rsst123
  • rsst123
If A,B,C,D are all invertible nxn matrices,solve for B AB^-1=D? I understand to solve for C you need to multiply each side with the inverse, but I dont understand the premultiply and postmultiply rules
Linear Algebra
katieb
  • katieb
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rsst123
  • rsst123
Sorry i typed that wrong its AB^-1C=D solve for B
hartnn
  • hartnn
Hello! :) in AB^-1 C = D, you need to isolate B, right?? Also, do you understand that in Matrix algebra, AB and BA are different and not equal everytime?
hartnn
  • hartnn
lets take a simpler problem XY = Z isolate Y you would pre multiply by X or post multiply by X?

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rsst123
  • rsst123
you would multiply each side with X^-1 correct?
rsst123
  • rsst123
pre multiply
hartnn
  • hartnn
yes, i meant X^-1 sorry, pre multiply or post multiply?
hartnn
  • hartnn
ok good
hartnn
  • hartnn
if you had post multiplied it, XYX^-1 = Z X^-1 you could not have combined X and X^-1 to get I (the Identity matrix)
hartnn
  • hartnn
now lets go to our problem, AB^-1 C = D pre multiply by C or post? :)
hartnn
  • hartnn
damn, i mean C^-1 :P
rsst123
  • rsst123
haha post right?
hartnn
  • hartnn
yes! do it, and post some steps?
rsst123
  • rsst123
AB^-1CC^-1=DC^-1 A^-1AB^-1=A^-1DC^-1 I get to B^-1=A^-1DC^-1 but im not sure what i should do from here
hartnn
  • hartnn
very good! you solved more than half of it! now take inverse on both sides...
hartnn
  • hartnn
and use this recursively, \(\large (PQ)^{-1} = Q^{-1}P^{-1}\)
rsst123
  • rsst123
perfect so the answer would be B=D^-1CA ?
hartnn
  • hartnn
how did u get that ?
hartnn
  • hartnn
\([A^{-1}D C^{-1}]^{-1} = (C^{-1})^{-1} D^{-1} (A^{-1})^{-1} = CD^{-1}A \)
rsst123
  • rsst123
oh sorry my bad I was looking at the wrong work I was doing before this is my first time using this site and it's very helpful thank you for your help!!
hartnn
  • hartnn
hey, most welcome ^_^ happy to help :) feel free to message me if you wanna ask anything about this site :) \(\Huge \mathcal{\text{Welcome To OpenStudy}\ddot\smile} \)

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