Let and A = [ 1 2 1 0 1 0 1 2 ] . If det ( adj ( 2 A - A T ) ) · adj ( A - 2 A T ) = 2 8 , then ( det ( A ) ) 2 is equal to [2024]

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Solution

Q.
Let $α \in (0, \infty)$ and $A = [\begin{matrix} 1 & 2 & α \\ 1 & 0 & 1 \\ 0 & 1 & 2 \end{matrix}] .$ If $\det (adj (2 A - A^{T})) \cdot adj (A - 2 A^{T}) = 2^{8},$ then $(\det {(A))}^{2}$ is equal to [2024]

1 1

2 16

3 36

4 49

Ans.
(2)

We have, $| adj (2 A - A^{T}) \cdot adj (A - 2 A^{T}) | = 2^{8}$

$\Rightarrow | adj (A - 2 A^{T}) \cdot (2 A - A^{T}) | = 2^{8}$ $(∵ adj (A B) = adj B \cdot adj A)$

$\Rightarrow | adj {- ((2 A^{T} - A) (2 A - A^{T}))} | = 2^{8}$

$\Rightarrow | adj {- ({(2 A - A^{T})}^{T} (2 A - A^{T}))} | = 2^{8}$

$\Rightarrow | adj {- (P^{T} P)} | = 2^{8}, where P = 2 A - A^{T}$

$\Rightarrow | - P^{T} \cdot P |^{2} = 2^{8} \Rightarrow | - P^{T} |^{2} | P |^{2} = 2^{8}$

$\Rightarrow | P |^{4} = 4^{4}$ $(∵ | P^{T} | = | P |)$

$\Rightarrow | P | = \pm 4 \Rightarrow | 2 A - A^{T} | = \pm 4 \Rightarrow$ $| \begin{matrix} 1 & 3 & 2 α \\ 0 & 1 & 1 \\ - α & 1 & 2 \end{matrix} |$ $= \pm 4$

$\Rightarrow 1 (0 - 1) - 3 (0 + α) + 2 α (0) = \pm 4 \Rightarrow - 1 - 3 α = \pm 4$

$\Rightarrow α = 1, \frac{- 5}{3} \Rightarrow α = 1$ $[∵ α \in (0, \infty)]$

Now, $| A | = 1 (0 - 1) - 2 (2 - 0) + α (1 - 0) = α - 5$

$∴ | A |^{2} = 16$

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