Given $A, B, C$ are $n \times n$ matrices, $B, C$ are nonsingular and $b$ is an $n$-vector, and $$x=B^{-1}(2A+I)(C^{-1}+A)b$$ This is a question coming from my computational mathematics course, and right now we are learning LU factorization. However, I don't see it can be used to solve this question. I also tried to move the elements to the other side, but things were getting too complicated to solve. Any answers will be appreciated!
How to implement the formula without computing any matrix inverse?
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matrices
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0In general, I don't think you can avoid finding the inverse matrix. Are there any other specific assumptions on $A,B,C$? – 2017-02-27
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0Please read [this post](http://meta.math.stackexchange.com/a/9960) and the others there for information on writing a good question for this site. In particular, people will be more willing to help if you [edit] your question to include some motivation, and an explanation of your own attempts. Ref: http://meta.math.stackexchange.com/a/11168/290189 – 2017-02-27
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0Unfortunately no. – 2017-02-27
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0Then unfortunately there's no short path. – 2017-02-27
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0I have already answered a very similar problem http://math.stackexchange.com/questions/2149656/avoiding-matrix-inversion/2149665#2149665 You can alter the answer to get the process for yours. – 2017-02-28
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I actually got the answer myself. Since $$x=B^{-1}(2A+I)(C^{-1}+A)b$$ By multiplying $B$ on both sides and opening the brackets, we get $$Bx=2AC^{-1}b+2AAb+C^{-1}b+Ab$$ Let $k$ be an n-vector, $k=C^{-1}b$, then $Ck=b$. By LU factorization, we can get $k$. Plug $k$ into the equation above, we get $$Bx=2Ak+2AAb+k+Ab$$ Let $y=2Ak+2AAb+k+Ab$, $y$ is an n-vector. Thus we get$$Bx=y$$Again, by LU factorization, we can find out $x$. QED.