# Solving purely symbolic systems of linear equations

• Aug 1st 2010, 01:01 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:

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

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

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

I next tried setting up the equations in matrix format:

$

\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:

$
\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, 02:19 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, 10:03 AM
jfortiv
Hi Vlasev,