Match List-I with List-II
| List-I (Matrix) | List-II (Inverse) | ||
| (A) | \(\begin{bmatrix}2 & 0\\ 0 & 4\end{bmatrix}\) | (I) | \(\begin{bmatrix}2 & -1\\ -5 & 3\end{bmatrix}\) |
| (B) | \(\begin{bmatrix}1 & 3\\ 0 & 1\end{bmatrix}\) | (II) | \(\begin{bmatrix}0 & 1\\ 1 & 0\end{bmatrix}\) |
| (C) | \(\begin{bmatrix}0 & 1\\ 1 & 0\end{bmatrix}\) | (III) | \(\begin{bmatrix}\frac{1}{2} & 0\\ 0 & \frac{1}{4}\end{bmatrix}\) |
| (D) | \(\begin{bmatrix}3 & 1\\ 5 & 2\end{bmatrix}\) | (IV) | \(\begin{bmatrix}1 & -3\\ 0 & 1\end{bmatrix}\) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A)-(III), (B)-(IV), (C)-(II), (D)-(I)
For a 2 × 2 matrix, \(\displaystyle A^{-1} = \frac{1}{|A|}\,\mathrm{adj}\,A\), where the adjoint swaps the diagonal entries and changes the signs of the other two.
(A) \(\begin{bmatrix}2 & 0\\ 0 & 4\end{bmatrix}\)
This matches (III) \(\begin{bmatrix}\frac{1}{2} & 0\\ 0 & \frac{1}{4}\end{bmatrix}\).
\( \displaystyle |A| = 8,\quad A^{-1} = \frac{1}{8}\begin{bmatrix}4 & 0\\ 0 & 2\end{bmatrix} = \begin{bmatrix}\frac{1}{2} & 0\\ 0 & \frac{1}{4}\end{bmatrix} \)
(B) \(\begin{bmatrix}1 & 3\\ 0 & 1\end{bmatrix}\)
This matches (IV) \(\begin{bmatrix}1 & -3\\ 0 & 1\end{bmatrix}\).
\( \displaystyle |A| = 1,\quad A^{-1} = \begin{bmatrix}1 & -3\\ 0 & 1\end{bmatrix} \)
(C) \(\begin{bmatrix}0 & 1\\ 1 & 0\end{bmatrix}\)
This matches (II) \(\begin{bmatrix}0 & 1\\ 1 & 0\end{bmatrix}\).
\( \displaystyle |A| = -1,\quad A^{-1} = \frac{1}{-1}\begin{bmatrix}0 & -1\\ -1 & 0\end{bmatrix} = \begin{bmatrix}0 & 1\\ 1 & 0\end{bmatrix} \)
This matrix is its own inverse: multiplying it by itself gives I.
(D) \(\begin{bmatrix}3 & 1\\ 5 & 2\end{bmatrix}\)
This matches (I) \(\begin{bmatrix}2 & -1\\ -5 & 3\end{bmatrix}\).
\( \displaystyle |A| = 6 - 5 = 1,\quad A^{-1} = \begin{bmatrix}2 & -1\\ -5 & 3\end{bmatrix} \)