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anonymous

  • one year ago

Applied Linear Algebra Question, I'm posting question give me one second

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  1. anonymous
    • one year ago
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  2. anonymous
    • one year ago
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    okay so I got to the point where I dont know what to do next

  3. anonymous
    • one year ago
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    I have to find values of "a" where the system has no solution, Unique Solution, and infinte solutions

  4. anonymous
    • one year ago
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    I got the system to row echelon form

  5. anonymous
    • one year ago
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    lt me draw what i got out

  6. anonymous
    • one year ago
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    |dw:1442684208867:dw|

  7. anonymous
    • one year ago
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    Do I need to put this into reduce echelon form?

  8. anonymous
    • one year ago
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    Actually yea I think that would help

  9. anonymous
    • one year ago
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    let me do that on paper, first brb

  10. anonymous
    • one year ago
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    oh wait made a mistake in the drawing, the "3a+10" has a denominator of "2a-4"

  11. anonymous
    • one year ago
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    |dw:1442685001288:dw|

  12. anonymous
    • one year ago
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    If a is not equal to 2, then we have a unique solution Plugging in a = 2 in the second we get 1 -2 3 3 0 -4+2a 3-3a 10-3a 0 0 1 -1 the middle row second term is non zero, so you can solve for it, and that produces a triangular system with a unique solution. Try a number.

  13. IrishBoy123
    • one year ago
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    find the determinant and set it to zero and that might save a lot of typing

  14. IrishBoy123
    • one year ago
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    ie set to zero and solve for a

  15. anonymous
    • one year ago
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    I assume you mean the determinant of the coefficient matrix.

  16. anonymous
    • one year ago
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    the determinant of the coefficient matrix is 12 - 6a the matrix has a unique solution as long as the determinant is non zero. 12 - 6a = 0 12 = 6a 12/6 = a if a ≠ 2 then the system has a unique solution (because the determinant is non-zero)

  17. anonymous
    • one year ago
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    if a = 2 you have a contradiction, can you see why?

  18. anonymous
    • one year ago
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    Case 1: "a" does not equal to 2, give a uniqure solution case 2: "a" equals 2, gives no Solution Case 3: there is never a infinite solution is that correct?

  19. IrishBoy123
    • one year ago
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    yeah, work that way :p if \(a \ne 2\) you have a solution so set a to 2 and do your reduction, see where you go

  20. anonymous
    • one year ago
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    It will be no solution if u get set a = 2

  21. anonymous
    • one year ago
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    correct

  22. anonymous
    • one year ago
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    So would it be right to say that this system does not have infinite solutions?

  23. anonymous
    • one year ago
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    So it has no solution when a=2, a unique solution a does not =2, but not infinite solutions right? or when a does not equal to 2 it will have infinite solution as well?

  24. anonymous
    • one year ago
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    anyone there?

  25. anonymous
    • one year ago
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    it will never have infinite solutions

  26. anonymous
    • one year ago
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    what I'm confused about is that if a does not equal to 2, will it have infinite solutions

  27. anonymous
    • one year ago
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    oh ok, can you tell me why though?

  28. anonymous
    • one year ago
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    if a does not equal to 2 it will have a unique solution if a is equal to 2 it will have no solution (because of contradiction) there is no case of 'infinite solutions'

  29. anonymous
    • one year ago
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    oooh I see, lol thanks, I blanked out for a second there. Thank you so much!

  30. anonymous
    • one year ago
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    Your welcome.

  31. IrishBoy123
    • one year ago
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    |dw:1442690538127:dw|

  32. IrishBoy123
    • one year ago
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    |dw:1442690850922:dw|

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