# Gaussian Elimination Method

• October 17th 2007, 08:37 AM
flippin4u
Gaussian Elimination Method
Help me if you can: Here are the questions.

1) Sovle the system of equations by the Gaussian elimination method:

{ 2x + y - 3z = 1
{ 3x - y + 4z = 6
{ x + 2y - z = 9

2) Solver the system of equations by the Gaussian elimination method:

{ x - y + z = 17
{ x + y - z = -11
{ x - y - z = 9

Thanks in advance for any help!!!

Kasey
• October 17th 2007, 10:23 AM
Jhevon
I'll do the easier one, the other is similar.
Quote:

Originally Posted by flippin4u
2) Solver the system of equations by the Gaussian elimination method:

{ x - y + z = 17
{ x + y - z = -11
{ x - y - z = 9

Step 1: Create an augmented matrix:

.... $x$.... $y$..... $z$
$\left( \begin{array}{ccc|c} 1 & -1 & 1 & 17 \\ 1 & 1 & -1 & -11 \\ 1 & -1 & -1 & 9 \end{array} \right)$

Step 2:
Now, Run Gauss-Jordan elimination on it to bring it to row-echelon form (or reduced row echelon form, i will do this).

.... $x$.... $y$..... $z$
$\left( \begin{array}{ccc|c} 1 & -1 & 1 & 17 \\ 1 & 1 & -1 & -11 \\ 1 & -1 & -1 & 9 \end{array} \right)$
----------------------
$\left( \begin{array}{ccc|c} 1 & -1 & -1 & 9 \\ 0 & 0 & 2 & 8 \\ 0 & 2 & 0 & -20 \end{array} \right)$
----------------------
$\left( \begin{array}{ccc|c} 1 & -1 & -1 & 9 \\ 0 & 0 & 2 & 8 \\ 0 & 2 & 0 & -20 \end{array} \right)$
----------------------
$\left( \begin{array}{ccc|c} 1 & -1 & -1 & 9 \\ 0 & 1 & 0 & -10 \\ 0 & 0 & 1 & 4 \end{array} \right)$
----------------------
$\left( \begin{array}{ccc|c} 1 & 0 & -1 & -1 \\ 0 & 1 & 0 & -10 \\ 0 & 0 & 1 & 4 \end{array} \right)$
--------------------
$\left( \begin{array}{ccc|c} 1 & 0 & 0 & 3 \\ 0 & 1 & 0 & -10 \\ 0 & 0 & 1 & 4 \end{array} \right)$
------------------

$x = 3, y = -10 \mbox { and } z = 4$

Now try the first one
• October 17th 2007, 01:27 PM
Soroban
Hello, Kasey!

Do you anything about Gaussian elimination?
. . The second one practically falls apart . . .

Quote:

$2)\;\begin{array}{ccc}x - y + z & = & 17 \\
x + y - z & = & \text{-}11 \\
x - y - z & = & 9\end{array}$

We have: . $\begin{bmatrix}1 & \text{-}1 & 1 &|& 17 \\
1 & 1 & \text{-}1 &| &\text{-}11 \\
1 & \text{-}1 & \text{-}1 &|& 9 \end{bmatrix}$

. $\begin{array}{c}\\ R_2-R_1 \\ R_3-R_1\end{array}\;
\begin{bmatrix}1 & \text{-}1 & 1 &|& 17 \\
0 & 2 & \text{-}2 &|& \text{-}28 \\
0 & 0 & \text{-}2 &|& \text{-}8 \end{bmatrix}$

. . . $\begin{array}{c} \\ \frac{1}{2}R_2 \\ \text{-}\frac{1}{2}R_3\end{array}\;
\begin{bmatrix}1 & \text{-}1 & 1 &|& 17 \\
0 & 1 & \text{-}1 &|& \text{-}14 \\
0 & 0 & 1 &|& 4 \end{bmatrix}$

. $\begin{array}{c}R_1+R_2 \\ R_2+R_3 \\ \\ \end{array}\;
\begin{bmatrix}1 & 0 & 0 &|& 3 \\
0 & 1 & 0 &|& \text{-}10 \\
0 & 0 & 1 &|& 4 \end{bmatrix}$