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JEE Main 202324 JanuaryEvening Shift
Exam
JEE Main
Year
2023
Date
24 January
Shift
Evening Shift
Subject
Mathematics
Chapter
Matrices

JEE Main 2023 · 24 January · Evening Shift · Mathematics · Matrices

Let A be a 3 × 3 matrix such that |adj(adj(adj . A))| = 12 ^ 4 ....

Let AA be a 3×33 \times 3 matrix such that \left|adj\left(adj\left(adj . A\right)\right)\right| = {12}^{4}.Then∣A−1adjA∣. Then \left|{A}^{- 1}adjA\right| is equal to

Options

  1. A232\sqrt{3}
  2. B6\sqrt{6}
  3. C1212
  4. D11

Correct answer

A. 232\sqrt{3}

Step-by-step solution

Given equation is

∣adj(adj(adj.A))∣=124\left|adj\left(adj\left(adj . A\right)\right)\right| = {12}^{4}

⇒∣A∣(n−1)3=124⇒ {\left|A\right|}^{( n - 1{)}^{3}} = {12}^{4}

∵n=3∵ n = 3

⇒∣A∣(3−1)3=124⇒ {\left|A\right|}^{( 3 - 1{)}^{3}} = {12}^{4}

⇒∣A∣8=124⇒ {\left|A\right|}^{8} = {12}^{4}

⇒∣A∣2=12⇒ {\left|A\right|}^{2} = 12

⇒∣A∣=23⇒ \left|A\right| = 2\sqrt{3}

Now, \left|{A}^{- 1}adjA\right| = \left|{A}^{- 1}\right| . \left|adjA\right| = \dfrac{1}{\left|A\right|} \times {\left|A\right|}^{3 - 1}</p><p></p><p>= \left|A\right| = 2\sqrt{3}$

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