Match List-I with List-II
| List-I | List-II | ||
| (Matrix A) | (Determinant of adj A) | ||
| (A) | \(\begin{bmatrix}2 & 1\\-1 & 2\end{bmatrix}\) | (I) | 3 |
| (B) | \(\begin{bmatrix}3 & 4\\3 & 6\end{bmatrix}\) | (II) | 6 |
| (C) | \(\begin{bmatrix}3 & 7\\-2 & -4\end{bmatrix}\) | (III) | 5 |
| (D) | \(\begin{bmatrix}4 & 3\\3 & 3\end{bmatrix}\) | (IV) | 2 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A)-(III), (B)-(II), (C)-(IV), (D)-(I)
For a square matrix of order n, \(|\mathrm{adj}\,A| = |A|^{n-1}\). For a 2 × 2 matrix, n − 1 = 1, so \(|\mathrm{adj}\,A| = |A|\). Each item only needs its determinant.
(A) \(\begin{bmatrix}2 & 1\\-1 & 2\end{bmatrix}\)
This matches (III) 5.
\( |A| = 4 - (-1) = 5 \)
(B) \(\begin{bmatrix}3 & 4\\3 & 6\end{bmatrix}\)
This matches (II) 6.
\( |A| = 18 - 12 = 6 \)
(C) \(\begin{bmatrix}3 & 7\\-2 & -4\end{bmatrix}\)
This matches (IV) 2.
\( |A| = -12 - (-14) = 2 \)
(D) \(\begin{bmatrix}4 & 3\\3 & 3\end{bmatrix}\)
This matches (I) 3.
\( |A| = 12 - 9 = 3 \)
So the matching is (A)-(III), (B)-(II), (C)-(IV), (D)-(I).