Let P = [ cases 3 & - 1 & - 2 2 & 0 & α 3 & - 5 & 0 cases ], where α ∈ R . Suppose Q = [ q _ ij ] is a matrix satisfying PQ = k I _ 3 for some non-zero k ∈ R . If q _ 23 = - (k)/(8) and |Q| = k ^ 2 2 , then α ^ 2 + k ^

Correct answer: 0 PQ = kI |P| . |Q| = k ^ 3 ⇒ |P| = 2k ≠ 0 ⇒ P is an invertible matrix ∵ PQ = kI ∴ Q = k P ^ - 1 I ∴ Q = (adj . P)/(2) ∵ q _ 23 = - (k)/(8) ∴ (- (3α + 4))/(2) = - (k)/(8) ⇒ k = 4 ∴ |P| = 2k ⇒ k = 10 + 6α (i) Put value of k in (i).. we get α = - 1 ⇒ α ^ 2 + k ^ 2 = 17

JEE Main 202124 FebruaryMorning Shift

Exam: JEE Main
Year: 2021
Date: 24 February
Shift: Morning Shift
Subject: Mathematics
Chapter: Matrices

JEE Main 2021 · 24 February · Morning Shift · Mathematics · Matrices

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

Let

P = \left[

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

\right],

where

\alpha \in R .

Suppose

Q = \left[{q}_{ij}\right]

is a matrix satisfying

PQ = k{I}_{3}

for
some non-zero

k \in R .

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

and \left|Q\right| = \dfrac{{k}^{2}}{2}

, then

{\alpha}^{2} + {k}^{2} is equal to_________.

Correct answer

Step-by-step solution

PQ = kI

<p>\left|P\right| . \left|Q\right| = {k}^{3}

$⇒ \left|P\right| = 2k \ne 0 ⇒ P$ is an invertible matrix

$∵ PQ = kI$

$∴ Q = k{P}^{- 1}I$

$∴ Q = \dfrac{adj . P}{2}$

$∵ {q}_{23} = - \dfrac{k}{8}$

$∴ \dfrac{- \left(3\alpha + 4\right)}{2} = - \dfrac{k}{8} ⇒ k = 4$

$∴ \left|P\right| = 2k ⇒ k = 10 + 6\alpha \ldots \left(i\right)$

Put value of $k in \left(i\right).. we get$ \alpha = - 1 $</p><p>$ ⇒ {\alpha}^{2} + {k}^{2} = 17 $</p>$

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