## Solve matrix equation without the inverses.

1. The problem statement, all variables and given/known data

If A, B, and C are nxn matrices, with B and C nonsingular, and b is an n-vector, how would you implement the formula
x = B$$^{-1}$$ (2A + I) (C$$^{-1}$$ + A)b

without computing any matrix inverses?

2. Relevant equations

Is there any identity for (2A+I)$$^{-1}$$ that is expressed without the inverse?

3. The attempt at a solution

x = B$$^{-1}$$ (2A + I) (C$$^{-1}$$ + A)b
Bx = (2A + I)(C$$^{-1}$$ + A)b

(2A + I)$$^{-1}$$Bx = C$$^{-1}$$b + Ab

If (2A+I)$$^{-1}$$ is expressed without the inverse, I would have proceeded as follows:

(2A + I)$$^{-1}$$Bx - Ab = C$$^{-1}$$b
C[(2A + I)$$^{-1}$$Bx - Ab] = CC$$^{-1}$$b
C[(2A + I)$$^{-1}$$Bx - Ab] = b
C(2A + I)$$^{-1}$$Bx - CAb = b
C(2A + I)$$^{-1}$$Bx = (CA+I)b

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 You can not be sure (2A + I)-1 exists! I would try to write it in the form ( . . . )(x - b) = 0.
 Ah , I see. Here's another try: x = B$$^{-1}$$(2A + I) (C$$^{-1}$$ + A)b Bx = (2A + I)(C$$^{-1}$$+ A)b Bx = (2A + I)C$$^{-1}$$b + (2A + I)Ab Bx - (2A + I)Ab = (2A + I)C$$^{-1}$$b From here, I don't know how to get rid of C$$^{-1}$$. Is it ok to postmultiply the matrices, with C?

## Solve matrix equation without the inverses.

As a continuation of the above solution, is it ok if I do the postmultiplication before the vector b with C on both sides?

Bx - (2A + I)Ab = (2A + I)C$$^{-1}$$b

==>

[Bx - (2A + I)Ab] C = (2A + I)C$$^{-1}$$C b
BCx - (2A + I)AC b = (2A + I) b