If $A$ and $B$ are invertible matrices of order $3$ then match List-I with List-II

List-I	List-II
(A) $\text{adj}(A)$	(I) $B^{-1} A^{-1}$
(B) $(AB)^{-1}$	(II) $\vert A\vert ^{-1}$
(C) $\vert A^{-1}\vert$	(III) $\vert A\vert ^2$
(D) $\vert \text{adj} A\vert$	(IV) $\vert A\vert A^{-1}$

Choose the correct answer from the options given below:

For an invertible matrix $A$ of order $n$:

$A^{-1} = \dfrac{\text{adj}(A)}{|A|}$, i.e., $\text{adj}(A) = |A| \cdot A^{-1}$

$(AB)^{-1} = B^{-1}A^{-1}$

$|A^{-1}| = |A|^{-1}$

$|\text{adj}(A)| = |A|^{n-1}$

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$\text{adj}(A)$

From the inverse formula: $A^{-1} = \dfrac{\text{adj}(A)}{|A|}$

$\text{adj}(A) = |A| \cdot A^{-1}$

$(A) \longrightarrow (IV)$

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$(AB)^{-1}$

$(AB)^{-1} = B^{-1}A^{-1}$

$(B) \longrightarrow (I)$

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$|A^{-1}|$

Taking determinant on both sides of $A \cdot A^{-1} = I$:

$|A| \cdot |A^{-1}| = |I| = 1$

$|A^{-1}| = \dfrac{1}{|A|}$

$= |A|^{-1}$

$(C) \longrightarrow (II)$

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$|\text{adj}(A)|$

Since $\text{adj}(A) = |A| \cdot A^{-1}$, taking determinant of both sides with $n = 3$:

$|\text{adj}(A)| = \big||A| \cdot A^{-1}\big|$

$= |A|^3 \cdot |A^{-1}|$

$= |A|^3 \cdot \dfrac{1}{|A|}$

$= |A|^2$

$(D) \longrightarrow (III)$

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$(A) \rightarrow (IV)$, $(B) \rightarrow (I)$, $(C) \rightarrow (II)$, $(D) \rightarrow (III)$