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jcd2012

  • 2 years ago

Linear Algebra: Find (if possible) values a, b, and c such that the system of linear equations has: a) no solution b) exactly one solution c) infinitely many solutions 2x - y + z = a x +y +2z = b 3y + 3z = c

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  1. jcd2012
    • 2 years ago
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    |dw:1348174076973:dw|

  2. inkyvoyd
    • 2 years ago
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    Try turning that into eschelon form.

  3. TuringTest
    • 2 years ago
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    I am thinking a bit differently...

  4. inkyvoyd
    • 2 years ago
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    Yeah, I skimmed through a textbok about 30 mins ago, I'm haven't exactly even taken this ;). I know a solution though, it's probably not the most efficient way...

  5. TuringTest
    • 2 years ago
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    I'm not sure myself either...

  6. TuringTest
    • 2 years ago
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    For any \(n\times n\) matrix \(A\) the system\[A\vec x=\vec b\]has exactly one solution for each \(n\times1\) matrix \(\vec b\) if and only if \(\det A\neq0\)

  7. TuringTest
    • 2 years ago
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    so what does a,b, and c have to do with the answer to your question I'm not even sure, the answer seems to be dependent on the matrix \(A\)....

  8. jcd2012
    • 2 years ago
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    a, b, and c are any real number such that each condition is met

  9. TuringTest
    • 2 years ago
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    right, any matrix \(\vec b\)

  10. jcd2012
    • 2 years ago
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    |dw:1348174993843:dw|

  11. jcd2012
    • 2 years ago
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    now, do I look for conditions in which the three cases can be met?

  12. TuringTest
    • 2 years ago
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    if this is always true then there are infinite solutions, so that would suggest that a=-3, c=3, and b is whatever (easy to figure out) how to test the other two conditions I'm not so sure

  13. jcd2012
    • 2 years ago
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    I'll start with that then. these kinds of questions happen to be the even numbered problems. thanks

  14. TuringTest
    • 2 years ago
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    sure, wish I could help more

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