The problem provides matrix $A$ and an equation involving the inverse of matrix products to find matrix $B$. Let's break it down:

• Matrix $A$ is given as:

$A = \begin{pmatrix} 3 & -1 \\ -2 & 1 \end{pmatrix}$

• The equation is:

$\left(A \cdot B^{-1}\right)^{-1} = \begin{pmatrix} 2 & 1 \\ 4 & 3 \end{pmatrix}$

To solve for matrix $B$, we can use the following steps:

1. Rewrite the equation: $A \cdot B^{-1} = \left(\begin{pmatrix} 2 & 1 \\ 4 & 3 \end{pmatrix}\right)^{-1}$

2. Calculate the inverse of the matrix $\begin{pmatrix} 2 & 1 \\ 4 & 3 \end{pmatrix}$.

3. Multiply both sides of the equation by $A^{-1}$ to solve for $B^{-1}$, then find $B$.

I'll begin with the calculations now.The matrix $B$ is:

$B = \begin{pmatrix} 4 & -1 \\ 6 & -1 \end{pmatrix}$

Would you like more details or have any questions?

Here are some related questions you might explore:

1. How is the inverse of a matrix calculated?
2. What are the properties of matrix multiplication and inversion?
3. How can you verify the solution by multiplying matrices?
4. What happens if a matrix is singular (non-invertible)?
5. How does the determinant affect the invertibility of a matrix?

Tip: Always verify your result by substituting it back into the original equation to ensure correctness.