Question 9.3: Use Gauss elimination to solve 3x1 − 0.1x2 − 0.2x3 = 7.85 (E...

The first part of the procedure is forward elimination. Multiply Eq. (E9.3.1) by 0.1∕3 and subtract the result from Eq. (E9.3.2) to give
7.00333x_2 − 0.293333x_3 = −19.5617
Then multiply Eq. (E9.3.1) by 0.3∕3 and subtract it from Eq. (E9.3.3). After these operations, the set of equations is
3x_1 − 0.1x_2 − 0.2x_3 = 7.85       (E9.3.4)
7.00333x_2 − 0.293333x_3 = −19.5617                                          (E9.3.5)
− 0.190000x_2 + 10.0200x_3 = 70.6150                     (E9.3.6)
To complete the forward elimination, x_2 must be removed from Eq. (E9.3.6). To accomplish this, multiply Eq. (E9.3.5) by −0.190000∕7.00333 and subtract the result from Eq. (E9.3.6). This eliminates x_2 from the third equation and reduces the system to an upper triangular form, as in
3x_1 − 0.1x_2 − 0.2x_3 = 7.85               (E9.3.7)
7.00333x_2 − 0.293333x_3 = −19.5617                                             (E9.3.8)
10.0120x_3 = 70.0843                            (E9.3.9)
We can now solve these equations by back substitution. First, Eq. (E9.3.9) can be solved for
x_3 = \frac{70.0843}{10.0120} = 7.00003
This result can be back-substituted into Eq. (E9.3.8), which can then be solved for
x_2 = \frac{−19.5617 +0.293333(7.00003)}{7.00003} = −2.50000
Finally, x_3 = 7.00003 \text{ and } x_2 = −2.50000 can be substituted back into Eq. (E9.3.7), which can be solved for
x_1 = \frac{7.85+0.1(−2.50000) +0.2(7.00003)}{3} = 3.00000
Although there is a slight round off error, the results are very close to the exact solution of x_1 = 3, x_2 = −2.5, \text{ and }x_3 = 7. This can be verified by substituting the results into the original equation set:
3(3) − 0.1(−2.5) − 0.2(7.00003) = 7.84999 ≅ 7.85
0.1(3) + 7(−2.5) − 0.3(7.00003) = −19.30000 = −19.3
0.3(3) − 0.2(−2.5) + 10(7.00003) = 71.4003 ≅ 71.4