Match List-I with List-II
| List-I | List-II | ||
| (A) | \( \begin{bmatrix}3 & -1\\4 & 1\end{bmatrix}^{-1} \) | (I) | \( \begin{bmatrix}36 & 0\\-9 & 18\end{bmatrix} \) |
| (B) | \( \begin{bmatrix}-1 & 2\\1 & -1\end{bmatrix}^{T} \) | (II) | \( \begin{bmatrix}36 & -9\\0 & 18\end{bmatrix} \) |
| (C) | \( 9\begin{bmatrix}4 & -1\\0 & 2\end{bmatrix} \) | (III) | \( \displaystyle\frac{1}{7}\begin{bmatrix}1 & 1\\-4 & 3\end{bmatrix} \) |
| (D) | \( \left(9\begin{bmatrix}4 & -1\\0 & 2\end{bmatrix}\right)^{T} \) | (IV) | \( \begin{bmatrix}-1 & 1\\2 & -1\end{bmatrix} \) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A) - (III), (B) - (IV), (C) - (II), (D) - (I)
(A) \( \begin{bmatrix}3 & -1\\4 & 1\end{bmatrix}^{-1} \)
This matches (III) \( \frac{1}{7}\begin{bmatrix}1 & 1\\-4 & 3\end{bmatrix} \). For a 2 × 2 matrix, swap the diagonal entries, change the signs of the other two and divide by the determinant:
\( \displaystyle |\mathrm{M}|=3(1)-(-1)(4)=7 \)
\( \displaystyle \mathrm{M}^{-1}=\frac{1}{7}\begin{bmatrix}1 & 1\\-4 & 3\end{bmatrix} \)
(B) \( \begin{bmatrix}-1 & 2\\1 & -1\end{bmatrix}^{T} \)
This matches (IV) \( \begin{bmatrix}-1 & 1\\2 & -1\end{bmatrix} \). The transpose turns rows into columns.
(C) \( 9\begin{bmatrix}4 & -1\\0 & 2\end{bmatrix} \)
This matches (II) \( \begin{bmatrix}36 & -9\\0 & 18\end{bmatrix} \). Every entry is multiplied by 9.
(D) \( \left(9\begin{bmatrix}4 & -1\\0 & 2\end{bmatrix}\right)^{T} \)
This matches (I) \( \begin{bmatrix}36 & 0\\-9 & 18\end{bmatrix} \). It is the transpose of the matrix in (C).
So the matching is (A)-(III), (B)-(IV), (C)-(II), (D)-(I).