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If A and B are invertible matrices of the same order, then,

Match List-I with List-II

LIST-ILIST-II
A. ∣AT∣\left\vert A^T \right\vertI. B−1A−1B^{-1}A^{-1}
B. A(adj A)A(\text{adj }A)II. adj A\text{adj }A
C. A−1∣A∣A^{-1}\left\vert A \right\vertIII. (adj A)A(\text{adj }A)A
D. (AB)−1(AB)^{-1}IV. ∣A∣\left\vert A \right\vert

Choose the correct answer from the options given below:

Solution

✅ Correct Option: 2

Transposing leaves the determinant unchanged, so ∣AT∣=∣A∣|A^T|=|A| (IV).

Also A(adj A)=(adj A)AA(\text{adj }A)=(\text{adj }A)A (III), ∣A∣A−1=adj A|A|A^{-1}=\text{adj }A (II),

and (AB)−1=B−1A−1(AB)^{-1}=B^{-1}A^{-1} (I).

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