|Mar11-12, 11:43 PM||#1|
First off i'm gonna start by saying, that i hate my algebra class. I dunno if its the teacher or the material, but i find this stuff so hard ! With calc i get most of it but just some hard questions i need to ask on here... with algebra i get like NONE of the theory/proofs behind it which is why i can never solve proofs in class. I can get regular numerical questions though lol. Heres the problem:
Suppose (A-B)D=0 where A,B,D are matrices of appropriate sizes and D is invertible. Prove that A=B.
So of course A-B is zero, because 0 times D would of course yield 0. But i have nooo idea at all how to prove this using all the theorems we learned in class. This is an intro Lin Alg. course so we haven't done anything to complicated, just did the basics behind inverses and all that. Also, i lost a page of my notes, which i think might be important, because there aren't too many theorems that i have in my notes right now to work with :(
Any help would be really appreciated. THANKS :)
I know somehow you'd have to eliminate the D so that you are left with A-B= 0 which of course leads to A=B, but i have no clue how to do that :(
|Mar12-12, 12:43 AM||#2|
Because D is invertible you can post-multiply both sides by D^-1 since it exists. This gives you (A-B)DD^-1 = 0 x D^-1 = 0 => (A-B)I = 0 => A-B = 0 >= A = B.
You can only do this is D is invertible because if it isn't then since D^-1 (inverse of D) doesn't exist, you can't gaurantee that for example you don't get cases like D = 0 or some other condition that you would get in say a system that has linearly dependent basis vectors (which means it's not really a basis).
|Mar12-12, 01:03 AM||#3|
OH GOT IT!
k i didn't realise you could just multiply both sides of the equation by D^-1.
Thanks so much :)
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