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A sum in determinants

  1. Jun 10, 2012 #1
    1. The problem

    Prove that

    | (a+b-c) (-c+a-b) (a+b+c) |
    | (a-c) (c-a) (b-a) | = (a+b-c)(-c+a-b)(a-c)
    | (a-b) (a-c) (a+b) |

    using properties of determinants without expanding a determinant

    2. The attempt at a solution

    I tried a lot of ways like with following steps C(3) -> C(3) + C(2) , R(1) -> R(1) + R(2) ,
    R(1) -> R(1) + R(2) - R(3).....but i could get nothing...its not my homework sum....i am just a 7th grade maths enthusiast.....i saw this sum somewhere.....i tried to solve this but couldnt get.......can anyone help me up to solve this......or atleast guide me.......where C(1) means column 1 and R(1) means Row 1 and similar things to that
     
    Last edited: Jun 10, 2012
  2. jcsd
  3. Jun 10, 2012 #2

    HallsofIvy

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    (a+b-c) (-c+a-b) (a+b+c) |
    | (a-c) (c-a) (b-a) | = (a+b-c)(-c+a-b)(a-c)
    | (a-b) (a-c) (a+b) |
    [tex]\left|\begin{array}{ccc}a+b- c & -c+a-b & a+b+c \\ a-c & c-a & b-a \\ a-b & a-c & a+b\end{array}\right|[/tex]

    "C(3) -> C(3) + C(2)". Okay, that gives
    [tex]\left|\begin{array}{ccc}a+b- c & -c+a-b & 2a \\ a-c & c-a & b+c-2a \\ a-b & a-c & 2a+b-c\end{array}\right|[/tex]

    But may I ask why you did that? How does it improve the situation?

    Notice that the right side, (a+b-c)(-c+a-b)(a-c), is a product of three numbers, one of which is the upper left element of the determinant. The determinant of a diagonal or triangular matrix is just the product of the numbers on the main diagonal so I would recommend using row operations to change all numbers below the main diagonal to 0. For example, changing R2 to R2- (a-c)/(a+b-c)R1 will make the second row of the first column equal to 0.
     
  4. Jun 10, 2012 #3
    that was one of the steps i did....i tried with many properties meddled with this a lot.....but couldnt get a thing......instead of what you said i would just add R(2) and R(3) and get a 0..(R2 -> R2 + R3)....so thats aint the problem.....u will get zeros.....bu the problem gets more complicated i u proceed.....so what i want is...a stock algorithm to get the answer.....
     
    Last edited: Jun 10, 2012
  5. Jun 10, 2012 #4

    Ray Vickson

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    The claimed result is false. Let D denote the determinant and R the expression on the right. When a=1, b=2, c=3 we have D = 16 but R = 0. When a=-1, b=0, c=2 we have D = 24 but R = -27.

    There must be something wrong with the expressions you wrote.

    RGV
     
  6. Jun 11, 2012 #5

    HallsofIvy

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    Thanks, Ray. I didn't check.
     
  7. Jun 12, 2012 #6
    Oh thanks......
     
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