JEE Advanced 2025 — Paper 2 — Question 21

Given $QR = RP$ with $R = \begin{pmatrix} r_1 & r_2 \\ r_3 & r_4 \end{pmatrix}$ and all entries non-zero. Compute $QR = \begin{pmatrix} x & y \\ z & 4 \end{pmatrix} \begin{pmatrix} r_1 & r_2 \\ r_3 & r_4 \end{pmatrix} = \begin{pmatrix} x r_1 + y r_3 & x r_2 + y r_4 \\ z r_1 + 4 r_3 & z r_2 + 4 r_4 \end{pmatrix}$. Compute $RP = \begin{pmatrix} r_1 & r_2 \\ r_3 & r_4 \end{pmatrix} \begin{pmatrix} 2 & 0 \\ 0 & 3 \end{pmatrix} = \begin{pmatrix} 2 r_1 & 3 r_2 \\ 2 r_3 & 3 r_4 \end{pmatrix}$. Equating entries gives four equations: $x r_1 + y r_3 = 2 r_1$, $x r_2 + y r_4 = 3 r_2$, $z r_1 + 4 r_3 = 2 r_3$, $z r_2 + 4 r_4 = 3 r_4$. From the first and third: $(x-2) r_1 + y r_3 = 0$ and $z r_1 + 2 r_3 = 0$.

Since $r_1, r_3 \neq 0$, the ratios give $\frac{r_3}{r_1} = \frac{2-x}{y} = -\frac{z}{2}$, so $(x-2)(2) = yz$, i.e., $2x - 4 = yz$. From the second and fourth: $(x-3) r_2 + y r_4 = 0$ and $z r_2 + r_4 = 0$.

Since $r_2, r_4 \neq 0$, the ratios give $\frac{r_4}{r_2} = \frac{3-x}{y} = -z$, so $x-3 = yz$. Equating the two expressions for $yz$: $2x - 4 = x - 3 \Rightarrow x = 1$. Then $yz = 1 - 3 = -2$. Thus $Q = \begin{pmatrix} 1 & y \\ z & 4 \end{pmatrix}$ with $yz = -2$. Characteristic polynomial: $\det(Q - \lambda I) = (1-\lambda)(4-\lambda) - yz = \lambda^2 - 5\lambda + 4 - (-2) = \lambda^2 - 5\lambda + 6$. For option A: $\det(Q - 2I) = 2^2 - 5(2) + 6 = 4 - 10 + 6 = 0$. True. For option B: $\det(Q - 6I) = 6^2 - 5(6) + 6 = 36 - 30 + 6 = 12$. True. For option C: $\det(Q - 3I) = 3^2 - 5(3) + 6 = 9 - 15 + 6 = 0$, not 15. False. For option D: $yz = -2$, not 2. False. Hence the correct options are A and B.