The given equation is:

$(7A)^{-1} = \begin{bmatrix} -3 & 7 \\ 1 & -2 \end{bmatrix}$

We need to find matrix $A$.

Step-by-step solution:

1. Take the inverse of both sides to eliminate the inverse on the left-hand side: $7A = \left( \begin{bmatrix} -3 & 7 \\ 1 & -2 \end{bmatrix} \right)^{-1}$

2. Find the inverse of the given matrix: $\text{Matrix} = \begin{bmatrix} -3 & 7 \\ 1 & -2 \end{bmatrix}$

   The formula to find the inverse of a 2x2 matrix $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$ is: $\frac{1}{ad - bc} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$ For the matrix $\begin{bmatrix} -3 & 7 \\ 1 & -2 \end{bmatrix}$, we compute the determinant first: $\text{Determinant} = (-3)(-2) - (7)(1) = 6 - 7 = -1$

   So, the inverse is: $\frac{1}{-1} \begin{bmatrix} -2 & -7 \\ -1 & -3 \end{bmatrix} = \begin{bmatrix} 2 & 7 \\ 1 & 3 \end{bmatrix}$

3. Substitute the inverse into the equation: $7A = \begin{bmatrix} 2 & 7 \\ 1 & 3 \end{bmatrix}$

4. Solve for $A$ by dividing both sides by 7 (or multiplying by $\frac{1}{7}$): $A = \frac{1}{7} \begin{bmatrix} 2 & 7 \\ 1 & 3 \end{bmatrix} = \begin{bmatrix} \frac{2}{7} & 1 \\ \frac{1}{7} & \frac{3}{7} \end{bmatrix}$

Thus, the matrix $A$ is:

$A = \begin{bmatrix} \frac{2}{7} & 1 \\ \frac{1}{7} & \frac{3}{7} \end{bmatrix}$

Let me know if you need further details!

Related questions:

1. How do you calculate the determinant of a 2x2 matrix?
2. What is the general formula for the inverse of a 2x2 matrix?
3. How does the determinant affect the existence of an inverse matrix?
4. Can the method of matrix inversion be applied to larger matrices (3x3 or 4x4)?
5. Why do we multiply by the inverse of 7 in this particular problem?

Tip: Always check if the determinant is non-zero before attempting to find the inverse of a matrix!