If \(A =\begin{bmatrix}1&2\\4&5\end{bmatrix}\), then
Match List-I with List-II
| List-I | List-II | ||
|---|---|---|---|
| (A) | \(\det (A)\) | (I) | \(-\displaystyle\frac{1}{3}\) |
| (B) | \(\det (A^{-1})\) | (II) | -12 |
| (C) | \(\det (2A)\) | (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)
A is of order 2, so \(\det(kA) = k^2\det(A)\).
(A) \(\det (A)\)
This matches (III) −3:
\(\det(A) = 1 \times 5 - 2 \times 4 = -3\)
(B) \(\det (A^{-1})\)
This matches (I) −1/3, since \(\det(A^{-1}) = \frac{1}{\det(A)}\):
\(\displaystyle \det(A^{-1}) = \frac{1}{-3} = -\frac{1}{3}\)
(C) \(\det (2A)\)
This matches (II) −12:
\(\det(2A) = 2^2 \times (-3) = -12\)
(D) \(\det (3A^T)\)
This matches (IV) −27, since \(\det(A^T) = \det(A)\):
\(\det(3A^T) = 3^2 \times (-3) = -27\)
So the matching is (A)-(III), (B)-(I), (C)-(II), (D)-(IV).