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Sistem preoblikujemo na
1x + 1y + 1z + 1w = 10
-1x + 1y - 1z + 1w = 2
1x - 1y - 1z + 1w = 0
-1x - 1y + 1z + 1w = 4
in si naredimo tabelo
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| x | y | z | w | 1 |
faktor |
pomnožimo => |
x | y | z | w | 1 |
odštejemo => |
|---|
| 1 | 1 | 1 | 1 | 10 |
/ |
1 | 1 | 1 | 1 | 10 |
| -1 | 1 | -1 | 1 | 2 |
-1 |
1 | -1 | 1 | -1 | -2 |
| 1 | -1 | -1 | 1 | 0 |
/ |
1 | -1 | -1 | 1 | 0 |
| -1 | -1 | 1 | 1 | 4 |
-1 |
1 | 1 | -1 | -1 | -4 |
|
| x | y | z | w | 1 |
odštejemo => |
x | y | z | w | 1 |
faktor |
pomnožimo => |
|---|
| 1 | 1 | 1 | 1 | 10 |
1 | 1 | 1 | 1 | 10 |
/ |
| 0 | 2 | 0 | 2 | 12 |
0 | 2 | 0 | 2 | 12 |
/ |
| 0 | 2 | 2 | 0 | 10 |
0 | 0 | -2 | 2 | 2 |
/ |
| 0 | 0 | 2 | 2 | 14 |
0 | 0 | 2 | 2 | 14 |
-1 |
|
| x | y | z | w | 1 |
odštejemo => |
x | y | z | w | 1 |
dobimo => |
|---|
| 1 | 1 | 1 | 1 | 10 |
1 | 1 | 1 | 1 | 10 |
| 0 | 2 | 0 | 2 | 12 |
0 | 2 | 0 | 2 | 12 |
| 0 | 2 | -2 | 2 | 2 |
0 | 0 | -2 | 2 | 2 |
| 0 | 0 | -2 | -2 | -14 |
0 | 0 | 0 | 4 | 16 |
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Dobimo w = 4, z = 3, y = 2 in x = 1.
Nazaj
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