Matice
| Reaktanty | Produkty | |||||
| As4O6 | Zn | H2SO4 | AsH3 | ZnSO4 | H2O | |
| a | b | c | p | q | r | |
| As | 4 | 1 | ||||
| O | 6 | 4 | 4 | 1 | ||
| Zn | 1 | 1 | ||||
| H | 2 | 3 | 2 | |||
| S | 1 | 1 | ||||
| náboj | ||||||
Bilance prvků
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|
+ 4·a | = | + 1·p |
|
|
+ 6·a + 4·c | = | + 4·q + 1·r |
|
|
+ 1·b | = | + 1·q |
|
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+ 2·c | = | + 3·p + 2·r |
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+ 1·c | = | + 1·q |
Bilance elektronů (náboje)
Zadání pro program Mathematica
eqns = {
+ 4*a== + 1*p,
+ 6*a + 4*c== + 4*q + 1*r,
+ 1*b== + 1*q,
+ 2*c== + 3*p + 2*r,
+ 1*c== + 1*q,
+0*a +0*b +0*c== +0*p +0*q +0*r}
Solve[eqns]
Neznámých koeficientů je: 6, počet nezávislých rovnic je: 5. počet nezávislých rovnic je: 6 - 5 = 1. Jedno z možných řešení je:
a = 1; b = 12; c = 12; p = 4; q = 12; r = 6Zadání (program Octave/Matlab) reaction_id-5-12.m
% % Jiri Jirat % Prague Institute of Chemical Technology % % % matice - 1. sloupec naboj, dalsi sloupce prvky % a = [ 0,4,0,6,0,0; 0,0,0,0,0,1; 0,0,2,4,1,0; 0,1,3,0,0,0; 0,0,0,4,1,1; 0,0,2,1,0,0] hodnost = rank(a) % hodnost matice = pocet nezavislych rovnic b = a' % transpozice matice c = null(b) % nalezeni baze nuloveho prostoru matice b reseni = rref(c') % upravy na "row reduced echelon form"
Řešení (program Octave/Matlab)
a =
0 4 0 6 0 0
0 0 0 0 0 1
0 0 2 4 1 0
0 1 3 0 0 0
0 0 0 4 1 1
0 0 2 1 0 0
hodnost = 5
b =
0 0 0 0 0 0
4 0 0 1 0 0
0 0 2 3 0 2
6 0 4 0 4 1
0 0 1 0 1 0
0 1 0 0 1 0
c =
0.045408
0.544892
0.544892
-0.181631
-0.544892
-0.272446
reseni =
1.0000 12.0000 12.0000 -4.0000 -12.0000 -6.0000
Zadání (program Mathematica)
m = {
{0,4,0,6,0,0},
{0,0,0,0,0,1},
{0,0,2,4,1,0},
{0,1,3,0,0,0},
{0,0,0,4,1,1},
{0,0,2,1,0,0}}
NullSpace[Transpose[m]]