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Determinant's solving without expansion

  1. Oct 20, 2013 #1
    Solve without expanding the determinant having elements which are listed as follows row-wise = {1^2,2^2,3^2,2^2,3^2,4^2,3^2,4^2,5^2} where , the determinant is of the order 3

    I have tried some combinations of operations performed row or column wise that could produce 2 zeros either in one of the rows or columns and then expanding the determinant along that row or column; but is there any simpler way out?
     
    Last edited: Oct 20, 2013
  2. jcsd
  3. Oct 20, 2013 #2

    UltrafastPED

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    1 4 9
    4 9 16
    9 16 25

    Just put it into upper triangular form ... then the product of the elements along the main diagonal will equal the determinant of the original matrix.
     
  4. Oct 21, 2013 #3

    epenguin

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    Maybe you are expected to use (a2 - b2) = (a - b)(a + b) = (a + b) in this case.
     
  5. Oct 22, 2013 #4

    HallsofIvy

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    I would think not! In general (a- b)(a+ b) is NOT equal to a+ b.
     
  6. Oct 22, 2013 #5

    epenguin

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    Not in general but in this case when subtracting certain rows from the next where (a -b) = 1 .
     
  7. Oct 24, 2013 #6
    How do i do that? Can you please tell me the operations?
     
  8. Oct 24, 2013 #7

    Mark44

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    No.
    Per the Physics Forums rules (https://www.physicsforums.com/showthread.php?t=414380) that you agreed to when you signed up for an account, we don't do your work for you. Here is from the Homework Help Guidelines section of the rules.
    Show us what you've done, and we'll steer you in the right direction.
     
  9. Oct 25, 2013 #8
    1 4 9
    4 9 16
    9 16 25
    2 times the column 1 - column 2 and taking the negative sign out of the determinant from row 1 and row 2
    2 -4 -9
    1 -9 -16
    2 16 25
    row 1 - 2 times row 2 and row 3 - 2 times row 2 and the expanding along column 1
    0 -16 -23
    1 -9 -16
    0 34 57
     
    Last edited: Oct 25, 2013
  10. Oct 25, 2013 #9

    epenguin

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    I suggest you modify that post to state what operations you performed, which is not obvious.

    Do your operations use in any way the systematic special features of the numbers in the problem or are they just what you would do for any generic 3 X 3 determinant?
     
  11. Oct 25, 2013 #10

    LCKurtz

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    I don't think any of the above suggestions would count as "not expanding" the determinant. I don't see the trick myself and would certainly like for Kartik to present the solution if this is a classroom problem for which he eventually is given the "answer".

    You all probably have seen the trick for expanding the Vandermonde determinant$$
    \left | \begin{array}{ccc}
    1&a&a^2\\
    1&b&b^2\\
    1&c&c^2
    \end{array}\right|$$where you replace the ##a## by ##x## giving$$
    P(x)=\left | \begin{array}{ccc}
    1&x&x^2\\
    1&b&b^2\\
    1&c&c^2
    \end{array}\right|$$Then you observe ##P(b)=P(c)=0## so ##P(x)=K(x-b)(x-c)## and ##K## is the coefficient of ##x^2##. That can be read from the determinant as ##(c-b)##. Therefore ##P(x) = (c-b)(x-b)(x-c)## and the original determinant is ##P(a)=(c-b)(a-b)(a-c)##.

    I would think they are looking for some trick like that, which I think would count as "not expanding" it. But what trick??
     
  12. Oct 25, 2013 #11

    HallsofIvy

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    Since you made no attempt to do this yourself it is difficult to say what kind of help you can use (whether you know what "row operations" are) or exactly what you mean by "expansion".

    One method I learned, way back in secondary school, for a three by three determinant, is to copy the first two rows again:
    [tex]\begin{vmatrix}1 & 4 & 9 \\ 4 & 9 & 16 \\ 9 & 16 & 25 \end{vmatrix} \begin{vmatrix}1 & 4 \\ 4 & 9 \\ 9 & 16 \end{vmatrix}[/tex]

    Now, sum the products of the three "upper left to lower right" diagonals: 1(9)(25)+ 4(16)(9)+ 9(4)(16), sum the products of the three "upper right to lower left" diagonals: 9(9)(9)+ 16(16)(1)+ 25(4)(4). Now subtract the second from the first.
     
    Last edited by a moderator: Oct 25, 2013
  13. Oct 25, 2013 #12

    LCKurtz

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    Heh heh. You're showing your age Halls. My favorite way for a quick answer too...
     
  14. Oct 26, 2013 #13

    epenguin

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    Yes I was reminded if the Vandermonde determinant, but this is not the Vandermonde determinant, I haven't looked into an argument of the type of Kurtz', but I have suggested an approach which comes out easily in a couple of lines. Whether it is within the 'without expansion' requirement is not clear, but it is 'mathematical', to the extent it enables solving the more general determinant extending that one in the obvious way - for all order >3 they are all 0 I think.

    Karting seems to have done an ordinary general expansion using the numbers without using their particular forms given.
     
  15. Oct 27, 2013 #14

    pasmith

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    I obtained the answer by considering
    [tex]
    P(x,y,z) = \left| \begin{array}{ccc}
    x^2 & (x+1)^2 & (x+2)^2 \\
    y^2 & (y+1)^2 & (y+2)^2 \\
    z^2 & (z+1)^2 & (z+2)^2
    \end{array}\right|
    [/tex]
    We are looking for [itex]P(1,2,3)[/itex].
     
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