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We are tasked with solving for the matrix \( X \) in the equation:

\[
A X (D + B X)^{-1} = C
\]

Assume all matrices are \( n \times n \) and invertible where necessary.

### Step-by-Step Solution:

1. **Multiply both sides by \( (D + BX) \) from the right:**
   To eliminate the inverse, multiply both sides of the equation on the right by \( D + BX \):
   \[
   A X (D + B X)^{-1}(D + BX) = C (D + BX)
   \]
   Simplifying the left-hand side:
   \[
   A X = C (D + BX)
   \]

2. **Expand the right-hand side:**
   Distribute \( C \) over \( (D + BX) \):
   \[
   A X = C D + C B X
   \]

3. **Group the terms with \( X \):**
   Move the term with \( X \) on the right-hand side to the left-hand side:
   \[
   A X - C B X = C D
   \]
   Factor out \( X \) on the left-hand side:
   \[
   (A - C B) X = C D
   \]

4. **Solve for \( X \):**
   Assuming \( A - C B \) is invertible, multiply both sides by \( (A - C B)^{-1} \) from the left:
   \[
   X = (A - C B)^{-1} C D
   \]

### Final Expression for \( X \):

\[
X = (A - C B)^{-1} C D
\]

Would you like further details or clarification on any step? Here are some related questions:

1. How do we determine if \( A - CB \) is invertible?
2. What if \( A - CB \) is not invertible? How would the solution change?
3. Can this method be generalized to other matrix equations involving inverses?
4. How would the process differ if \( B \) or \( D \) were non-square matrices?
5. What real-world applications might involve solving matrix equations like this?

**Tip:** When working with matrix inverses, always ensure that the matrix in question is invertible by checking its determinant or other criteria for non-singularity.

Solve for the matrix X if A X (D + B X)^{-1} = C. Assume that all matrices are n x n and invertible as needed.

Learn how to solve for the matrix X in the equation AX(D + BX)^{-1} = C with a step-by-step solution. This problem explores key concepts in matrix algebra, including matrix inversion and linear equations, suitable for college-level students.

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