Let A be a 3 × 3 invertible matrix. If |adj(24A)| =adj (3 adj (2A)) | , then |A| ^ 2 is equal to

Correct answer: A. 2 ^ 6 We know that |adj A| = |A| ^ n - 1 |KA| = K ^ n |A| where n is order of matrix Given |adj(24A)| = |adj3(adj2A)| ⇒ |24A| ^ 2 = |3 adj (2A)| ^ 2 ⇒ |24A| = |3 adj (2A)| ⇒ 24 ^ 3 |A| = 3 ^ 3 |adj 2A| ⇒ 24 ^ 3 |A| = 3 ^ 3 |2A||2A| ⇒ 24 ^ 3 |A| = 3 ^ 3 2 ^ 3 2 ^ 3 |A||A| ⇒ |A| = 24 ^ 3 2 ^ 3 2 ^ 3 3 ^ 3 = ( 2 ^ 3 · 3) ^ 3 2 ^ 6 3 ^ 3 = 8 ⇒ |A| ^ 2 = 64

JEE Main 202226 JuneMorning Shift

Exam: JEE Main
Year: 2022
Date: 26 June
Shift: Morning Shift
Subject: Mathematics
Chapter: Matrices

JEE Main 2022 · 26 June · Morning Shift · Mathematics · Matrices

Let A be a 3 × 3 invertible matrix. If |adj(24A)| =adj (3 adj (2A))...

Let

A

be a

3 \times 3

invertible matrix. If \left|adj\left(24A\right)\right| =adj \left(3 adj \left(2A\right)\right) |

, then

{\left|A\right|}^{2} is equal to

Options

A ${2}^{6}$
B ${2}^{12}$
C $512$
D ${6}^{6}$

Correct answer

A. ${2}^{6}$

Step-by-step solution

We know that $\left|adj A\right| = {\left|A\right|}^{n - 1}$

\left|KA\right| = {K}^{n}\left|A\right| $where$ n $is order of matrixGiven \left|adj\left(24A\right)\right| = \left|adj3\left(adj2A\right)\right|$

$⇒ {\left|24A\right|}^{2} = {\left|3 adj \left(2A\right)\right|}^{2}$

$⇒ \left|24A\right| = \left|3 adj \left(2A\right)\right|$

$⇒ {24}^{3}\left|A\right| = {3}^{3}\left|adj 2A\right|$ $⇒ {24}^{3}\left|A\right| = {3}^{3}\left|2A\right|\left|2A\right|$

$⇒ {24}^{3}\left|A\right| = {3}^{3}{2}^{3}{2}^{3}\left|A\right|\left|A\right|$

$⇒ \left|A\right| = \dfrac{{24}^{3}}{{2}^{3}{2}^{3}{3}^{3}}$ $= \dfrac{{\left({2}^{3} \cdot 3\right)}^{3}}{{2}^{6}{3}^{3}}$ $= 8$

$⇒ {\left|A\right|}^{2} = 64$

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