Match List-I with List-II
|
List-I (Matrix A) |
List-II (Determinant of Adjoint of A) |
|
(A) $\begin{bmatrix}3&1\\4&2 |
(I) 9 |
|
(B) $\begin{bmatrix}5&-1\\4&2 |
(II) 8 |
|
(C) $\begin{bmatrix}6&-1\\2&1 |
(III) 14 |
|
(D) $\begin{bmatrix}4&1\\3&3 |
(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 Adjoint of A) |
|
(A) $\begin{bmatrix}3&1\\4&2 |
(IV) 2 |
|
(B) $\begin{bmatrix}5&-1\\4&2 |
(III) 14 |
|
(C) $\begin{bmatrix}6&-1\\2&1 |
(II) 8 |
|
(D) $\begin{bmatrix}4&1\\3&3 |
(I) 9 |
Given: Determinant of adj(A) for a 2×2 matrix is given by:
$\det(\text{adj}(A)) = (\det(A))^{n-1}, \; n=2 \;\;\Rightarrow\;\; \det(\text{adj}(A)) = \det(A)$
Now, compute determinants:
(A) $A=\begin{bmatrix}3 & 1 \\ 4 & 2\end{bmatrix}, \;\det(A)=3(2)-1(4)=6-4=2$ $\;\;\Rightarrow \det(\text{adj}(A))=2$
(B) $A=\begin{bmatrix}5 & -1 \\ 4 & 2\end{bmatrix}, \;\det(A)=5(2)-(-1)(4)=10+4=14$ $\;\;\Rightarrow \det(\text{adj}(A))=14$
(C) $A=\begin{bmatrix}6 & -1 \\ 2 & 1\end{bmatrix}, \;\det(A)=6(1)-(-1)(2)=6+2=8$ $\;\;\Rightarrow \det(\text{adj}(A))=8$
(D) $A=\begin{bmatrix}4 & 1 \\ 3 & 3\end{bmatrix}, \;\det(A)=4(3)-1(3)=12-3=9$ $\;\;\Rightarrow \det(\text{adj}(A))=9$