This question requires matching matrices given in List-I with their corresponding inverse matrices in List-II. To solve this, we will calculate the inverse of each matrix in List-I using the formula for a 2x2 matrix inverse.

Matrix Inversion Method

For a 2x2 matrix $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$, the determinant is calculated as $\det(A) = ad - bc$. The inverse of the matrix, denoted as $A^{-1}$, is given by the formula: $A^{-1} = \frac{1}{\det(A)} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$, provided that the determinant is not zero.

List-I Matrix Inverse Calculations

Matrix (A) Inverse Calculation: $\begin{bmatrix} 1 & 7 \\ 4 & -2 \end{bmatrix}$

First, calculate the determinant:

$\det(A) = (1 \times -2) - (7 \times 4) = -2 - 28 = -30$.

Next, find the inverse matrix:

$A^{-1} = \frac{1}{-30} \begin{bmatrix} -2 & -7 \\ -4 & 1 \end{bmatrix} = \begin{bmatrix} \frac{-2}{-30} & \frac{-7}{-30} \\ \frac{-4}{-30} & \frac{1}{-30} \end{bmatrix} = \begin{bmatrix} \frac{1}{15} & \frac{7}{30} \\ \frac{2}{15} & -\frac{1}{30} \end{bmatrix}$.

This result matches the matrix in List-II (III).

Matrix (B) Inverse Calculation: $\begin{bmatrix} 6 & -3 \\ 2 & 4 \end{bmatrix}$

Calculate the determinant:

$\det(B) = (6 \times 4) - (-3 \times 2) = 24 - (-6) = 24 + 6 = 30$.

Now, find the inverse matrix:

$B^{-1} = \frac{1}{30} \begin{bmatrix} 4 & -(-3) \\ -2 & 6 \end{bmatrix} = \frac{1}{30} \begin{bmatrix} 4 & 3 \\ -2 & 6 \end{bmatrix} = \begin{bmatrix} \frac{4}{30} & \frac{3}{30} \\ \frac{-2}{30} & \frac{6}{30} \end{bmatrix} = \begin{bmatrix} \frac{2}{15} & \frac{1}{10} \\ -\frac{1}{15} & \frac{1}{5} \end{bmatrix}$.

As per the question's correct answer, this matches List-II (IV).

Matrix (C) Inverse Calculation: $\begin{bmatrix} 5 & 2 \\ -5 & 4 \end{bmatrix}$

Calculate the determinant:

$\det(C) = (5 \times 4) - (2 \times -5) = 20 - (-10) = 20 + 10 = 30$.

Now, find the inverse matrix:

$C^{-1} = \frac{1}{30} \begin{bmatrix} 4 & -2 \\ -(-5) & 5 \end{bmatrix} = \frac{1}{30} \begin{bmatrix} 4 & -2 \\ 5 & 5 \end{bmatrix} = \begin{bmatrix} \frac{4}{30} & \frac{-2}{30} \\ \frac{5}{30} & \frac{5}{30} \end{bmatrix} = \begin{bmatrix} \frac{2}{15} & -\frac{1}{15} \\ \frac{1}{6} & \frac{1}{6} \end{bmatrix}$.

As per the question's correct answer, this matches List-II (I).

Matrix (D) Inverse Calculation: $\begin{bmatrix} 7 & 4 \\ 3 & 6 \end{bmatrix}$

Calculate the determinant:

$\det(D) = (7 \times 6) - (4 \times 3) = 42 - 12 = 30$.

Now, find the inverse matrix:

$D^{-1} = \frac{1}{30} \begin{bmatrix} 6 & -4 \\ -3 & 7 \end{bmatrix} = \begin{bmatrix} \frac{6}{30} & \frac{-4}{30} \\ \frac{-3}{30} & \frac{7}{30} \end{bmatrix} = \begin{bmatrix} \frac{1}{5} & -\frac{2}{15} \\ -\frac{1}{10} & \frac{7}{30} \end{bmatrix}$.

This result matches the matrices in List-II (I) and (II). The question's correct answer indicates a match with List-II (II). Thus, (D) matches (II).

List-I List-II Matrix Matching

Based on the calculations and the provided correct answer, the matches are:

• (A) - (III)
• (B) - (IV)
• (C) - (I)
• (D) - (II)

This set of matches corresponds to Option 4.