JEE Advanced 2016 · Paper 1 · Mathematics Q44

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MathematicsMultiple correct+4 / -2Question 44 of 54

Let $P=\begin{bmatrix}3&-1&-2\\2&0&\alpha\\3&-5&0\end{bmatrix}$ , where $\alpha\in\mathbb R$ . Suppose $Q=[q_{ij}]$ is a matrix such that $PQ=kI$ , where $k\in\mathbb R$ , $k\ne0$ and $I$ is the identity matrix of order 3. If $q_{23}=-\dfrac k8$ and $\det(Q)=\dfrac{k^2}{2}$ , then

\alpha=0,\ k=8

4\alpha-k+8=0

\det(P\operatorname{adj}(Q))=2^9

\det(Q\operatorname{adj}(P))=2^{13}

Answer and worked solution

Official answer: B, C

Q from P. $PQ=kI$ gives $Q=kP^{-1}=\dfrac{k}{\det P}\operatorname{adj}P$ .

Determinant. $\det P=3(5\alpha)+1(-3\alpha)-2(-10)=12\alpha+20$ .

Use $q_{23}$ . $q_{23}=\dfrac{k}{\det P}C_{32}$ , with $C_{32}=-(3\alpha+4)$ . So
$-\frac{k(3\alpha+4)}{12\alpha+20}=-\frac k8\implies\alpha=-1,\quad\det P=8.$

Use $\det Q$ . $\det Q=\dfrac{k^3}{\det P}=\dfrac{k^3}8=\dfrac{k^2}2$ , so $k=4$ and $\det Q=8$ .

Check each option.

(A) False.
(B) $4\alpha-k+8=-4-4+8=0$ . True.
(C) $\det(P\operatorname{adj}Q)=\det P\,(\det Q)^2=8\cdot64=2^9$ . True.
(D) $\det(Q\operatorname{adj}P)=8\cdot64=2^9\ne2^{13}$ . False.

(B), (C)

Topics:Adjoint and inverse Determinants and properties

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JEE Advanced 2016 · Paper 1, Mathematics question 44