If A is a square matrix of order 3 such that det ( A ) = 3 and det ( adj ( - 4 adj ( - 3 adj ( 3 adj ( ( 2 A ) - 1 ) ) ) ) ) = 2 m 3 n , then m + 2 n is equal to : [2024]

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Solution

Q.
If $A$ is a square matrix of order 3 such that $\det (A) = 3$ and $\det (adj (- 4 adj (- 3 adj (3 adj ({(2 A)}^{- 1}))))) = 2^{m} 3^{n},$ then $m + 2 n$ is equal to : [2024]

1 6

2 2

3 3

4 4

Ans.
(4)

We have, $| A | = 3$

Now, $|adj(- 4 adj (- 3 adj (3 adj {(2 A)}^{- 1}))) |$

$=$ $| - 4 adj (- 3 adj (3 adj {(2 A)}^{- 1})) |^{2}$ $[∵ | adj A | = | A |^{n - 1}, n is order of A]$

$= ({{(- 4)}^{3})}^{2}$ $| adj (- 3 adj (3 adj {(2 A)}^{- 1})) |^{2}$

$= 4^{6}$ $| - 3 adj (3 adj {(2 A)}^{- 1}) |^{4}$

$= 4^{6} 3^{12}$ $| adj (3 adj {(2 A)}^{- 1}) |^{4}$

$= 4^{6} 3^{12}$ $| 3 adj {(2 A)}^{- 1} |^{8}$

$= 4^{6} 3^{12} 3^{24}$ $| {(2 A)}^{- 1} |^{16}$

$= 4^{6} 3^{36} 2^{- 48}$ $| A^{- 1} |^{16}$

$= \frac{4^{6} 3^{36}}{2^{48} \cdot 3^{16}} = 2^{- 36} 3^{20}$

$\Rightarrow m = - 36, n = 20$

$\Rightarrow m + 2 n = - 36 + 40 = 4$

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