Match List-I with List-II
|
List-I (Matrix A) |
List-II (Determinant of adj A) |
|
(A) $\begin{bmatrix}2&1\\0&-1 |
(I) 6 |
|
(B) $\begin{bmatrix}0&1\\4&-1 |
(II) 5 |
|
(C) $\begin{bmatrix}1&2\\-3&-1 |
(III) -4 |
|
(D) $\begin{bmatrix}4&-2\\3&0 |
(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)
|
List-I (Matrix A) |
List-II (Determinant of adj A) |
|
(A) $\begin{bmatrix}2&1\\0&-1 |
(IV) - 2 |
|
(B) $\begin{bmatrix}0&1\\4&-1 |
(III) -4 |
|
(C) $\begin{bmatrix}1&2\\-3&-1 |
(II) 5 |
|
(D) $\begin{bmatrix}4&-2\\3&0 |
(I) 6 |
For a 2×2 matrix $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$, $\det(\text{adj }A) = \det(A)$ because $\text{adj }A$ is 2×2 and $\det(\text{adj }A) = \det(A)^{n-1} = \det(A)^{2-1} = \det(A)$
Compute determinants:
(A) $A = \begin{bmatrix}2 & 1 \\ 0 & -1\end{bmatrix}$ → $\det(A) = 2 \cdot (-1) - 0 \cdot 1 = -2$ → $\det(\text{adj }A) = -2$ → (IV)
(B) $A = \begin{bmatrix}0 & 1 \\ 4 & -1\end{bmatrix}$ → $\det(A) = 0\cdot(-1) - 4\cdot1 = -4$ → $\det(\text{adj }A) = -4$ → (III)
(C) $A = \begin{bmatrix}1 & 2 \\ -3 & -1\end{bmatrix}$ → $\det(A) = 1\cdot(-1) - (-3\cdot2) = -1 + 6 = 5$ → $\det(\text{adj }A) = 5$ → (II)
(D) $A = \begin{bmatrix}4 & -2 \\ 3 & 0\end{bmatrix}$ → $\det(A) = 4\cdot0 - 3\cdot(-2) = 0 + 6 = 6$ → $\det(\text{adj }A) = 6$ → (I)