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Matrix determinant

  1. Jul 27, 2006 #1

    Bob

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    Evaluate the following determinant. Write your answer as a polynomial in x.

    [tex]\begin{array}{|lcr|}a-x&b&c\\1&-x&0\\0&1&-x\end{array}[/tex]


    Please help me! thanks.
     
  2. jcsd
  3. Jul 28, 2006 #2
    Just find the determinant as usual. A cofactor expansion from one of the rows or columns containing a zero is probably the easiest way.
     
  4. Jul 28, 2006 #3

    HallsofIvy

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    You do realize, don't you, that you are expected to show us what you have tried so we can suggest changes? What durt suggested is very general but without knowing where you are having trouble we can't be more specific. It won't help you for someone else to do it for you.

    (I confess that I find the answer amusing!)
     
  5. Jul 28, 2006 #4
    clever problem :rofl:
     
  6. Jul 28, 2006 #5

    Bob

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    The answer is [tex]-x^3+ax^2+(b-a)x-a+c[/tex]
     
  7. Jul 28, 2006 #6
    not quite :smile:
     
  8. Jul 28, 2006 #7
    Show us the work!!!! How did you get that wrong answer?
     
  9. Jul 29, 2006 #8

    Bob

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    :eek:

    [tex]-x^3+ax^2+bx+c[/tex]
     
  10. Jul 30, 2006 #9

    HallsofIvy

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    What part of "Show us the work!!!! How did you get that wrong answer?" did you not understand?
     
  11. Jul 30, 2006 #10

    Bob

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    :bugeye: I am sorry.


    [tex](a-x)(x*x-1)-1(-bx-c)+0\\=ax^2-a-x^3+x+bx+c\\=-x^3+ax^2-(a-b)x-a+c[/tex]
     
  12. Jul 30, 2006 #11
    Recheck your first term, (a-x)(x*x-1) isn't quite right.
     
  13. Jul 31, 2006 #12

    HallsofIvy

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    Okay, you are expanding by the first column:
    [tex](a-x)\left|\begin{array}{cc}-x & 0 \\1 & -x\end{array}\right|- (1)\left|\begin{array}{cc}b & c \\ 1 & -x\end{array}\right|[/tex]
    As dleet said, check that first number. 0*1 is not 1!
     
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