Match List-I with List-II
| List-I | List-II | ||
| (Matrix A) | (Determinant of adj A) | ||
| (A) | \(\begin{bmatrix}2&1\\0&-1\end{bmatrix}\) | (I) | 6 |
| (B) | \(\begin{bmatrix}0&1\\4&-1\end{bmatrix}\) | (II) | 5 |
| (C) | \(\begin{bmatrix}1&2\\-3&-1\end{bmatrix}\) | (III) | −4 |
| (D) | \(\begin{bmatrix}4&-2\\3&0\end{bmatrix}\) | (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 order 2 this gives \(|\mathrm{adj}\,A| = |A|\), so each List-II value is simply the determinant of A.
(A) \(\begin{bmatrix}2&1\\0&-1\end{bmatrix}\)
This matches (IV) −2.
\(|A| = 2(-1) - 1(0) = -2\)
(B) \(\begin{bmatrix}0&1\\4&-1\end{bmatrix}\)
This matches (III) −4.
\(|A| = 0(-1) - 1(4) = -4\)
(C) \(\begin{bmatrix}1&2\\-3&-1\end{bmatrix}\)
This matches (II) 5.
\(|A| = 1(-1) - 2(-3) = -1 + 6 = 5\)
(D) \(\begin{bmatrix}4&-2\\3&0\end{bmatrix}\)
This matches (I) 6.
\(|A| = 4(0) - (-2)(3) = 6\)
So the matching is (A)-(IV), (B)-(III), (C)-(II), (D)-(I).