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Having trouble solving using properties of determinants ...

  1. Sep 1, 2015 #1
    1. The problem statement, all variables and given/known data
    dY0tvZ6.png
    I'm a bit at a loss - I thought the last row with '1's would be useful, but it just gave me:
    (b2c - bc2) - (a2c - ac2) + (a2b - ab2)
    and
    bc(b - c) - ac(a - c) + ab(a - b)

    But then it is a dead end. I am probably doing something stupid again ...

    Any help appreciated.
     
  2. jcsd
  3. Sep 1, 2015 #2

    DEvens

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    I will give you a couple hints. First, what are the properties of determinants that you might possibly choose?

    For example: What will you get if a = b? In particular, notice what happens to the columns when a = b. What does that tell you about the form of the determinant?

    For example: What do you get if you swap the first and second columns? That is, what do you get if you swap a for b? What does that tell you about the form of the determinant?

    For example: What do you get if you multiply each element of the matrix by a constant K? What does that tell you about the form of the determinant?
     
  4. Sep 1, 2015 #3

    ehild

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    Remember you can add or subtract a column from an other one. With that, you can make all but one element in a row equal to zero.
     
  5. Sep 1, 2015 #4

    Ray Vickson

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    In addition to the hint in #3, remember that you do not change the value of the determinant when you add a multiple of one row onto another row, or a multiple of one column onto another column.
     
  6. Sep 1, 2015 #5
    Well thanks guys, I have managed to get rid of the bottom row with your methods. Again, not too sure where to go from there. As the first row is kind of a multiple of the second row (R1 = R22) I thought I could go from there, but it didn't seem to pan out. Am I on the right track, or have I completely misunderstood?

    Again, thanks a ton.
     
  7. Sep 1, 2015 #6

    Mark44

    Staff: Mentor

    Then you have probably made a mistake. If a row or column of a square matrix consists only of 0 entries, its determinant is 0. From the problem statement, the determinant is zero only under certain conditions. It would be helpful if you showed what you did to get rid of the bottom row.
     
  8. Sep 1, 2015 #7
    I basically subtracted the columns from each other until the row of 1s disappeared.
    z9sAg02.jpg
    sigh... this is really wrong isn't it?
     
  9. Sep 1, 2015 #8

    Mark44

    Staff: Mentor

    The first two determinants look OK, but not the third. If you subtract C1 from C3, which is what your notation says, you don't get 0 in the lower right corner.
     
  10. Sep 1, 2015 #9

    Ray Vickson

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    You went too far: you could stop at the second line and evaluate the determiant
    [tex] \left| \begin{array}{ccc}
    a^2 - b^2 & b^2 - c^2 & c^2 \\
    a - b & b - c & c \\
    0 & 0 & 1
    \end{array} \right|
    [/tex]

    If you "expand" this along the 3rd column, you just end up needing to evaluate a ##2 \times 2## determinant, which is pretty easy.
     
  11. Sep 1, 2015 #10
    Wouldn't the 2x2 determinant be (a2 - b2)(b - c) - (a - b)(b2 - c2)? Because I do not think this gives the correct answer - it gives me (a2 + bc)(b - c). Maybe I could manipulate this somehow to give me what the answer is looking for ...
     
  12. Sep 1, 2015 #11

    Ray Vickson

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    Sure: just remember that ##a^2-b^2 = (a-b)(a+b)## and ##b^2 - c^2 = (b-c)(b+c)##, and pull out common factors.
     
  13. Sep 2, 2015 #12
    YES!
    Thank you! I think I got it!
    bxRfFhO.jpg

    Really appreciate your patience and cleverness guys! Keep up the good work.
     
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