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I'm learning how to factorize determinants of a square matrix in school, but we haven't learnt a general method to do that, besides 'creating zeros'. So I thought maybe I'll ask here if someone does know a method that generalizes the factorization of a square matrix.

An example
I had this determinant to factorize on a test:

$$\left| \begin{array}{ccc} x & y & 1 \\ x^2 & y^2 & 1 \\ x^3 & y^3 & 1 \end{array} \right| $$

So I started and got the following, quite easy:

$$ \begin{align} \left| \begin{array}{ccc} x & y & 1 \\ x^2 & y^2 & 1 \\ x^3 & y^3 & 1 \end{array} \right| &= xy \cdot \left| \begin{array}{ccc} 1 & 1 & 1 \\ x & y & 1 \\ x^2 & y^2 & 1 \\ \end{array} \right| & &(1) \end{align} $$

At this point I was stuck. I had now idea how to go on from this point. But when I got my test back, it was corrected by the teacher:

$$ \begin{align*} xy \cdot \left| \begin{array}{ccc} 1 & 1 & 1 \\ x & y & 1 \\ x^2 & y^2 & 1 \\ \end{array} \right| &= xy \cdot \left| \begin{array}{ccc} 0 & 0 & 1 \\ x - 1 & y -1 & 1 \\ x^2 -1 & y^2 - 1 & 1 \\ \end{array} \right| \\ &= xy \cdot (x-1)(y-1) \left| \begin{array}{ccc} 0 & 0 & 1 \\ 1 & 1 & 1 \\ x + 1 & y + 1 & 1 \end{array} \right| & \text{because } x^2 - 1^2 = (x-1)(x+1)\\ &=xy \cdot (x-1)(y-1) \cdot 1 \cdot(1 \cdot (y+1) - 1 \cdot (x + 1)) & \text{Laplace with row 1}\\ &= xy \cdot (x-1)(x+1)(y-x) \end{align*} $$

Alright, so I had to see myself at (1) that I had to subtract row 3 from row 1 and 2. Then I had to see I had to use the formula $a^2 - b^2 = (a-b)(a+b)$. I'm guessing a lot of you didn't see you had to do that. So my question to you: is there an easier way?

PS: I'll accept 'no' for an answer!

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Your final result is wrong. The effectively $2\times2$ determinant that is left over evaluates to $y-x$, not $(x+1)(y+1)$ (subtraction is not multiplication). –  Marc van Leeuwen Dec 17 '11 at 11:58
    
I edited my mistake –  user3.1415 Dec 17 '11 at 14:44

1 Answer 1

up vote 1 down vote accepted

There is generally no easy way. Concise (closed) formula for the determinants of a general matrix is rare. For your example, it is too special, the determiant is equal to $\left| \begin{array}{cccc}1 & 1 & 1 & 1\\ x & y & 1 & 0\\ x^2 & y^2 & 1 &0\\ x^3 & y^3 & 1&0 \end{array} \right|$, where the matrix is a Vandermonde matrix whose determiant has a closed expression.

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