Let and A = . If B = is the matrix of cofactors of the elements of A , then d e t ( A B ) is equal to : [2024]

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Solution

Q.
Let $α β \neq 0$ and $A = [\begin{matrix} β & α & 3 \\ α & α & β \\ - β & α & 2 α \end{matrix}] .$ If $B = [\begin{matrix} 3 α & - 9 & 3 α \\ - α & 7 & - 2 α \\ - 2 α & 5 & - 2 β \end{matrix}]$ is the matrix of cofactors of the elements of $A,$ then $d e t (A B)$ is equal to : [2024]

1 125

2 64

3 343

4 216

Ans.
(4)

Clearly $B = {(adj A)}^{T}$

Now, $| A B |$ $= | A |$ $| {(adj A)}^{T} |$

   $= | A | | adj A |$

$= | A | | A |^{2}$

   $= | A |^{3}$

Now, $A = [\begin{matrix} β & α & 3 \\ α & α & β \\ - β & α & 2 α \end{matrix}]$ and ${(adj A)}^{T} = [\begin{matrix} 3 α & - 9 & 3 α \\ - α & 7 & - 2 α \\ - 2 α & 5 & - 2 β \end{matrix}]$

$⇒Cofactor of a_{11} = (2 α^{2} - α β) = 3 α$

$Cofactor of a_{21} = (2 α^{2} - 3 α) = {(- 1)}^{2 + 1} (- α)$

$\Rightarrow 2 α^{2} - 4 α = 0 \Rightarrow α = 0, 2$

$\Rightarrow β = 1 for α = 2$

$∴$    $| A |$ $= [\begin{matrix} 1 & 2 & 3 \\ 2 & 2 & 1 \\ - 1 & 2 & 4 \end{matrix}] = 6 - 18 + 18 = 6$

$\Rightarrow | A B | = 6^{3} = 216$

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