L— 9 J
1
1
—9-
P
17
Zo voortgaande komen we tenslotte aan de laatste vergelijking
p2\ x2 ~f" P31 ^3 Pni xn
(22)
Substitutie van xn
Voorbeeld
x3, x2 in (22) geeft tenslotte xv
3—1 6
x1
I
1 2—3
0
2 —3 —1
x3
Orthogonaliseren van de basis geeft het volgende resultaat
3 1/2 10/3
r*ll
I
i 5/2 —10/3
0
to
to
H
Co"
Hieruit volgt: X1 15/14; X2 37/21; X3 1.
X3 Tfg I.
Pil
I
O
P31
p32
I
I
O
L O
-1/2
I
O
x2= X2 5/21 x3 37/21 5/21 2.
Xl X1+ 1/2 *2 13/14 x3 —15/14
I3/I4
—5/21
1
13/14 i-
Summary. The solution of n equations with n unknowns is based on the
following operations:
1. Orthogonalisation of the basis vectors av a2, an and calculation of
the matrix P with the aid of the formulas (7), (10), (12).
2. Calculation of Xn n 1, 2, x according to formula (15).
3. Calculation of xn, *„_2, xt according to the formulas (19),
(20), (21), (22).
The technique described is illustrated by an example.
