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anonymous

  • 5 years ago

For which values of t does the following homogeneous linear system have non-trivial solutions? 12x1-x2+x3=0 tx1 +x3=0 x2+tx3=0

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  1. anonymous
    • 5 years ago
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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...

  2. anonymous
    • 5 years ago
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    linearly* independent...

  3. anonymous
    • 5 years ago
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    Okay, but how do you get the t-value?

  4. anonymous
    • 5 years ago
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    there are no t-values, this is not a dependent system, it is inconssistent, the only solution is the trivial solution (0,0,0)...

  5. anonymous
    • 5 years ago
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    oh, do you mean the t in front of the x1 at r2,c1?

  6. anonymous
    • 5 years ago
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    There are actually two t's the other one at r3,c3

  7. anonymous
    • 5 years ago
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    (I'm in the same class so we're learning about this, so I'm curious as to how to solve this too).

  8. anonymous
    • 5 years ago
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    ok, sorry, misread it...

  9. anonymous
    • 5 years ago
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    I know t = -4 or 3 from the back of the book

  10. anonymous
    • 5 years ago
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    ok, i may nead a second...if that's ok...

  11. anonymous
    • 5 years ago
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    take your time

  12. anonymous
    • 5 years ago
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    is this your first semester of linear algebra?

  13. anonymous
    • 5 years ago
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    yes

  14. anonymous
    • 5 years ago
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    ok...

  15. anonymous
    • 5 years ago
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    Mine too

  16. anonymous
    • 5 years ago
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    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 \[t(1+t)-12\neq0\]

  17. anonymous
    • 5 years ago
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    so that is a row in your matrix?

  18. anonymous
    • 5 years ago
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    yes one row ended up to be [0 0 t(1+t)-12| 0]

  19. anonymous
    • 5 years ago
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    ok perfect, thank you!

  20. anonymous
    • 5 years ago
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    Happy to help ^^

  21. anonymous
    • 5 years ago
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    Cool! I actually got that once I pulled out my whiteboard and tried to do row echelon form. Thanks guys!

  22. anonymous
    • 5 years ago
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    Hey quantish you should post more linear algebra problems so it forces me to study. :D

  23. anonymous
    • 5 years ago
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    Haha, you should talk to applekiwi...the solution that I was working on was more backward and difficult ; )

  24. anonymous
    • 5 years ago
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    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.

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