Example:
Consider the system of equations
\[x_1+x_2+2x_3=9\]
\[2x_1+4x_2-3x_3=1\]
\[3x_1+6x_2-5x_3=0\]
Solve the system to 2 decimal places.
The augmented matrix associated with this system is
\[ \left( \begin{array}{ccc} 1 & 1 & 2 \\ 2 & 4 & -3 \\ 3 & 6 & -5 \end{array} \right| \left| \begin{array}{c} 9 \\ 1 \\ 0 \end{array} \right)\]
We need to change the order of the rows so that the largest element by magnitude is in the first row. We do this by interchanging the first and third row.
\[ \left( \begin{array}{ccc} 3 & 6 & -5 \\ 2 & 4 & -3 \\ 1 & 1 & 2 \end{array} \right| \left| \begin{array}{c} 0 \\ 1 \\ 9 \end{array} \right)\]
Divide every element in the first row by 3 so the leading entry is 1.
\[ \left( \begin{array}{ccc} 1 & 2 & -1.67 \\ 2 & 4 & -3 \\ 1 & 1 & 2 \end{array} \right| \left| \begin{array}{c} 0 \\ 1 \\ 9 \end{array} \right)\]
Subtract the twice the first row from the second and subtract the first row from the third row so the first entry in these rows is zero.
\[ \left( \begin{array}{ccc} 1 & 2 & -1.67 \\ 0 & 0 & 0.34 \\ 0 & -1 & 3.67 \end{array} \right| \left| \begin{array}{c} 0 \\ 1 \\ 9 \end{array} \right)\]
Interchange the second and third rows, multiplying the third row by -1.
\[ \left( \begin{array}{ccc} 1 & 2 & -1.67 \\ 0 & 1 & -3/67 \\ 0 & 0 & 0.34 \end{array} \right| \left| \begin{array}{c} 0 \\ -9 \\ 1 \end{array} \right)\]
Divide the third row by 0.34 so the leading term becomes 1.
\[ \left( \begin{array}{ccc} 1 & 2 & -1.67 \\ 0 & 1 & -3/67 \\ 0 & 0 & 1 \end{array} \right| \left| \begin{array}{c} 0 \\ -9 \\ 2.94 \end{array} \right)\]
Then
\[x_3=2.94\]
\[x_2-3,67x_3=-9 \rightarrow x_2=-9+3.67 x_3 = -9 +3.67 \times 2.94 =1.79\]
\[x_1+2x_2-1.67x_3=0 \rightarrow x_1=-2x_2+1.67x_3 = -2 \times 1.79+1.67 \times 2.94=1.33\]
Gaussian elimination with partial pivoting reducing rounding errors in the calculations when all calculations are to a set number of decimal places or significant figures if ccoefficients are large.