Solve for \( X \) from the matrix equation below. Here \( I \) is the identity matrix and \( \operatorname{det}(B) \neq 0 \) and \( \operatorname{det}(A) \neq 0 \). \[ B\left(X^{-1}-I\right) A+B=A-B X^{-1} \]

To solve for \( X \) in the matrix equation \[ B\left(X^{-1}-I\right) A+B=A-B X^{-1}, \] we start by rearranging the equation. First, we can simplify the left-hand side: \[ B(X^{-1} - I)A + B = A - B X^{-1}. \] This can be rewritten as: \[ B(X^{-1} - I)A = A - B X^{-1} - B. \] Next, we can simplify the right-hand side: \[ A - B X^{-1} - B = A - B - B X^{-1}. \] Now, we can express the equation as: \[ B(X^{-1} - I)A = A - B - B X^{-1}. \] Next, we can isolate \( B(X^{-1} - I)A \): \[ B(X^{-1} - I)A = A - B - B X^{-1}. \] Now, we can factor out \( B \) from the right-hand side: \[ B(X^{-1} - I)A = A - B(1 + X^{-1}). \] To simplify further, we can multiply both sides by \( B^{-1} \) (since \( \operatorname{det}(B) \neq 0 \)): \[ (X^{-1} - I)A = B^{-1}A - (1 + X^{-1}). \] Next, we can rearrange this equation to isolate \( X^{-1} \): \[ X^{-1}A - IA = B^{-1}A - 1 - X^{-1}. \] Now, we can combine the \( X^{-1} \) terms: \[ X^{-1}A + X^{-1} = B^{-1}A - 1 + IA. \] Factoring out \( X^{-1} \): \[ X^{-1}(A + I) = B^{-1}A - 1. \] Now, we can solve for \( X^{-1} \): \[ X^{-1} = (B^{-1}A - 1)(A + I)^{-1}. \] Taking the inverse of both sides gives us \( X \): \[ X = (A + I)(B^{-1}A - 1)^{-1}. \] Thus, the solution for \( X \) is: \[ \boxed{(A + I)(B^{-1}A - I)^{-1}}. \]