JEE Advanced 2019 · Paper 2 · Mathematics Q2

← JEE Advanced 2019 · Paper 2

MathematicsMultiple correct+4 / -1Question 38 of 54

Let $x\in\mathbb R$ and let

$P = \begin{bmatrix}1 & 1 & 1\\ 0 & 2 & 2\\ 0 & 0 & 3\end{bmatrix},\quad Q = \begin{bmatrix}2 & x & x\\ 0 & 4 & 0\\ x & x & 6\end{bmatrix}\quad\text{and}\quad R = PQP^{-1}.$

Then which of the following options is/are correct?

There exists a real number

x

such that

PQ = QP

\det R = \det\begin{bmatrix}2 & x & x\\ 0 & 4 & 0\\ x & x & 5\end{bmatrix} + 8

, for all

x\in\mathbb R

For

x = 0

, if

R\begin{bmatrix}1\\ a\\ b\end{bmatrix} = 6\begin{bmatrix}1\\ a\\ b\end{bmatrix}

, then

a + b = 5

For

x = 1

, there exists a unit vector

\alpha\hat i + \beta\hat j + \gamma\hat k

for which

R\begin{bmatrix}\alpha\\ \beta\\ \gamma\end{bmatrix} = \begin{bmatrix}0\\ 0\\ 0\end{bmatrix}

Answer and worked solution

Official answer: B, C

(A) Compare the products:

row 2 of $PQ$ is $(2x,\,8+2x,\,12)$ ;
row 2 of $QP$ is $(0,\,8,\,8)$ .

They differ for every $x$ (12 ≠ 8). False.

(B) Similar matrices have equal determinants:
$\det R=\det Q=2(24)-x(0)+x(0-4x)=48-4x^2.$
The given matrix has determinant $40-4x^2$ , and adding 8 gives $48-4x^2$ . True.

(C) For $x=0$ , $Q=\operatorname{diag}(2,4,6)$ , so $R$ 's eigenvector for 6 is $Pe_3=(1,2,3)^T$ . Hence $a=2$ , $b=3$ and $a+b=5$ . True.

(D) For $x=1$ , $\det R=44\ne0$ , so $R$ has no nonzero null vector. False.

(B), (C)

Topics:Determinants and properties Matrix algebra

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JEE Advanced 2019 · Paper 2, Mathematics question 2