Let $P=\left[\begin{array}{ccc}3 & -1 & -2 \\ 2 & 0 & \alpha \\ 3 & -5 & 0\end{array}\right]$, where $\alpha \in \mathbb{R}$. Suppose $Q=\left[q_{i j}\right]$ is a matrix such that $P Q=k I$, where $k \in \mathbb{R}, k \neq 0$ and $I$ is the identity matrix of order 3. If $q_{23}=-\frac{k}{8}$ and $\operatorname{det}(Q)=\frac{k^{2}}{2}$, then

(B) $4 \alpha-k+8=0$; (C) $\operatorname{det}(P \operatorname{adj}(Q))=2^{9}$

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

Matrices — JEE Advanced · 2016 · 22 May 2016 · Paper 1 Question with Solution

Let

P = 323 - 1 0 - 5 - 2 α 0

, where

α \in R

. Suppose

Q = [q_{ij}]

is a matrix such that

P Q = k I

, where

k \in R, k \neq = 0

and

I

is the identity matrix of order 3. If

q_{23} = - \frac{k}{8}

and

det (Q) = \frac{k ^{2}}{2}

, then

Answer: (B), (C)

From

P Q = k I

, we have

Q = k P^{- 1}

, so

q_{ij} = k (P^{- 1})_{ij}

.

Given

q_{23} = - \frac{k}{8}

, we have

(P^{- 1})_{23} = - \frac{1}{8}

.

Using the cofactor formula:

(P^{- 1})_{23} = \frac{C _{32}}{d e t P}

where

C_{32}

is the cofactor.

C_{32} = - det [32 - 2 α] = - (3 α + 4)

det P = 3 (5 α) - (- 1) (0 - 3 α) + (- 2) (- 10) = 15 α + 3 α + 20 = 18 α + 20

.

So

\frac{- ( 3 α + 4 )}{18 α + 20} = - \frac{1}{8}

, giving

8 (3 α + 4) = 18 α + 20

, thus

24 α + 32 = 18 α + 20

, so

6 α = - 12

, hence

α = - 1

.

Then

det P = 18 (- 1) + 20 = 2

. Wait, let me recalculate:

det P = 20 + 12 α = 20 - 12 = 8

.

From

det (P Q) = det P \cdot det Q = k^{3}

, we have

8 \cdot \frac{k ^{2}}{2} = k^{3}

, so

4 k^{2} = k^{3}

, giving

k = 4

.

(B)

4 α - k + 8 = 4 (- 1) - 4 + 8 = 0

. True.

(C)

det (P adj Q) = det P \cdot (det Q)^{2} = 8 \cdot 8^{2} = 2^{9}

. True.

JEE Advanced · 2016 · 22 May 2016 · Paper 1 · One or more correct options