Let I=( array ll 1 & 0 0 & 1 array ) and P=( array ll 2 & 0 0 & 3 array ) . Let Q=( array ll x & y z & 4 array ) for some non-zero real numbers x, y , and z , for which there is 2 × 2 matrix R with all entries being

Correct answer: A. The determinant of Q-2 I is zero aligned & QR=RP & P=( array ll 2 & 0 0 & 3 array ) Q=( array ll x & y z & 4 array ) & ( array ll x & y z & 4 array )( array ll r_1 & r_2 r_3 & r_4 array )=( array ll r_1 & r_2 r_3 & r_4 array )( array ll 2 & 0 0 & 3 array ) & . array l xr_1+yr_3=2 r_1 zr_1+4 r_3=2 r_3 array \ arrow x-2 -y = r_3 r_1 = z -2 aligned 2 x-4=y z . array l xr_2+yr_4=3 r_2 zr_2+4 r_4=3 r_4 array \ arrow x-3 -y = r_4 r_2 =-z (x-3=yz) 2 x-4=x-3 ( x=1 ~\&~ y z=-2) aligned & Q-λ I=( array cc x-λ & y z & 4-λ array ) & |Q-λ I|=(λ-x)(λ-4)-yz & =λ^2-(x+4) λ+4 x-yz & |Q-λ I|=λ^2-5 λ+6 aligned Now verify the option

JEE Advanced 2025Paper 2

Exam: JEE Advanced
Year: 2025
Paper: Paper 2
Subject: Mathematics
Chapter: Matrices

JEE Advanced 2025 · Paper 2 · Mathematics · Matrices

Let I=( array ll 1 & 0 0 & 1 array ) and P=( array ll 2 & 0 0 & 3...

Let

I=\left(

\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}

\right)

and

P=\left(

\begin{array}{ll}2 & 0 \\ 0 & 3\end{array}

\right)

. Let

Q=\left(

\begin{array}{ll}\mathrm{x} & \mathrm{y} \\ \mathrm{z} & 4\end{array}

\right)

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

Q R=R P

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

Options

AThe determinant of $Q-2 I$ is zero
BThe determinant of $\mathrm{Q}-6 I$ is 12
CThe determinant of $Q-3 I$ is 15
D $y z=2$

Correct answer

A. The determinant of $Q-2 I$ is zero

Step-by-step solution

\begin{aligned} & \mathrm{QR}=\mathrm{RP} \\ & P=\left(\begin{array}{ll} 2 & 0 \\ 0 & 3 \end{array}\right) \quad Q=\left(\begin{array}{ll} x & y \\ z & 4 \end{array}\right) \\ & \left(\begin{array}{ll} x & y \\ z & 4 \end{array}\right)\left(\begin{array}{ll} r_1 & r_2 \\ r_3 & r_4 \end{array}\right)=\left(\begin{array}{ll} r_1 & r_2 \\ r_3 & r_4 \end{array}\right)\left(\begin{array}{ll} 2 & 0 \\ 0 & 3 \end{array}\right) \\ & \left.\begin{array}{l} \mathrm{xr}_1+\mathrm{yr}_3=2 \mathrm{r}_1 \\ \mathrm{zr}_1+4 \mathrm{r}_3=2 \mathrm{r}_3 \end{array}\right\} \rightarrow \frac{\mathrm{x}-2}{-\mathrm{y}}=\frac{\mathrm{r}_3}{\mathrm{r}_1}=\frac{\mathrm{z}}{-2} \end{aligned}

2 x-4=y z

\left.

\begin{array}{l}\mathrm{xr}_2+\mathrm{yr}_4=3 \mathrm{r}_2 \\ \mathrm{zr}_2+4 \mathrm{r}_4=3 \mathrm{r}_4\end{array}

\right\} \rightarrow \frac{\mathrm{x}-3}{-\mathrm{y}}=\frac{\mathrm{r}_4}{\mathrm{r}_2}=-\mathrm{z} \quad

(\mathrm{x}-3=\mathrm{yz})

\Rightarrow 2 x-4=x-3

(\Rightarrow x=1 ~\&~ y z=-2)

\begin{aligned} & \Rightarrow \mathrm{Q}-\lambda \mathrm{I}=\left(\begin{array}{cc} \mathrm{x}-\lambda & \mathrm{y} \\ \mathrm{z} & 4-\lambda \end{array}\right) \\ & |\mathrm{Q}-\lambda \mathrm{I}|=(\lambda-\mathrm{x})(\lambda-4)-\mathrm{yz} \\ & =\lambda^2-(\mathrm{x}+4) \lambda+4 \mathrm{x}-\mathrm{yz} \\ & |\mathrm{Q}-\lambda \mathrm{I}|=\lambda^2-5 \lambda+6 \end{aligned}

$ Now verify the option

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