Let \(A\) be a non-singular square matrix of order \(n\), then
Match List-I with List-II
| List-I | List-II | ||
| (A) | \(A(\mathrm{adj}\ A)\) | (I) | \(\displaystyle\frac{1}{|A|}\) |
| (B) | \(|\mathrm{adj}\ A|\) | (II) | \(|A|^n\) |
| (C) | \(|A^{-1}|\) | (III) | \(|A|I\) |
| (D) | \(|A(\mathrm{adj}\ A)|\) | (IV) | \(|A|^{n-1}\) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A)-(III), (B)-(IV), (C)-(I), (D)-(II)
All four items follow from the basic property of the adjoint, \(A(\mathrm{adj}\ A) = (\mathrm{adj}\ A)A = |A|I\), and from \(|kI| = k^n\) for a matrix of order n.
(A) \(A(\mathrm{adj}\ A)\)
This matches (III) \(|A|I\), since \(A(\mathrm{adj}\ A) = |A|I\).
(B) \(|\mathrm{adj}\ A|\)
This matches (IV) \(|A|^{n-1}\). Taking determinants in \(A(\mathrm{adj}\ A) = |A|I\):
\(|A|\,|\mathrm{adj}\ A| = |A|^n\)
A is non-singular, so \(|A| \ne 0\), and dividing by \(|A|\) gives \(|\mathrm{adj}\ A| = |A|^{n-1}\).
(C) \(|A^{-1}|\)
This matches (I) \(\frac{1}{|A|}\). Taking determinants in \(AA^{-1} = I\):
\(|A|\,|A^{-1}| = |I| = 1\)
\( \displaystyle |A^{-1}| = \frac{1}{|A|} \)
(D) \(|A(\mathrm{adj}\ A)|\)
This matches (II) \(|A|^n\):
\(|A(\mathrm{adj}\ A)| = \big|\,|A|I\,\big|\)
\(= |A|^n\,|I| = |A|^n\)
So the matching is (A)-(III), (B)-(IV), (C)-(I), (D)-(II).