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If A=[1111031−21]A=\begin{bmatrix}1 & 1 & 1\\1 & 0 & 3\\1 & -2 & 1\end{bmatrix}, then

Match List-I with List-II

List-IList-II
(A) ∣A∣\lvert A\rvert(I) 48
(B) ∣2A∣\lvert 2A\rvert(II) 36
(C) ∣adj⁡(A)∣\left\lvert\operatorname{adj}(A)\right\rvert(III) 12
(D) 2∣A∣2\lvert A\rvert(IV) 6

Choose the correct answer from the options given below:

Solution

✅ Correct Option: 1

∣A∣=1(0+6)−1(1−3)+1(−2−0)=6|A|=1(0+6)-1(1-3)+1(-2-0)=6, so (A)-(IV).

For order 33, ∣2A∣=23∣A∣=48|2A|=2^3|A|=48, so (B)-(I); ∣adjA∣=∣A∣2=36|adjA|=|A|^{2}=36, so (C)-(II); and 2∣A∣=122|A|=12, so (D)-(III).

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