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coefficient matrix = 12, -1, 1; 1,0,1; 0,1,1... augmented matrix = 12,-1, 1,0; 1,0,1,0; 0,1,1,0.... RREF---> gives identity matrix, so they are all liniar independent...
Okay, but how do you get the t-value?
there are no t-values, this is not a dependent system, it is inconssistent, the only solution is the trivial solution (0,0,0)...
oh, do you mean the t in front of the x1 at r2,c1?
There are actually two t's the other one at r3,c3
(I'm in the same class so we're learning about this, so I'm curious as to how to solve this too).
ok, sorry, misread it...
I know t = -4 or 3 from the back of the book
ok, i may nead a second...if that's ok...
take your time
is this your first semester of linear algebra?
You learn how to turn that system into matrix right? It is easier if you did. With that technique you will end up with t(1+t)-12 to get nontrivial solution, that expression must not equal to 0 so you just solve \[t(1+t)-12\neq0\]
so that is a row in your matrix?
yes one row ended up to be [0 0 t(1+t)-12| 0]
ok perfect, thank you!
Happy to help ^^
Cool! I actually got that once I pulled out my whiteboard and tried to do row echelon form. Thanks guys!
Hey quantish you should post more linear algebra problems so it forces me to study. :D
Haha, you should talk to applekiwi...the solution that I was working on was more backward and difficult ; )
I have a test coming up and even though I look over the material in the chapters, make flashcards full of definitions and theorems, and try to pratice and review problems, I am very slow.