If $A =\begin{bmatrix}1&2\\4&5\end{bmatrix}$, then
Match List-I with List-II
|
List-I |
List-II |
|
(A) $\text{det (A)}$ |
(I) $-\frac{1}{3}$ |
|
(B) $det (A^{-1})$ |
(II) -12 |
|
(C) $\text{det (2A)}$ |
(III) -3 |
|
(D) $det (3A^T)$ |
(IV) -27 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A)-(III), (B)-(I), (C)-(II), (D)-(IV)
|
List-I |
List-II |
|
(A) $\text{det (A)}$ |
(III) -3 |
|
(B) $det (A^{-1})$ |
(I) $-\frac{1}{3}$ |
|
(C) $\text{det (2A)}$ |
(II) -12 |
|
(D) $det (3A^T)$ |
(IV) -27 |
Given:
$A = \begin{bmatrix} 1 & 2 \\ 4 & 5 \end{bmatrix}$
$\det(A) = (1)(5) - (2)(4) = 5 - 8 = -3$
⇒ (A) $\det(A) = -3$ → (III)
$\det(A^{-1}) = \frac{1}{\det(A)} = \frac{1}{-3} = -\frac{1}{3}$
⇒ (B) $\det(A^{-1}) = -\frac{1}{3}$ → (I)
$\det(2A) = 2^2 \det(A) = 4(-3) = -12$
⇒ (C) $\det(2A) = -12$ → (II)
$\det(3A^T) = 3^2 \det(A^T) = 9\det(A) = 9(-3) = -27$
⇒ (D) $\det(3A^T) = -27$ → (IV)
Final Matching:
(A) → (III), (B) → (I), (C) → (II), (D) → (IV)