Question 9.4: Use Gauss elimination to solve 0.0003x1 + 3.0000x2 = 2.0001 ...

Multiplying the first equation by 1∕(0.0003) yields
x_1 + 10,000x_2 = 6667
which can be used to eliminate x_1 from the second equation:
−9999x_2 = −6666
which can be solved for x_2 = 2∕3. This result can be substituted back into the first equation to evaluate x_1:
x_1 =\frac{2.00001 −3( 2∕3)}{0.0003}                                                   (E9.4.1)
Due to subtractive cancellation, the result is very sensitive to the number of significant figures carried in the computation:

Note how the solution for x_1 is highly dependent on the number of significant figures. This is because in Eq. (E9.4.1), we are subtracting two almost-equal numbers.

x_1 =\frac{2.00001 −3( 2∕3)}{0.0003}                                                   (E9.4.1)
On the other hand, if the equations are solved in reverse order, the row with the larger pivot element is normalized. The equations are
1.0000x_1 + 1.0000x_2 = 1.0000
0.0003x_1 + 3.0000x_2 = 2.0001
Elimination and substitution again yields x_2 = 2∕3. For different numbers of significant figures, x_1 can be computed from the first equation, as in
x_1 = \frac{1−(2∕3)}{1}
This case is much less sensitive to the number of significant figures in the computation:

Thus, a pivot strategy is much more satisfactory.