What is the best way to find triple roots in a 6x6 matrix?

[Irid]

I was solving a mechanics problem, and eventually obtained this matrix:

$$\begin{vmatrix}<br /> 1-\Omega^2 & -1 & -1/2 & 0 & 0 & 0\\<br /> -1 & 3/2 -\Omega^2 & 0 & 0 & -1/2 & 1/2\\<br /> 0 & -1/2 & 1/2-\Omega^2 & 0 & 1/2 & -1/2\\<br /> 0 & 0 & 0 & 1-\Omega^2 & 0 & -1\\<br /> 0 & -1/2 & 1/2 & 0 & 1/2-\Omega^2 & -1/2\\<br /> 0 & 1/2 & -1/2 & -1 & -1/2 & 3/2-\Omega^2<br /> \end{vmatrix} = 0$$

I must find the roots $$\Omega^2$$. The textbook gives a hint that I should rearrange columns to show that there is a triple root $$\Omega^2 = 0$$, and then reduce the matrix to 3x3 and then the determinant is easy from there. I'm able to switch columns back and forth, but I don't know how to find any triple roots from that. I'm pretty new to matrices. Could you give me some useful properties or anything else that would be helpful?

Oh, and the matrix may be wrong, but not too much wrong, maybe just a few elements... Notice that it's symmetric.

 

[gabbagabbahey]

I think the idea is to use elementary row and column operations to get a matrix of the form:
$$\begin{vmatrix}<br /> \Omega^2 & 0 & 0 & 0 & 0 & 0\\<br /> 0 & \Omega^2 & 0 & 0 & 0 & 0\\<br /> 0 & 0 & \Omega^2 & 0 & 0 & 0\\<br /> 0 & 0 & 0 & \ldots & \ldots & \ldots \\<br /> 0 & 0 & 0 & \ldots & \ldots & \ldots \\<br /> 0 & 0 & 0 & \ldots & \ldots & \ldots<br /> \end{vmatrix} = 0$$

Each time you interchange two rows or columns in a matrix, you change the determinant by a factor of -1. Each time you add a multiple of a row or column to another row or column, the determinant is unchanged.

In this case you can use either of these operations without affecting the result. (-1*0=0)