Solving for a matrix in equation form Solve for $X$ assuming all matrices are n x n and invertible as needed.
$$B(X+A)^{-1}=C$$
I solved this the following way:


*

*Multiply both sides by $(X+A)$

*Multiply both sides by the inverse of $C$

*Isolate $X$


My final answer is $$X=BC^{-1}-A$$ but the book got the answer $$X=C^{-1}B-A$$
I know that this problem comes from multiplication.  When I multiply both sides by any value, how do I know which side to put it on?
For instance, if I have the equation:
$A=XB$ and I want to solve for $X$, I have to multiply both sides by $B^{-1}$
But which side do I put the new multiple on?  Is it $AB^{-1}=XBB^{-1}$ or is it $B^{-1}A=B^{-1}XB$
 A: Since matrix multiplication is non-commuting, you can only say $AA^{-1}=1$, but you can't say $AXA^{-1}=X$ since $AXA^{-1}$ isn't generally the same as $AA^{-1}X$.
In the case of $B(X+A)^{-1}=C$, we get:
$B(X+A)^{-1}=C\\
B(X+A)^{-1}(X+A)=B=C(X+A)\\C^{-1}C(X+A)=X+A=C^{-1}B\\X=C^{-1}B - A$
A: You can multiply by $B^{-1}$ you can multiply left or right side. $AB^{-1}=XBB^{-1}$ is correct and $B^{-1}A=B^{-1}XB$ is also correct. But if you want to use identity $B^{-1}B=BB^{-1}=I$ you should have $B$ "near" $B^{-1}$. You can simplify $XB^{-1}B=XI=X$, but not $B^{-1}XB$.
A: If the matrix you are 'getting rid of' is the left factor, multiply on the left by its inverse (since the inverse and the matrix must be next to each other).  Similarly if the matrix you're getting rid of is the right factor, multiply by its inverse on the right.
There is another thing going on in your problem, which is for invertible matrices $M, N$, we have $(MN)^{-1}=N^{-1}M^{-1}$, since $MNN^{-1}M^{-1}=MIM^{-1}=MM^{-1}=I$.
So in your problem you would get $X+A=(B^{-1}C)^{-1}=C^{-1}(B^{-1})^{-1}=C^{-1}B$
