick every correct option, then check.
Reveal answer + step-by-step solution

Correct answer:A, B

Solution

$QR = RP$ with $R$ invertible (entries non-zero, plus $R$ must be invertible for similarity since diag $P$ is diagonalizable): $Q$ is similar to $P$. Hence $Q$ has the same eigenvalues $2$ and $3$, so $\text{trace}(Q) = 5$ and $\det(Q) = 6$. Trace $= x + 4 = 5$ $\to$ $x = 1$. $\det = 4 \cdot 1 - yz = 6$ $\to$ $yz = -2$. (A) $\det(Q - 2I) = (x-2)(4-2) - yz = (-1)(2) - (-2) = 0$. TRUE. (B) $\det(Q - 6I) = (1-6)(4-6) - (-2) = 10 + 2 = 12$. TRUE. (C) $\det(Q - 3I) = (1-3)(4-3) - (-2) = -2 + 2 = 0$, not $15$. FALSE. (D) $yz = -2$, not $2$. FALSE. Correct: (A), (B).

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