Gaussian Elimination Example By Rcantua

considered the following equations:
x - 3y + z = 4
2x - 8y + 8z = -2
-6x + 3y - 15z = 9

We can convert the equations into an augmented matrix:

(1)
\begin{align} \left(\begin{array}{cc} 1 & -3 & 1 & 4 \\ 2 & -8 & 8 & -2 \\-6 & 3 & -15 & 9\end{array}\right) \end{align}

When we look at the start of the left column, we find 1 which is a non-zero number, which means that this works as a pivot.

Next we try to get entry (2,1) to equal zero. To do this we can take one-third times the 3rd row and add it to the 2nd row. (1/3)R3+ R2 = R2. (*we could have also done -2R1+R2=R2)

Now our matrix is:

(2)
\begin{align} \left(\begin{array}{cc} 1 & -3 & 1 & 4 \\ 0 & -7 & 3 & 1 \\-6 & 3 & -15 & 9\end{array}\right) \end{align}

Now we can try to get a zero in entry (3,1). We can do (6)R1+R3 = R3

Now our matrix is:

(3)
\begin{align} \left(\begin{array}{cc} 1 & -3 & 1 & 4 \\ 0 & -7 & 3 & 1 \\0 & -15 & -9 & 33\end{array}\right) \end{align}

Now we can try to get a zero in entry (3,2). We can do (-15/7)R2+R3 = R3

Now our matrix is:

(4)
\begin{align} \left(\begin{array}{cc} 1 & -3 & 1 & 4 \\ 0 & -7 & 3 & 1 \\0 & 0 & (-108/7) & (216/7)\end{array}\right) \end{align}

From this we can figure out that z = -2 as the third row means 0x + 0y + (-108/7)z = 216/7.

Now we can substitute Z back into the 2nd equation.
0x + -7y + 3(-2) = 1
y = -1

And finally we can substitute y and z into the 1st equation
1x + -3(-1) + 1(-2) = 4
x = 3

so we have
x = 3
y = -1
z = -2

page revision: 3, last edited: 01 Feb 2012 18:50