LINEARE GLEICHUNGSSYSTEME MIT 3 UNBEKANNTEN

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1. SCHRITT:𝑎1𝑥 + 𝑏1𝑦 + 𝑐1𝑧 = 𝑑1𝑎2𝑥 + 𝑏2𝑦 + 𝑐2𝑧 = 𝑑2𝑎3𝑥 + 𝑏3𝑦 + 𝑐3𝑧 = 𝑑3

→

𝑎1𝑥 + 𝑏1𝑦 + 𝑐1𝑧 = 𝑑1𝑏4𝑦 + 𝑐4𝑧 = 𝑑4𝑏5𝑦 + 𝑐5𝑧 = 𝑑5

x + 𝑦 − z = 9x + 2𝑦 − 4z = 15x + 3𝑦 − 9z = 23

𝐼𝐼𝐼𝐼𝐼𝐼

x + 𝑦 − z = 9−x − 2𝑦 + 4z = −15−x − 3𝑦 + 9z = −23

𝐼∙ (−1)∙ (−1)

x + 𝑦 − z = 9−𝑦 + 3z = −6

−2𝑦 + 8z = −14

𝐼𝐼 + 𝐼𝐼𝐼 + 𝐼𝐼𝐼

2. SCHRITT: 𝑎1𝑥 + 𝑏1𝑦 + 𝑐1𝑧 = 𝑑1

𝑏4𝑦 + 𝑐4𝑧 = 𝑑4𝑏5𝑦 + 𝑐5𝑧 = 𝑑5

→

𝑎1𝑥 + 𝑏1𝑦 + 𝑐1𝑧 = 𝑑1𝑏4𝑦 + 𝑐4𝑧 = 𝑑4

𝑐6𝑧 = 𝑑6

x + 𝑦 − z = 9−𝑦 + 3z = −6

−2𝑦 + 8z = −14

𝐼∙ (−2)𝐼𝐼𝐼

x + 𝑦 − z = 92𝑦 − 6z = 12

−2𝑦 + 8z = −14

𝐼𝐼𝐼𝐼𝐼𝐼

𝑥 + 𝑦 − 𝑧 = 92𝑦 − 6𝑧 = 12

2𝑧 = −2

𝐼𝐼𝐼

𝐼𝐼 + 𝐼𝐼𝐼

3. SCHRITT:𝑎1𝑥 + 𝑏1𝑦 + 𝑐1𝑧 = 𝑑1

𝑏4𝑦 + 𝑐4𝑧 = 𝑑4𝑐6𝑧 = 𝑑6

→

𝑎1𝑥 + 𝑏1𝑦 + 𝑐1𝑧 = 𝑑1𝑏4𝑦 + 𝑐4𝑧 = 𝑑4

𝑧 = 𝑑7

x + 𝑦 − z = 92𝑦 − 6z = 12

2z = −2

𝐼𝐼𝐼𝐼𝐼𝐼

x + 𝑦 − z = 92𝑦 − 6z = 12

z = −1

𝐼𝐼𝐼: 2

4. SCHRITT: 𝑎1𝑥 + 𝑏1𝑦 + 𝑐1𝑧 = 𝑑1

𝑏4𝑦 + 𝑐4𝑧 = 𝑑4𝑧 = 𝑑7

→

𝑎1𝑥 + 𝑏1𝑦 + 𝑐1𝑧 = 𝑑1𝑦 = 𝑑8𝑧 = 𝑑7

x + 𝑦 − z = 92𝑦 − 6z = 12

z = −1

𝐼𝐼𝐼𝐼𝐼𝐼

x + 𝑦 − z = 92𝑦 − 6 ∙ (−1) = 12

z = −1

𝐼𝐼𝐼𝐼 𝑖𝑛 𝐼𝐼 𝑒𝑖𝑛𝑔𝑒𝑠𝑒𝑡𝑧𝑡

𝐼𝐼𝐼

x + 𝑦 − z = 92𝑦 = 6z = −1

𝐼−6𝐼𝐼𝐼

x + 𝑦 − z = 9𝑦 = 3z = −1

𝐼: 2𝐼𝐼𝐼

5. SCHRITT: 𝑎1𝑥 + 𝑏1𝑦 + 𝑐1𝑧 = 𝑑1

𝑦 = 𝑑8𝑧 = 𝑑7

→

𝑥 = 𝑑9𝑦 = 𝑑8𝑧 = 𝑑7

x + 𝑦 − z = 9𝑦 = 3z = −1

𝐼𝐼𝐼𝐼𝐼𝐼

x + 3 − (−1) = 9𝑦 = 3z = −1

𝐼𝐼𝐼𝐼𝐼𝐼

x = 𝟓𝑦 = 𝟑z = −𝟏

−4𝐼𝐼𝐼𝐼𝐼

IL = {(5,3,−1)}