Solve matrix equation without the inverses.

[rapuy]

Homework Statement

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?

Homework Equations

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

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

 

[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$$^{-1}$$(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?

 

[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$$^{-1}$$b

==>

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