Let A =6 , where A is a 3 × 3 matrix. If adj (3 adj ( A 2 . adj (2 A ) ) ) =2 m .3 n , m , n in N , then m + n is equal to \ \ \ \ .

Mathematics · Matrices

JEE Main 2026 — 23 January, Morning Shift — Question 21

Let $|\mathrm{A}|=6$ , where A is a $3 \times 3$ matrix. If $\left|\operatorname{adj}\left(3 \operatorname{adj}\left(\mathrm{~A}^{2} . \operatorname{adj}(2 \mathrm{~A})\right)\right)\right|=2^{\mathrm{m}} .3^{\mathrm{n}}, \mathrm{m}, \mathrm{n} \in \mathbf{N}$ , then $\mathrm{m}+\mathrm{n}$ is equal to $\_\_\_\_$ .

Answer: 62

Numerical answer — enter this value.

Step-by-step solution

Given $|A| = 6$ , where $A$ is a $3 \times 3$ matrix.

We need to evaluate:

\left|\operatorname{adj}\left(3 \operatorname{adj}\left(A^{2}\cdot \operatorname{adj}(2A)\right)\right)\right|

Properties used (for $3 \times 3$ matrix):

|\operatorname{adj}(A)| = |A|^{2}

|kA| = k^{3}|A|

\operatorname{adj}(kA) = k^{2}\operatorname{adj}(A)

|AB| = |A||B|

Step 1: Evaluate $\operatorname{adj}(2A)$

\operatorname{adj}(2A) = 2^{2}\operatorname{adj}(A) = 4\operatorname{adj}(A)

Step 2: Compute determinant of $A^{2}\operatorname{adj}(2A)$

A^{2}\operatorname{adj}(2A) = A^{2} \cdot 4\operatorname{adj}(A) = 4A^{2}\operatorname{adj}(A)

|4A^{2}\operatorname{adj}(A)| = 4^{3} |A^{2}| |\operatorname{adj}(A)|

= 4^{3} \cdot |A|^{2} \cdot |A|^{2}

= 4^{3} \cdot 6^{4}

= (2^{2})^{3} (2 \cdot 3)^{4}

= 2^{6} \cdot 2^{4} \cdot 3^{4}

= 2^{10} 3^{4}

Step 3: Determinant of adj

\left|\operatorname{adj}\left(A^{2}\operatorname{adj}(2A)\right)\right| = (2^{10}3^{4})^{2} = 2^{20}3^{8}

Step 4: Multiply by 3

|3X| = 3^{3}|X|

= 3^{3} \cdot 2^{20}3^{8} = 2^{20}3^{11}

Step 5: Final adj

\left|\operatorname{adj}\left(3 \operatorname{adj}\left(A^{2}\operatorname{adj}(2A)\right)\right)\right| = (2^{20}3^{11})^{2} = 2^{40}3^{22}

m = 40, \quad n = 22

m+n = \boxed{62}

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 23 January, Morning Shift
Subject: Mathematics
Chapter: Matrices
Topic: Adjoint of a Square Matrix