Let I=(\beginarrayll1 & 0 0 & 1\endarray) and P=(\beginarrayll2 & 0 0 & 3\endarray) . Let Q=(\beginarrayllx & y z & 4\endarray) for some non-zero real numbers x, y , and z , for which there is a 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?

The determinant of Q - 2I is zero, The determinant of Q - 6I is 12

The determinant of Q - 2I is zero

The determinant of Q - 6I is 12

The determinant of Q - 3I is 15

Matrices and Determinants PYQ — JEE Advanced 2025 · Shift 2 · Q21 with Solution

View written solutionFree

Correct answer: A, B

Given matrices

P=\begin{pmatrix}2&0\\0&3\end{pmatrix}, \qquad Q=\begin{pmatrix}x&y\\ z&4\end{pmatrix}

and there exists a matrix

R=\begin{pmatrix}a&b\\ c&d\end{pmatrix}

such that all of $a,b,c,d$ are non-zero real numbers and

QR=RP.

We use this condition to determine constraints on $Q$ .

Compute both sides of $QR=RP$

First,

QR= \begin{pmatrix}x&y\\ z&4\end{pmatrix} \begin{pmatrix}a&b\\ c&d\end{pmatrix} = \begin{pmatrix} xa+yc & xb+yd \\ za+4c & zb+4d \end{pmatrix}.

Next,

RP= \begin{pmatrix}a&b\\ c&d\end{pmatrix} \begin{pmatrix}2&0\\0&3\end{pmatrix} = \begin{pmatrix} 2a & 3b \\ 2c & 3d \end{pmatrix}.

Equating corresponding entries:

xa+yc=2a \tag{1}

xb+yd=3b \tag{2}

za+4c=2c \tag{3}

zb+4d=3d \tag{4}

Since $a,b,c,d\neq 0$ , we can use these equations effectively.

Extract relations from (3) and (4)

From (3):

za+4c=2c \implies za=-2c \implies c=-\frac{za}{2}.

From (4):

zb+4d=3d \implies zb=-d \implies d=-zb.

Now substitute into (1) and (2).

From (1):

xa+y\left(-\frac{za}{2}\right)=2a

Since $a\neq 0$ ,

x-\frac{yz}{2}=2. \tag{5}

From (2):

xb+y(-zb)=3b

Since $b\neq 0$ ,

x-yz=3. \tag{6}

Subtract (6) from (5):

\left(x-\frac{yz}{2}\right)-(x-yz)=2-3

\frac{yz}{2}=-1

yz=-2.

Then from (6):

x-(-2)=3 \implies x+2=3 \implies x=1.

So,

Q=\begin{pmatrix}1&y\\ z&4\end{pmatrix}, \qquad yz=-2.

Now test each option

For

Q-\lambda I=\begin{pmatrix}1-\lambda & y\\ z & 4-\lambda\end{pmatrix},

we have

\det(Q-\lambda I)=(1-\lambda)(4-\lambda)-yz.

Since $yz=-2$ ,

\det(Q-\lambda I)=(1-\lambda)(4-\lambda)+2.

Option A: $\det(Q-2I)=0$

\det(Q-2I)=(1-2)(4-2)+2=(-1)(2)+2=0.

So A is true.

Option B: $\det(Q-6I)=12$

\det(Q-6I)=(1-6)(4-6)+2=(-5)(-2)+2=10+2=12.

So B is true.

Option C: $\det(Q-3I)=15$

\det(Q-3I)=(1-3)(4-3)+2=(-2)(1)+2=0.

So C is false.

Option D: $yz=2$

We found

yz=-2,

so D is false.

Final answer

The true statements are:

\boxed{A,\ B}

Comparison with stored answer

Stored correct answer: A, B

Our derived answer: A, B

Hence, the answer agrees with the stored correct answer.

More from Matrices and Determinants

View written solutionFree

Correct answer: A, B

Given matrices

P=\begin{pmatrix}2&0\\0&3\end{pmatrix}, \qquad Q=\begin{pmatrix}x&y\\ z&4\end{pmatrix}

and there exists a matrix

R=\begin{pmatrix}a&b\\ c&d\end{pmatrix}

such that all of $a,b,c,d$ are non-zero real numbers and

QR=RP.

We use this condition to determine constraints on $Q$ .

Compute both sides of $QR=RP$

First,

QR= \begin{pmatrix}x&y\\ z&4\end{pmatrix} \begin{pmatrix}a&b\\ c&d\end{pmatrix} = \begin{pmatrix} xa+yc & xb+yd \\ za+4c & zb+4d \end{pmatrix}.

Next,

RP= \begin{pmatrix}a&b\\ c&d\end{pmatrix} \begin{pmatrix}2&0\\0&3\end{pmatrix} = \begin{pmatrix} 2a & 3b \\ 2c & 3d \end{pmatrix}.

Equating corresponding entries:

xa+yc=2a \tag{1}

xb+yd=3b \tag{2}

za+4c=2c \tag{3}

zb+4d=3d \tag{4}

Since $a,b,c,d\neq 0$ , we can use these equations effectively.

Extract relations from (3) and (4)

From (3):

za+4c=2c \implies za=-2c \implies c=-\frac{za}{2}.

From (4):

zb+4d=3d \implies zb=-d \implies d=-zb.

Now substitute into (1) and (2).

From (1):

xa+y\left(-\frac{za}{2}\right)=2a

Since $a\neq 0$ ,

x-\frac{yz}{2}=2. \tag{5}

From (2):

xb+y(-zb)=3b

Since $b\neq 0$ ,

x-yz=3. \tag{6}

Subtract (6) from (5):

\left(x-\frac{yz}{2}\right)-(x-yz)=2-3

\frac{yz}{2}=-1

yz=-2.

Then from (6):

x-(-2)=3 \implies x+2=3 \implies x=1.

So,

Q=\begin{pmatrix}1&y\\ z&4\end{pmatrix}, \qquad yz=-2.

Now test each option

For

Q-\lambda I=\begin{pmatrix}1-\lambda & y\\ z & 4-\lambda\end{pmatrix},

we have

\det(Q-\lambda I)=(1-\lambda)(4-\lambda)-yz.

Since $yz=-2$ ,

\det(Q-\lambda I)=(1-\lambda)(4-\lambda)+2.

Option A: $\det(Q-2I)=0$

\det(Q-2I)=(1-2)(4-2)+2=(-1)(2)+2=0.

So A is true.

Option B: $\det(Q-6I)=12$

\det(Q-6I)=(1-6)(4-6)+2=(-5)(-2)+2=10+2=12.

So B is true.

Option C: $\det(Q-3I)=15$

\det(Q-3I)=(1-3)(4-3)+2=(-2)(1)+2=0.

So C is false.

Option D: $yz=2$

We found

yz=-2,

so D is false.

Final answer

The true statements are:

\boxed{A,\ B}

Comparison with stored answer

Stored correct answer: A, B

Our derived answer: A, B

Hence, the answer agrees with the stored correct answer.

Matrices and Determinants question

Matrices and Determinants question

Option A: $\det(Q-2I)=0$

Option B: $\det(Q-6I)=12$

Option C: $\det(Q-3I)=15$

Option D: $yz=2$

More from Matrices and Determinants

Matrices and Determinants question

Matrices and Determinants question

Option A: $\det(Q-2I)=0$

Option B: $\det(Q-6I)=12$

Option C: $\det(Q-3I)=15$

Option D: $yz=2$

More from Matrices and Determinants

Option A: det⁡(Q−2I)=0\det(Q-2I)=0det(Q−2I)=0

Option B: det⁡(Q−6I)=12\det(Q-6I)=12det(Q−6I)=12

Option C: det⁡(Q−3I)=15\det(Q-3I)=15det(Q−3I)=15

Option D: yz=2yz=2yz=2

More from Matrices and Determinants

Option A: det⁡(Q−2I)=0\det(Q-2I)=0det(Q−2I)=0

Option B: det⁡(Q−6I)=12\det(Q-6I)=12det(Q−6I)=12

Option C: det⁡(Q−3I)=15\det(Q-3I)=15det(Q−3I)=15

Option D: yz=2yz=2yz=2

More from Matrices and Determinants

Option A: $\det(Q-2I)=0$

Option B: $\det(Q-6I)=12$

Option C: $\det(Q-3I)=15$

Option D: $yz=2$

Option A: $\det(Q-2I)=0$

Option B: $\det(Q-6I)=12$

Option C: $\det(Q-3I)=15$

Option D: $yz=2$