Let P = [ cases 3 & - 1 & - 2 2 & 0 & α 3 & - 5 & 0 cases ] , where α ∈ R, Suppose Q = [ q _ ij ] is a matrix such that PQ = kI , where k ∈ R, k ≠ 0 and I is the identity matrix of order 3. If q _ 23 = - k 8 and det (Q)

Correct answer: B. 4α - k + 8 = 0 PQ = kI |P||Q| = k ^ 3 ( as order is 3 ) ⇒ |P| = 2k ≠ 0 ⇒ P is an invertible matrix ∵ PQ = kI ∴ Q = k P ^ - 1 I = K(1)/(|P|)adj P ∴ Q = (adj . P)/(2) ∵ q _ 23 = - (k)/(8)( ∴ cofactor P _ 32 = - (3α + 4) ) ⇒ 12α + 16 = K and (adjP)/(2) = [ P _ 32 ] ^ 2 ^ T - q _ 23 ∴ undefined ...... (i) Also|P| = 2k ⇒ k = 10 + 6α ......(ii) Solving these two ⇒ 12α + 16 = 10 + 6α ⇒ α = - 1 & K = 4 ∴ 4α - k + 8 = 0 ( ⇒ 4 ( - 1 ) - ( 4 ) + 8 = 0 ⇒ - 4 - 4 + 8 = 0 ⇒ - 8 + 8 = 0 ) And det (P(adj (Q)) = |P||adj (Q)| = 2k . ( k ^ 2 2 ) ^ 2 = k ^ 5 2 = 2 ^ 9 Also det (Q (adj (P))) = |Q| |adj (P)| = k ^ 2 2 (2k) ^ 2 = 2k ^ 4 = 2 ^ 9

JEE Advanced 2016Paper 1

Exam: JEE Advanced
Year: 2016
Paper: Paper 1
Subject: Mathematics
Chapter: Matrices

JEE Advanced 2016 · Paper 1 · Mathematics · Matrices

Let P = [ cases 3 & - 1 & - 2 2 & 0 & α 3 & - 5 & 0 cases ] , where...

Let

\text{P} = \left[

\begin{cases}3 & - 1 & - 2 \\ 2 & 0 & \text{\alpha} \\ 3 & - 5 & 0\end{cases}

\right]

, where

\text{\alpha} \in \text{R},

Suppose

\text{Q} = \left[{\text{q}}_{\text{i}\text{j}}\right]

is a matrix such that

PQ = kI

, where

\text{k} \in \text{R}, \text{k} \ne 0

and

I

is the identity matrix of order 3. If

{\text{q}}_{23} = - \dfrac{\text{k}}{8}

and

\text{det} \left(\text{Q}\right) = \dfrac{{\text{k}}^{2}}{2}

then

Options

A $\alpha = 0, k = 8$
B $4\alpha - k + 8 = 0$
C $det \left(Padj\left(Q\right)\right) = {2}^{9}$
D $det \left(Q(adj\left(P)\right)\right) = {2}^{13}$

Correct answer

B. $4\alpha - k + 8 = 0$

Step-by-step solution

PQ = kI

<br />\left|P\right|\left|Q\right| = {k}^{3}

( as order is 3 )

⇒ \left|P\right| = 2k \ne 0 ⇒ P

is an invertible matrix

∵ PQ = kI

∴ Q = k{P}^{- 1}I = K\dfrac{1}{\left|P\right|}adj P

∴ Q = \dfrac{adj . P}{2}

∵ {q}_{23} = - \dfrac{k}{8}( ∴

cofactor

{P}_{32} = - \left(3\alpha + 4\right) )

⇒ 12\alpha + 16 = K \text{\text{ and } }\dfrac{adjP}{2} = {\dfrac{{\left[{P}_{32}\right]}^{}}{2}}^{T} - {q}_{23}

∴

undefined ...... (i)
Also\left|P\right| = 2k ⇒ k = 10 + 6\alpha

......(ii)<br />Solving these two ⇒ 12\alpha + 16 = 10 + 6\alpha ⇒ \alpha = - 1

K = 4

∴ 4\alpha - k + 8 = 0 ( ⇒ 4 ( - 1 ) - ( 4 ) + 8 = 0 ⇒ - 4 - 4 + 8 = 0 ⇒ - 8 + 8 = 0 )

And

det \left(P\left(adj (Q\right)\right) = \left|P\right|\left|adj (Q)\right| = 2k . {\left(\dfrac{{k}^{2}}{2}\right)}^{2} = \dfrac{{k}^{5}}{2} = {2}^{9}

Also

\text{det} \left(\text{Q} \left(\text{adj} \left(\text{P}\right)\right)\right) = \left|\text{Q}\right| \left|\text{adj} \left(\text{P}\right)\right| = \dfrac{{\text{k}}^{2}}{\text{2}} {\left(\text{2k}\right)}^{2} = {\text{2k}}^{\text{4}} = {\text{2}}^{9}

Practice Matrices on Quantrex Academy →

More from Matrices