Let A be a square matrix of order 3 such that det(A) = – 2 and det(3 adj(– 6 adj(3A)) = , m > n. Then 4m + 2n is equal to __________. [2025]

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Solution

Q.
Let A be a square matrix of order 3 such that det(A) = – 2 and det(3 adj(– 6 adj(3A))) = $2^{m + n} \cdot 3^{m n}$ , m > n. Then 4m + 2n is equal to __________. [2025]

Ans.
(34)

We have, $| A |$ = – 2 and det(3 adj(– 6 adj(3A))) = $2^{m + n} \cdot 3^{m n}$

$\Rightarrow 3^{3} \det (adj (- 6 adj (3 A))) = 2^{m + n} \cdot 3^{m n}$

$\Rightarrow 3^{3} | - 6 adj (3 A) |^{2} = 2^{m + n} \cdot 3^{m n}$

$\Rightarrow 3^{3} ({{(- 6)}^{3})}^{2}$ $| adj (3 A) |^{2}$ $= 2^{m + n} \cdot 3^{m n}$

$\Rightarrow 3^{3} {(6)}^{6}$ $| {(3 A)}^{2} |^{2}$ $= 2^{m + n} \cdot 3^{m n}$

$\Rightarrow 3^{3} \times 6^{6} \times {(3^{3})}^{4} | A |^{4} = 2^{m + n} \cdot 3^{m n}$

$\Rightarrow 3^{21} \times 2^{6} \times {(- 2)}^{4} = 2^{m + n} \cdot 3^{m n}$ $\Rightarrow 3^{21} \times 2^{10} = 2^{m + n} \cdot 3^{m n}$

On comparing the powers, we get m + n = 10 and mn = 21

$∴$ m = 7 and n = 3

$∴$ Value of $4 m + 2 n = 4 \times 7 + 2 \times 3 = 34$ .

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