Geoscience Reference
In-Depth Information
Putting these values in the normal equations gives:
11 436 9643
,
.
b
−
1171 4642
.
b
+
3458 8215
.
b
=
6428
.
7858
1
2
3
−
1171 4642
.
b
+
5998 9643
.
b
+
1789 6786
.
b
=
2
632 2143
.
1
2
3
3458 8215
.
b
−
1789 6786
.
b
+
2606 1072
.
b
=
3327 9286
.
1
2
3
These equations can be solved by any of the standard procedures for simultaneous equations. One
approach (applied to the above equations) is as follows:
1. Divide through each equation by the numerical coefficient of
b
1
.
b
−
0 102 427 897
.
,
,
b
+
0 302 424 788
.
,
,
b
=
0 562 10
.
,
5 960
,
1
2
3
b
−
5 120 911 334
.
,
,
b
−
1 527 727 949
.
,
,
b
=− ..
2
246 943 867
,
,
1
2
3
b
+
0 517 424 389
.
,
,
b
+
0 753 466 8
.
,
,
09
b
= .
0 962 156 792
,
,
1
2
2. Subtract the second equation from the first and the third from the first so as to leave two
equations in
b
2
and
b
3
.
5.018,483,437
b
2
+ 1.830,152,737
b
3
= 2.809,049,827
- 0.619,852,286
b
2
- 0.451,042,021
b
3
= -0.400,050,832
3. Divide through each equation by the numerical coefficient of
b
2
.
b
2
+ 0.364,682,430
b
3
= 0.559,740,779
b
2
+ 0.727,660,494
b
2
= 0.645,397,042
4. Subtract the second of these equations from the first, leaving one equation in
b
3
.
-0.362,978,064
b
3
= -0.085,656,263
5. Solve for
b
3
.
=
−
−
0 085 656 263
0 326 978 064
.
,
,
b
3
=
0 235 981 9
.
,
,27
.
,
,
6. Substitute this value of
b
3
in one of the equations (the first one, for example) of step 3 and
solve for
b
2
.
b
2
+ (0.364,682,43)(0.381,927) = 0.59,740,779
b
2
= 0.473,682,316
7. Substitute the solutions for
b
2
and
b
3
in one of the equations (the first one, for example) of
step 1, and solve for
b
1
.
b
1
- (0.102,427,897)(0.473,682,316) + (0.302,424,788)(0.235,981,927) = 0.562,105,960
b
1
= 0.539,257,459
Search WWH ::
Custom Search