# Identifying factors of higher order in a determinant

Consider a $n\times n$ matrix $A$ whose elements are some polynomials in the indeterminates $x_1, x_2,\ldots,x_m$. To calculate the determinant of such a matrix, one of the usual ways is to treat the determinant as a polynomial in $x_1,\ldots,x_m$ and identify its factors. The usual idea being if $x = y$ makes the determinant vanish then $x -y$ is one of the factors. What I however do not understand is how to identify its order, that is to identify the exact $k$ such that $(x - y)^k$ is the factor.

For example, consider the matrix

$\left( \begin{array}{ccc} (x + a_1)^2 & (x + a_1)*(y + a_1) & (y + a_1) \\ (x + a_2)^2 & (x + a_2)*(y + a_2) & (y + a_2) \\ (x + a_3)^2 & (x + a_1)*(y + a_3) & (y + a_3) \\ \end{array} \right)$. The determinant of this matrix is given by $(x - y)^2 * (a_1 - a_2) * (a_1 - a_3) * (a_2 - a_3).$ I want to know how to guess the order $2$ of $x-y$.

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You can see it as a polynomial in $x$ and fix $y$. If you compute the derivative, you get three determinant, which all vanish when $x=y$, but when you take the second derivative it may not be equal to $0$. – Davide Giraudo Jan 29 '12 at 17:16
If you set $x=y$, the resulting matrix has rank 1. – Kevin Jan 29 '12 at 19:22
The rank of the original matrix above is 3. After substituting $x = y$, the rank now becomes $2$. Am i missing something? – Suresh Jan 30 '12 at 3:12
Thank your very much Giraudo. Your suggestion seems interesting. I will work on this. – Suresh Jan 30 '12 at 3:13