For any integer $n\ge 2$ and two $n\times n$ matrices with real
entries $A,\; B$ that satisfy the equation
$$A^{-1}+B^{-1}=(A+B)^{-1}\;$$
prove that $\det (A)=\det(B)$. 

Does the same conclusion follow for matrices with complex entries?

(Proposed by Zbigniew Skoczylas, Wroclaw University of Technology)

1. Given the equation \( A^{-1} + B^{-1} = (A + B)^{-1} \), we start by multiplying both sides by \( A \) from the left and \( B \) from the right:
   \[
   A(A^{-1} + B^{-1})B = A(A + B)^{-1}B
   \]
2. Simplifying the left-hand side:
   \[
   A A^{-1} B + A B^{-1} B = A (A + B)^{-1} B
   \]
   \[
   B + A = A (A + B)^{-1} B
   \]
3. Now, multiply both sides by \( (A + B) \):
   \[
   (A + B)(B + A) = A B
   \]
4. Expanding the left-hand side:
   \[
   AB + A^2 + B^2 + BA = AB
   \]
   \[
   A^2 + B^2 + BA = 0
   \]
5. Rearrange the equation:
   \[
   A^2 + B^2 + BA = 0
   \]
6. To find the determinant, we use the property of determinants:
   \[
   \det(A^2 + B^2 + BA) = \det(0)
   \]
   \[
   \det(A^2) + \det(B^2) + \det(BA) = 0
   \]
7. Since \( \det(A^2) = (\det(A))^2 \) and \( \det(B^2) = (\det(B))^2 \), we have:
   \[
   (\det(A))^2 + (\det(B))^2 + \det(BA) = 0
   \]
8. Using the property \( \det(BA) = \det(B)\det(A) \), we get:
   \[
   (\det(A))^2 + (\det(B))^2 + \det(B)\det(A) = 0
   \]
9. Let \( x = \det(A) \) and \( y = \det(B) \), then:
   \[
   x^2 + y^2 + xy = 0
   \]
10. This can be factored as:
    \[
    (x + y)^2 - xy = 0
    \]
    \[
    (x + y)^2 = xy
    \]
11. Solving for \( x \) and \( y \):
    \[
    x = y
    \]
    Therefore, \( \det(A) = \det(B) \).

For matrices with complex entries, consider the matrices:
\[ 
A = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}, \quad B = \begin{bmatrix} \omega & 0 \\ 0 & \omega \end{bmatrix} 
\]
where \( \omega = e^{\frac{2 \pi i}{3}} \). These matrices satisfy the given equation but have different determinants:
\[ 
\det(A) = 1, \quad \det(B) = \omega^2 
\]
Thus, the conclusion does not hold for matrices with complex entries.

The final answer is \( \boxed{ \det(A) = \det(B) } \) for real matrices, and the conclusion does not hold for complex matrices.