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Solve matrix equation without the inverses.

  • Thread starter rapuy
  • Start date
  • #1
3
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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[tex]^{-1}[/tex] (2A + I) (C[tex]^{-1}[/tex] + A)b

without computing any matrix inverses?

Homework Equations



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

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
 

Answers and Replies

  • #2
124
0
You can not be sure (2A + I)-1 exists!

I would try to write it in the form ( . . . )(x - b) = 0.
 
  • #3
3
0
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?
 
  • #4
3
0
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
 

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