Let \(A = \begin{bmatrix}3 & 2\\ 5 & 4\end{bmatrix}\).
Match List-I with List-II
| List-I | List-II | ||
| (A) | \(\mathrm{adj}\,A\) | (I) | \(\begin{bmatrix}3 & 2\\ 5 & 4\end{bmatrix}\) |
| (B) | \(A^{-1}\) | (II) | \(\begin{bmatrix}4 & -2\\ -5 & 3\end{bmatrix}\) |
| (C) | \((A^{T})^{-1}\) | (III) | \(\begin{bmatrix}2 & -\frac{5}{2}\\ -1 & \frac{3}{2}\end{bmatrix}\) |
| (D) | \(\mathrm{adj}(\mathrm{adj}\,A)\) | (IV) | \(\begin{bmatrix}2 & -1\\ -\frac{5}{2} & \frac{3}{2}\end{bmatrix}\) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A)-(II), (B)-(IV), (C)-(III), (D)-(I)
The determinant of A:
\( |A| = 3 \times 4 - 2 \times 5 = 2 \)
(A) \(\mathrm{adj}\,A\)
This matches (II) \(\begin{bmatrix}4 & -2\\ -5 & 3\end{bmatrix}\).
Swap the diagonal entries 3 and 4, and change the signs of 2 and 5.
(B) \(A^{-1}\)
This matches (IV) \(\begin{bmatrix}2 & -1\\ -\frac{5}{2} & \frac{3}{2}\end{bmatrix}\).
\( \displaystyle A^{-1} = \frac{1}{2}\begin{bmatrix}4 & -2\\ -5 & 3\end{bmatrix} = \begin{bmatrix}2 & -1\\ -\frac{5}{2} & \frac{3}{2}\end{bmatrix} \)
(C) \((A^{T})^{-1}\)
This matches (III) \(\begin{bmatrix}2 & -\frac{5}{2}\\ -1 & \frac{3}{2}\end{bmatrix}\).
Since \((A^{T})^{-1} = (A^{-1})^{T}\), it is the transpose of \(A^{-1}\): the entries −1 and \(-\frac{5}{2}\) change places.
(D) \(\mathrm{adj}(\mathrm{adj}\,A)\)
This matches (I) \(\begin{bmatrix}3 & 2\\ 5 & 4\end{bmatrix}\).
Taking the adjoint of \(\begin{bmatrix}4 & -2\\ -5 & 3\end{bmatrix}\): swap 4 and 3, change the signs of −2 and −5. This gives back A, since for every 2 × 2 matrix adj(adj A) = A.