844 APPENDIX E
D 1 =∣ ∣ ∣ ∣ ∣ ∣
b 1 a 12 a 13
b 2 a 22 a 23
b 3 a 32 a 33∣ ∣ ∣ ∣ ∣ ∣=b 1∣
∣
∣
∣a 22 a 23
a 32 a 33∣
∣
∣
∣−a^12∣
∣
∣
∣b 2 a 23
b 3 a 33∣
∣
∣
∣+a^13∣
∣
∣
∣b 2 a 22
b 3 a 32∣
∣
∣
∣=b 1 (a 22 a 33 −a 23 a 32 )−a 12 (b 2 a 33 −a 23 b 3 )+a 13 (b 2 a 32 −a 22 b 3 ) (6b)D 2 =∣ ∣ ∣ ∣ ∣ ∣
a 11 b 1 a 13
a 21 b 2 a 23
a 31 b 3 a 33∣ ∣ ∣ ∣ ∣ ∣=a 11∣
∣
∣
∣b 2 a 23
b 3 a 33∣
∣
∣
∣−b^1∣
∣
∣
∣a 21 a 23
a 31 a 33∣
∣
∣
∣+a^13∣
∣
∣
∣a 21 b 2
a 31 b 3∣
∣
∣
∣=a 11 (b 2 a 33 −a 23 b 3 )−b 1 (a 21 a 33 −a 23 a 31 )+a 13 (a 21 b 3 −b 2 a 31 ) (6c)D 3 =∣ ∣ ∣ ∣ ∣ ∣
a 11 a 21 b 3
a 21 a 22 b 2
a 31 a 32 b 3∣ ∣ ∣ ∣ ∣ ∣=a 11∣
∣
∣
∣a 22 a 23
a 32 a 33∣
∣
∣
∣−a^12∣
∣
∣
∣a 21 b 2
a 31 b 3∣
∣
∣
∣+b^1∣
∣
∣
∣a 21 a 22
a 31 a 32∣
∣
∣
∣=a 11 (a 22 b 3 −b 2 a 32 )−a 12 (a 21 b 3 −b 2 a 31 )+b 1 (a 21 a 32 −a 22 a 31 ) (6d)As an alternative approach to the solution of three simultaneous equations, typically given
by Equations (4a), (4b), and (4c), one can always use one of the equations to express an unknown
in terms of the other two; substitution into the remaining two equations reduces the problem to
two equations with two unknowns. This method is most convenient when one does not need the
unknown eliminated and one of theb-terms equals zero to simplify the elimination process.GAUSS ELIMINATION
A simple technique, which is also used in digital computer methods, for solving linear simul-
taneous algebraic equations is the method of Gauss elimination. The basic idea is to reduce the
equations through manipulations to anequivalentform which istriangular; for example, the
equation set of Equation (4) would be simplified to its equivalent in a triangular form given
typically by
c 11 x 1 +c 12 x 2 +c 13 x 3 =d 1 (7a)
c 22 x 2 +c 23 x 3 =d 2 (7b)
c 33 x 3 =d 3 (7c)
The reduction of Equation (4) to that of Equation (7) is accomplished by the following tasks:- One can multiply any equation in the set by a nonzero number without changing the
solution. - One can also add or subtract any two equations and replace one of the two with the result.
Once the triangular form of Equation (7) is achieved, one can solve for the unknowns byback
substitution:
From Equation (7c),x 3 =d 3
c 33(8a)Substituting Equation (8a) in Equation (7b), one gets