If $A$ is an invertible matrix of order 3, then

Match List-I with List-II

List-I	List-II
(A) $\vert \text{adj} A\vert$	(I) $8\vert A\vert$
(B) $\vert A(\text{adj} A)\vert$	(II) $\vert A\vert ^2$
(C) $\vert 2A\vert$	(III) $\frac{1}{\vert A\vert }$
(D) $\vert A^{-1}\vert$	(IV) $\vert A\vert ^3$

Choose the correct answer from the options given below:

For an invertible matrix of order $n$:

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

$A(\text{adj } A) = |A| \cdot I$

$|kA| = k^n|A|$

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

Since $A$ is of order 3, $n = 3$.

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

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

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

$= |A|^2$

$(A) \rightarrow (II)$

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

Using the identity $A(\text{adj } A) = |A| \cdot I$, taking determinant on both sides:

$|A(\text{adj } A)| = \big||A| \cdot I\big|$

Since $|A|$ is a scalar and $|kI| = k^n$:

$= |A|^3 \cdot |I|$

$= |A|^3 \cdot 1$

$= |A|^3$

$(B) \rightarrow (IV)$

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(C) $|2A|$

$|kA| = k^n |A|$

$|2A| = 2^3 \cdot |A|$

$= 8|A|$

$(C) \rightarrow (I)$

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(D) $|A^{-1}|$

From $A \cdot A^{-1} = I$, taking determinant on both sides:

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

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

$(D) \rightarrow (III)$

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$(A) \rightarrow (II)$

$(B) \rightarrow (IV)$

$(C) \rightarrow (I)$

$(D) \rightarrow (III)$