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If A is a square matrix of order nn, then

Match the LIST-I with LIST-II

LIST-ILIST-II
A.∣A(adj A)∣\vert A(\text{adj }A) \vertI.∣A∣n−1\vert A \vert^{n-1}
B.adj(adj A)\text{adj}(\text{adj }A)II.∣A∣I\vert A \vert I, where II is the identity matrix of order nn
C.∣adj A∣\vert \text{adj }A \vertIII.∣A∣n−2A\vert A \vert^{n-2} A
D.A(adj A)A(\text{adj }A)IV.∣A∣n\vert A \vert^n

Choose the correct answer from the options given below:

Solution

✅ Correct Option: 1

For a square matrix of order nn: A(adj A)=∣A∣IA(\text{adj }A)=|A|I, so its determinant is ∣A∣n|A|^n. Also ∣adj A∣=∣A∣n−1|\text{adj }A|=|A|^{n-1} and adj(adj A)=∣A∣n−2A\text{adj}(\text{adj }A)=|A|^{n-2}A. Matching gives A-IV, B-III, C-I, D-II.

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