## Q. 9.3

Use Gauss elimination to solve

$3x_1 − 0.1x_2 − 0.2x_3 = 7.85$          (E9.3.1)

$0.1x_1 + 7x_2 − 0.3x_3 = −19.3$               (E9.3.2)

$0.3x_1 − 0.2x_2 + 10x_3 = 71.4$                    (E9.3.3)

## Verified Solution

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