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Sistem preoblikujemo na
1x + 1y + 0z = 5
0x - 1y + 1z = 1
3x + 0y - 1z = 2
in si naredimo tabelo
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| x | y | z | 1 | faktor |
pomnožimo => |
x | y | z | 1 |
odštejemo => |
|---|
| 1 | 1 | 0 | 5 | 3 |
3 | 3 | 0 | 15 |
| 0 | -1 | 1 | 1 | / |
0 | -1 | 1 | 1 |
| 3 | 0 | -1 | 2 | / |
3 | 0 | -1 | 2 |
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| x | y | z | 1 | faktor |
pomnožimo => |
x | y | z | 1 |
odštejemo => |
|---|
| 1 | 1 | 0 | 5 | / |
1 | 1 | 0 | 5 |
| 0 | -1 | 1 | 1 | -3 |
0 | 3 | -3 | -3 |
| 0 | 3 | 1 | 13 | / |
0 | 3 | 1 | 13 |
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| x | y | z | 1 |
dobimo => |
|---|
| 1 | 1 | 0 | 5 |
| 0 | 1 | -1 | -1 |
| 0 | 0 | -4 | -16 |
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Dobimo z = 4, y = 3 in x = 2.
Nazaj
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