Match List-I with List-II
| List-I | List-II | ||
| (Matrix A) | (Determinant of Adjoint of A) | ||
| (A) | \(\begin{bmatrix}3 & 1\\4 & 2\end{bmatrix}\) | (I) | 9 |
| (B) | \(\begin{bmatrix}5 & -1\\4 & 2\end{bmatrix}\) | (II) | 8 |
| (C) | \(\begin{bmatrix}6 & -1\\2 & 1\end{bmatrix}\) | (III) | 14 |
| (D) | \(\begin{bmatrix}4 & 1\\3 & 3\end{bmatrix}\) | (IV) | 2 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A)-(IV), (B)-(III), (C)-(II), (D)-(I)
For a square matrix of order n, \(|\mathrm{adj}\,A| = |A|^{n-1}\). For a matrix of order 2 this gives \(|\mathrm{adj}\,A| = |A|\), so each value in List-II is just the determinant of the matrix.
(A) \(\begin{bmatrix}3 & 1\\4 & 2\end{bmatrix}\)
This matches (IV) 2:
\(|A| = 3(2) - 1(4) = 6 - 4 = 2\)
(B) \(\begin{bmatrix}5 & -1\\4 & 2\end{bmatrix}\)
This matches (III) 14:
\(|A| = 5(2) - (-1)(4) = 10 + 4 = 14\)
(C) \(\begin{bmatrix}6 & -1\\2 & 1\end{bmatrix}\)
This matches (II) 8:
\(|A| = 6(1) - (-1)(2) = 6 + 2 = 8\)
(D) \(\begin{bmatrix}4 & 1\\3 & 3\end{bmatrix}\)
This matches (I) 9:
\(|A| = 4(3) - 1(3) = 12 - 3 = 9\)
So the matching is (A)-(IV), (B)-(III), (C)-(II), (D)-(I).