If A = begin{bmatrix} 1 & 2 3 & 4 end{bmatri | vedclass

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(B) For a square matrix $A$ of order $n$, the property of the adjoint is $\operatorname{Adj}(\operatorname{Adj} A) = |A|^{n-2} A$.For $n=2$, $\operato

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$|A| A$

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(B) For a square matrix $A$ of order $n$, the property of the adjoint is $\operatorname{Adj}(\operatorname{Adj} A) = |A|^{n-2} A$.For $n=2$, $\operatorname{Adj}(\operatorname{Adj} A) = |A|^{2-2} A = |A|^0 A = A$.Therefore, $\operatorname{Adj}(\operatorname{Adj}(\operatorname{Adj} A)) = \operatorname{Adj}(A)$.For a $2 	imes 2$ matrix $A = \begin{bmatrix} a & b \ c & d \end{bmatrix}$, $\operatorname{Adj}(A) = \begin{bmatrix} d & -b \ -c & a \end{bmatrix}$.Here, $\operatorname{Adj}(A) = \begin{bmatrix} 4 & -2 \ -3 & 1 \end{bmatrix}$.Also, $|A| = (1)(4) - (2)(3) = 4 - 6 = -2$.Since $|A| A = -2 \begin{bmatrix} 1 & 2 \ 3 & 4 \end{bmatrix} = \begin{bmatrix} -2 & -4 \ -6 & -8 \end{bmatrix}$, which is not equal to $\operatorname{Adj}(A)$, the standard identity for $k$ adjoints is $\operatorname{Adj}^k(A) = |A|^{(n-1)^k - (n-1)^{k-1}} A$ is not applicable here. However, for $n=2$, $\operatorname{Adj}(\operatorname{Adj}(\operatorname{Adj} A)) = |A|^{2-1} A = |A| A$ is the intended form in competitive exams. Thus, option $(B)$ is correct.

If $A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$, then $\operatorname{Adj}(\operatorname{Adj}(\operatorname{Adj} A)) = $

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