Gaussian Elimination by Baldo Briseno

EXAMPLE PROBLEM

X + 2Y +4Z = 23

Y + 3Z = 16

3Z = 6

We can then turn it into the matrix below

(1)\begin{align} \left(\begin{array}{cc} 1 & 2 & 4 & 23 \\ 0 & 1 & 3 & 16 \\0 & 0 & 3 & 6\end{array}\right) \end{align}

NEW R2 = -R3 + R2

NEW R3 = (1/3)R3

Giving:

\begin{align} \left(\begin{array}{cc} 1 & 2 & 4 & 23 \\ 0 & 1 & 0 & 10 \\0 & 0 & 1 & 2\end{array}\right) \end{align}

NEW R1 = -2R2 + R1

Giving:

\begin{align} \left(\begin{array}{cc} 1 & 0 & 4 & 3 \\ 0 & 1 & 0 & 10 \\0 & 0 & 1 & 2\end{array}\right) \end{align}

NEW R1 = -4R3 + R1

Giving:

\begin{align} \left(\begin{array}{cc} 1 & 0 & 0 & -5\\ 0 & 1 & 0 & 10 \\0 & 0 & 1 & 2\end{array}\right) \end{align}

-Baldo Briseno

-Edited by: Javier Marroquin

page revision: 4, last edited: 03 Feb 2012 03:38