Let \(A = \begin{bmatrix} 2 & 3 \\ 1 & 4 \end{bmatrix}\).
Match List-I with List-II
| List-I | List-II | ||
| (A) | \(|A^{-1}|\) | (I) | 5 |
| (B) | \(|\mathrm{adj}\,A|\) | (II) | \(-\displaystyle\frac{3}{5}\) |
| (C) | The sum of the diagonal elements of \(A^{-1}\) | (III) | \(\displaystyle\frac{1}{5}\) |
| (D) | The element in the first row and second column of \(A^{-1}\) | (IV) | \(\displaystyle\frac{6}{5}\) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A)-(III), (B)-(I), (C)-(IV), (D)-(II)
The determinant and the inverse of A are
\(|A| = 2 \times 4 - 3 \times 1 = 5\)
\( \displaystyle A^{-1} = \frac{1}{5}\begin{bmatrix} 4 & -3 \\ -1 & 2 \end{bmatrix} \)
(A) \(|A^{-1}|\)
This matches (III) \(\displaystyle\frac{1}{5}\). Since \(AA^{-1} = I\), \(|A||A^{-1}| = 1\):
\( \displaystyle |A^{-1}| = \frac{1}{|A|} = \frac{1}{5} \)
(B) \(|\mathrm{adj}\,A|\)
This matches (I) 5. For a matrix of order n, \(|\mathrm{adj}\,A| = |A|^{n-1}\), and here n = 2:
\(|\mathrm{adj}\,A| = 5^{1} = 5\)
(C) The sum of the diagonal elements of \(A^{-1}\)
This matches (IV) \(\displaystyle\frac{6}{5}\).
\( \displaystyle \frac{4}{5} + \frac{2}{5} = \frac{6}{5} \)
(D) The element in the first row and second column of \(A^{-1}\)
This matches (II) \(-\displaystyle\frac{3}{5}\), read directly from \(A^{-1}\).