Match List-I with List-II
| List-I | List-II | ||
| (A) | If A is a non-singular matrix of order n, then \(|A\,(\mathrm{adj}\,A)|\) is equal to | (I) | \(|A|^{n-1}\) |
| (B) | If A is a non-singular matrix of order n, then \(|\mathrm{adj}\,(\mathrm{adj}\,A)|\) is equal to | (II) | \(|A|^{n-2}A\) |
| (C) | If A is a non-singular matrix of order n, then \(\mathrm{adj}\,(\mathrm{adj}\,A)\) is equal to | (III) | \(|A|^{n}\) |
| (D) | If A is a non-singular matrix of order n, then \(|(\mathrm{adj}\,A)|\) is equal to | (IV) | \(|A|^{(n-1)^2}\) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A)-(III), (B)-(IV), (C)-(II), (D)-(I)
Everything follows from the basic identity \(A\,(\mathrm{adj}\,A) = |A|\,I_n\), with \(|A| \ne 0\). (D) is taken before (B), because (B) uses it.
(A) \(|A\,(\mathrm{adj}\,A)|\)
This matches (III) \(|A|^{n}\). Since \(|kI_n| = k^n\):
\(|A\,(\mathrm{adj}\,A)| = \big|\,|A|\,I_n\big| = |A|^{n}\)
(D) \(|(\mathrm{adj}\,A)|\)
This matches (I) \(|A|^{n-1}\). From (A), \(|A|\,|\mathrm{adj}\,A| = |A|^{n}\); dividing by \(|A|\):
\(|\mathrm{adj}\,A| = |A|^{n-1}\)
(B) \(|\mathrm{adj}\,(\mathrm{adj}\,A)|\)
This matches (IV) \(|A|^{(n-1)^2}\). The matrix adj A is also non-singular of order n, so the result of (D) applies to it:
\(|\mathrm{adj}\,(\mathrm{adj}\,A)| = |\mathrm{adj}\,A|^{n-1}\)
\(|\mathrm{adj}\,(\mathrm{adj}\,A)| = \left(|A|^{n-1}\right)^{n-1} = |A|^{(n-1)^2}\)
(C) \(\mathrm{adj}\,(\mathrm{adj}\,A)\)
This matches (II) \(|A|^{n-2}A\). The basic identity for the matrix adj A gives
\((\mathrm{adj}\,A)\,\mathrm{adj}\,(\mathrm{adj}\,A) = |\mathrm{adj}\,A|\,I_n = |A|^{n-1}I_n\)
Multiplying on the left by A and using \(A\,(\mathrm{adj}\,A) = |A|\,I_n\):
\(|A|\,\mathrm{adj}\,(\mathrm{adj}\,A) = |A|^{n-1}A\)
\(\mathrm{adj}\,(\mathrm{adj}\,A) = |A|^{n-2}A\)
So the matching is (A)-(III), (B)-(IV), (C)-(II), (D)-(I).