## rsst123 one year ago 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

Sorry i typed that wrong its AB^-1C=D solve for B

2. 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?

3. hartnn

lets take a simpler problem XY = Z isolate Y you would pre multiply by X or post multiply by X?

you would multiply each side with X^-1 correct?

pre multiply

6. hartnn

yes, i meant X^-1 sorry, pre multiply or post multiply?

7. hartnn

ok good

8. 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)

9. hartnn

now lets go to our problem, AB^-1 C = D pre multiply by C or post? :)

10. hartnn

damn, i mean C^-1 :P

haha post right?

12. hartnn

yes! do it, and post some steps?

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

14. hartnn

very good! you solved more than half of it! now take inverse on both sides...

15. hartnn

and use this recursively, $$\large (PQ)^{-1} = Q^{-1}P^{-1}$$

perfect so the answer would be B=D^-1CA ?

17. hartnn

how did u get that ?

18. hartnn

$$[A^{-1}D C^{-1}]^{-1} = (C^{-1})^{-1} D^{-1} (A^{-1})^{-1} = CD^{-1}A$$

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!!

20. hartnn

hey, most welcome ^_^ happy to help :) feel free to message me if you wanna ask anything about this site :) $$\Huge \mathcal{\text{Welcome To OpenStudy}\ddot\smile}$$