Solve matrix equation without the inverses.

rapuy

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[tex]^{-1}[/tex] (2A + I) (C[tex]^{-1}[/tex] + A)b

without computing any matrix inverses?

2. Relevant equations

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

3. The attempt at a solution

x = B[tex]^{-1}[/tex] (2A + I) (C[tex]^{-1}[/tex] + A)b
Bx = (2A + I)(C[tex]^{-1}[/tex] + A)b

(2A + I)[tex]^{-1}[/tex]Bx = C[tex]^{-1}[/tex]b + Ab

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

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

Outlined

You can not be sure (2A + I)^-1 exists!

I would try to write it in the form ( . . . )(x - b) = 0.

rapuy

Ah , I see.

Here's another try:

x = B[tex]^{-1}[/tex](2A + I) (C[tex]^{-1}[/tex] + A)b
Bx = (2A + I)(C[tex]^{-1}[/tex]+ A)b
Bx = (2A + I)C[tex]^{-1}[/tex]b + (2A + I)Ab
Bx - (2A + I)Ab = (2A + I)C[tex]^{-1}[/tex]b

From here, I don't know how to get rid of C[tex]^{-1}[/tex]. Is it ok to postmultiply the matrices, with C?

rapuy

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[tex]^{-1}[/tex]b

==>

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