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If AA is a square matrix of order 3×33 \times 3 and ∣adj A∣=4|adj\ A| = 4, then

Match List-I with List-II

List-IList-II
(A) ∣A−1∣\left\vert A^{-1} \right\vert(I) 108108
(B) ∣3 adjA∣\left\vert 3\,\text{adj}A \right\vert(II) ±12\pm\dfrac{1}{2}
(C) ∣2A∣\left\vert 2A \right\vert(III) ±8\pm8
(D) ∣A.adjA∣\left\vert A.\text{adj}A \right\vert(IV) ±16\pm16

Choose the correct answer from the options given below:

Solution

✅ Correct Option: 1

For order 3, ∣adj A∣=∣A∣2=4⇒∣A∣=±2|adj\,A| = |A|^2 = 4 \Rightarrow |A| = \pm 2.

(A) ∣A−1∣=1∣A∣=±12|A^{-1}| = \frac{1}{|A|} = \pm\frac{1}{2}.

(B) ∣3 adj A∣=33(4)=108|3\,adj\,A| = 3^3(4) = 108.

(C) ∣2A∣=23∣A∣=±16|2A| = 2^3|A| = \pm 16.

(D) ∣A⋅adj A∣=∣∣A∣I∣=∣A∣3=±8|A\cdot adj\,A| = ||A|I| = |A|^3 = \pm 8.

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