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Match the LIST-I with LIST-II

LIST-ILIST-II
A. If A is an invertible square matrix then ∣A−1∣=\lvert A^{-1}\rvert =I. 1∣A∣(adj A)\dfrac{1}{\lvert A\rvert}(\mathrm{adj}\,A)
B. For every square matrix (adj A) A =II. 1∣A∣\dfrac{1}{\lvert A\rvert}
C. For a square matrix A, if ∣A∣≠0\lvert A\rvert \neq 0, then A−1=A^{-1} =III. B−1A−1B^{-1}A^{-1}
D. If A and B are invertible matrices, then (AB)−1=(AB)^{-1} =IV. ∣A∣.I\lvert A\rvert.I, where I is the identity matrix of same order as A

Choose the correct answer from the options given below:

Solution

✅ Correct Option: 2

Standard results: ∣A−1∣=1∣A∣|A^{-1}|=\dfrac{1}{|A|} gives A-II.

(adj A)A=∣A∣I(\mathrm{adj}\,A)A = |A|I gives B-IV.

A−1=1∣A∣adj AA^{-1}=\dfrac{1}{|A|}\mathrm{adj}\,A gives C-I.

Reversal law (AB)−1=B−1A−1(AB)^{-1}=B^{-1}A^{-1} gives D-III.

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