Let A = [ 1 2 3 3 1 1 1 2 ] , | A | = 2 . If | 2 ? adj ( 2 ? adj ( 2 A ) ) | = 32 n , then 3 n + is equal to [2023]

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Solution

Q.
Let $A = [\begin{matrix} 1 & 2 & 3 \\ α & 3 & 1 \\ 1 & 1 & 2 \end{matrix}],$ $| A |$ $= 2 .$ If $| 2 adj (2 adj (2 A)) | = 32^{n},$ then $3 n + α$ is equal to [2023]

1 10

2 11

3 9

4 12

Ans.
(2)

We have, $| 2 (adj (2 adj 2 A)) |$

$=$ $| 2 (adj (8 adj A)) |$ $(∵ adj (k A) = k^{n - 1} adj (A))$

$=$ $| 2 (64 adj (adj A)) |$ $=$ $| 128 adj (adj A)) |$

$= {(128)}^{3}$ $| adj (adj A) |$ $= {(128)}^{3} | A |^{4}$

$\Rightarrow {(128)}^{3} {(2)}^{4} = 32^{n} \Rightarrow {(2^{7})}^{3} {(2)}^{4} = {(2^{5})}^{n}$

$\Rightarrow 2^{25} = 2^{5 n} \Rightarrow 25 = 5 n \Rightarrow n = 5$

Also, $| A |$ $= 2 \Rightarrow$ $| \begin{matrix} 1 & 2 & 3 \\ α & 3 & 1 \\ 1 & 1 & 2 \end{matrix} |$ $= 2$

$\Rightarrow 1 (6 - 1) - 2 (2 α - 1) + 3 (α - 3) = 2$

$\Rightarrow 5 - 4 α + 2 + 3 α - 9 = 2 \Rightarrow - α - 4 = 0 \Rightarrow α = - 4$

$So, 3 n + α = 3 (5) - 4 = 11$

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