# Solving purely symbolic systems of linear equations

• Aug 1st 2010, 12:01 AM
jfortiv
Solving purely symbolic systems of linear equations
Hello,

I'm working through a finite element text which provides a symbolic solution to a system of linear equations which I suspect might be incorrect based on some checks that I've done. I'm interested in double checking their math, but I'm finding it excessively complicated with the techniques that I know how to use. Here is the system:

$\displaystyle u_i=\alpha_1+\alpha_2 x_i + \alpha_3 y_i$
$\displaystyle u_j=\alpha_1+\alpha_2 x_j + \alpha_3 y_j$
$\displaystyle u_m=\alpha_1+\alpha_2 x_m + \alpha_3 y_m$

I'm trying to solve for $\displaystyle \alpha_1, \alpha_2, \alpha_3$.

First I tried substitution by solving the first equation for $\displaystyle \alpha_1$ and plugging into the second equation. Then I tried solving that for $\displaystyle \alpha_2$, etc. The math just got so messy that I gave up.

I next tried setting up the equations in matrix format:

$\displaystyle \left[\begin{matrix}1&x_i&y_i\\1&x_j&y_j\\1&x_m&y_m\end{ matrix}\right] \left[\begin{matrix}\alpha_1\\ \alpha_2\\ \alpha_3\end{matrix}\right] = \left[\begin{matrix}u_i\\ u_j\\ u_m\end{matrix}\right]$

I tried solving for the alpha vector by inverting the 3x3 matrix using Gauss-Jordan elimination on the following:

$\displaystyle \left[\begin{matrix}1&x_i&y_i&1&0&0\\1&x_j&y_j&0&1&0\\1& x_m&y_m&0&0&1\end{matrix}\right]$

This seems just as complicated. I noticed that the book made mention of a determinant in the solution (although it did not show a solution process. Is there an easier way of solving this system symbolically that I am not aware of?

Thanks,
James
• Aug 1st 2010, 01:19 AM
Vlasev
I think it's just Gaussian elimination. However, since you must divide by variables at some spots, you gotta make sure that you don't get zeroes in the denominator. The system has a unique solution if the matrix is invertible, right? So you can just take the determinant and see.

Hint: Once you solve for a1,a2 and a3, the denominators will be just the determinant of the coefficient matrix.

In fact, try the 2D case and then extend it.
• Aug 1st 2010, 09:03 AM
jfortiv
Hi Vlasev,

Thanks for your response!

I went through the 2D case and was able to prove to myself that the denominator is the determinant of the coefficient matrix. I would like to now understand how I can use that knowledge a priori when solving to somehow simplify the process. As I see it, it doesn't really eliminate any complexity if I still have to solve the whole system longhand and then pull out the determinant from the denominators of each entry after the fact. Is there some trick that I'm missing here?

Thanks,
James