If A is a non-singular matrix of order n and k is any constant, then
Match List-I with List-II
| List-I | List-II | ||
| (A) | |adj A| | (I) | kn|A|n-1 |
| (B) | |A adj A| | (II) | |A|n-1 |
| (C) | |k adj A| | (III) | \( |A|^{(n-1)^2} \) |
| (D) | |adj(adj A)| | (IV) | |A|n |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A) - (II), (B) - (IV), (C) - (I), (D) - (III)
Everything follows from the two standard results \( \mathrm{A}(\mathrm{adj\,A})=|\mathrm{A}|\,\mathrm{I} \) and \( |k\mathrm{B}|=k^n|\mathrm{B}| \) for a matrix B of order n.
(A) \( |\mathrm{adj\,A}| \)
This matches (II) |A|n-1.
\( \displaystyle |\mathrm{A}|\,|\mathrm{adj\,A}|=\big|\,|\mathrm{A}|\,\mathrm{I}\,\big|=|\mathrm{A}|^n \)
\( \displaystyle |\mathrm{adj\,A}|=|\mathrm{A}|^{n-1} \)
(B) \( |\mathrm{A}\,\mathrm{adj\,A}| \)
This matches (IV) |A|n.
\( \displaystyle |\mathrm{A}\,\mathrm{adj\,A}|=\big|\,|\mathrm{A}|\,\mathrm{I}\,\big|=|\mathrm{A}|^n \)
(C) \( |k\,\mathrm{adj\,A}| \)
This matches (I) kn|A|n-1.
\( \displaystyle |k\,\mathrm{adj\,A}|=k^n\,|\mathrm{adj\,A}| \)
\( \displaystyle =k^n|\mathrm{A}|^{n-1} \)
(D) \( |\mathrm{adj(adj\,A)}| \)
This matches (III) |A|(n-1)2.
The rule \( |\mathrm{adj\,B}|=|\mathrm{B}|^{n-1} \), used for B = adj A (also of order n), gives
\( \displaystyle |\mathrm{adj(adj\,A)}|=|\mathrm{adj\,A}|^{n-1} \)
\( \displaystyle =\left(|\mathrm{A}|^{n-1}\right)^{n-1} \)
\( \displaystyle =|\mathrm{A}|^{(n-1)^2} \)
So the matching is (A)-(II), (B)-(IV), (C)-(I), (D)-(III).