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Interesting Method of Finding a Determinant

Date: 08/01/2008 at 09:59:58
From: Daya
Subject: Prove the problem

I need to show that the right side is the determinant of the given matrix:

  X  X^2  YZ
  Y  Y^2  ZX  =  (X-Y)(Y-Z)(Z-X)(XY + YZ + ZX)
  Z  Z^2  XY



Date: 08/01/2008 at 22:17:01
From: Doctor Ali
Subject: Re: Prove the problem

Hi Daya!

Thanks for writing to Dr. Math.

We want to evaluate:

  | x  x^2  yz |
  | y  y^2  xz |
  | z  z^2  xy |

Surely we can prove the formula by using long expansion of 
determinants.

But here I'd like to use a method that you are maybe familiar with,
which can simplify the process a little bit.  You may think more and
come up with a simpler method.

Firstly, subtract the second row from the first row.  We know that 
the determinant will not change .  So, we can say that the above 
expression (required determinant) equals,

  | x-y    x^2 - y^2   z(y-x) |
  |  y        y^2        xz   |
  |  z        z^2        xy   |

As you see, from the first row, we can factor (x-y):

        |  1    x+y    -z |
  (x-y) |  y    y^2    xz |
        |  z    z^2    xy |

Again we can subtract the third row from the second row:

        |  1       x + y         -z   |
  (x-y) | y-z    y^2 - z^2     x(z-y) |
        |  z        z^2          xy   |

Again we can factor (y-z) from the second row:

             |  1    x+y    -z |
  (x-y)(y-z) |  1    y+z    -x |
             |  z    z^2    xy |

Now, we can subtract the second row from the first.  We'll have:

             |  0    x-z    x-z |
  (x-y)(y-z) |  1    y+z     -x |
             |  z    z^2     xy |

We can now factor (x-z) from the first row:

                  |  0     1     1 |
  (x-y)(y-z)(x-z) |  1    y+z   -x |
                  |  z    z^2   xy |

I know that you may prefer to expand the remaining determinant, since 
we have came up with most of the answer!  But I'm personally 
interested in simplifying determinants as much as possible using these
methods; I think that it's more fun!  Let's continue:

We can now subtract z times the first column from the second column:  

                  |  0    1    1 |
  (x-y)(y-z)(x-z) |  1    y   -x |
                  |  z    0   xy |

As you see, I try to make as many zeros as possible, since they make 
some parts of the expansion vanish.

Now, it is a good time to expand the determinant.  See:

  |  0    1    1 |
  |  1    y   -x | = 0 ( ... ) - 1 (xy) + z(-x-y) = -(xy+xz+yz)
  |  z    0   xy |

So all in all, the determinant will be:

  (x-y)(y-z)(x-z)(-(xy+xz+yz))

Let's apply the negative sign in (x-z) and write it as (z-x).  We'll 
have

  (x - y)(y - z)(z - x)(xy + xz + yz),

which is what we were supposed to prove.

Let me tell you something at the end.  You may ask:  How did all of 
these steps come into your mind and how did you know that you should 
do each step that way?  The rough answer is that I tried to make zeros 
as much as possible.  But note that if you solve some similar problems 
yourself, you'll get the idea!  It is a matter of practicing more and 
more, not a matter of talent or whatever you may call it.

Does that make sense?

Please write back if you still have any difficulties.

- Doctor Ali, The Math Forum
  http://mathforum.org/dr.math/ 
Associated Topics:
High School Linear Algebra

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