Matice
| Reaktanty | Produkty | ||||||
| As2S3 | NaNO3 | Na2CO3 | Na3AsO4 | Na2SO4 | CO2 | NaNO2 | |
| a | b | c | p | q | r | s | |
| As | 2 | 1 | |||||
| S | 3 | 1 | |||||
| Na | 1 | 2 | 3 | 2 | 1 | ||
| N | 1 | 1 | |||||
| O | 3 | 3 | 4 | 4 | 2 | 2 | |
| C | 1 | 1 | |||||
| náboj | |||||||
Bilance prvků
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+ 2·a | = | + 1·p |
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+ 3·a | = | + 1·q |
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+ 1·b + 2·c | = | + 3·p + 2·q + 1·s |
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+ 1·b | = | + 1·s |
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+ 3·b + 3·c | = | + 4·p + 4·q + 2·r + 2·s |
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+ 1·c | = | + 1·r |
Bilance elektronů (náboje)
Zadání pro program Mathematica
eqns = {
+ 2*a== + 1*p,
+ 3*a== + 1*q,
+ 1*b + 2*c== + 3*p + 2*q + 1*s,
+ 1*b== + 1*s,
+ 3*b + 3*c== + 4*p + 4*q + 2*r + 2*s,
+ 1*c== + 1*r,
+0*a +0*b +0*c== +0*p +0*q +0*r +0*s}
Solve[eqns]
Neznámých koeficientů je: 7, počet nezávislých rovnic je: 6. počet nezávislých rovnic je: 7 - 6 = 1. Jedno z možných řešení je:
a = 1; b = 14; c = 6; p = 2; q = 3; r = 6; s = 14Zadání (program Octave/Matlab) reaction_id-1-5.m
% % Jiri Jirat % Prague Institute of Chemical Technology % % % matice - 1. sloupec naboj, dalsi sloupce prvky % a = [ 0,2,0,0,0,0,3; 0,0,0,1,1,3,0; 0,0,1,0,2,3,0; 0,1,0,0,3,4,0; 0,0,0,0,2,4,1; 0,0,1,0,0,2,0; 0,0,0,1,1,2,0] 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 2 0 0 0 0 3
0 0 0 1 1 3 0
0 0 1 0 2 3 0
0 1 0 0 3 4 0
0 0 0 0 2 4 1
0 0 1 0 0 2 0
0 0 0 1 1 2 0
hodnost = 6
b =
0 0 0 0 0 0 0
2 0 0 1 0 0 0
0 0 1 0 0 1 0
0 1 0 0 0 0 1
0 1 2 3 2 0 1
0 3 3 4 4 2 2
3 0 0 0 1 0 0
c =
0.045739
0.640345
0.274434
-0.091478
-0.137217
-0.274434
-0.640345
reseni =
1.0000 14.0000 6.0000 -2.0000 -3.0000 -6.0000 -14.0000
Zadání (program Mathematica)
m = {
{0,2,0,0,0,0,3},
{0,0,0,1,1,3,0},
{0,0,1,0,2,3,0},
{0,1,0,0,3,4,0},
{0,0,0,0,2,4,1},
{0,0,1,0,0,2,0},
{0,0,0,1,1,2,0}}
NullSpace[Transpose[m]]