Elementarmathematik Beispiele

$5 x - y - z = 11$ , $7 x + 5 y + 5 z = 9$ , $- 4 x - 9 y - 9 z = 2$

Schritt 1

Write the system as a matrix.

$[\begin{array}{c} 5 & - 1 & - 1 & 11 \\ 7 & 5 & 5 & 9 \\ - 4 & - 9 & - 9 & 2 \end{array}]$

Schritt 2

Ermittele die normierte Zeilenstufenform.

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Schritt 2.1

Multiply each element of $R_{1}$ by $\frac{1}{5}$ to make the entry at $1, 1$ a $1$ .

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Schritt 2.1.1

Multiply each element of $R_{1}$ by $\frac{1}{5}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} \frac{5}{5} & \frac{- 1}{5} & \frac{- 1}{5} & \frac{11}{5} \\ 7 & 5 & 5 & 9 \\ - 4 & - 9 & - 9 & 2 \end{array}]$

Schritt 2.1.2

Vereinfache $R_{1}$ .

$[\begin{array}{c} 1 & - \frac{1}{5} & - \frac{1}{5} & \frac{11}{5} \\ 7 & 5 & 5 & 9 \\ - 4 & - 9 & - 9 & 2 \end{array}]$

Schritt 2.2

Perform the row operation $R_{2} = R_{2} - 7 R_{1}$ to make the entry at $2, 1$ a $0$ .

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Schritt 2.2.1

Perform the row operation $R_{2} = R_{2} - 7 R_{1}$ to make the entry at $2, 1$ a $0$ .

$[\begin{array}{c} 1 & - \frac{1}{5} & - \frac{1}{5} & \frac{11}{5} \\ 7 - 7 \cdot 1 & 5 - 7 (- \frac{1}{5}) & 5 - 7 (- \frac{1}{5}) & 9 - 7 (\frac{11}{5}) \\ - 4 & - 9 & - 9 & 2 \end{array}]$

Schritt 2.2.2

Vereinfache $R_{2}$ .

$[\begin{array}{c} 1 & - \frac{1}{5} & - \frac{1}{5} & \frac{11}{5} \\ 0 & \frac{32}{5} & \frac{32}{5} & - \frac{32}{5} \\ - 4 & - 9 & - 9 & 2 \end{array}]$

Schritt 2.3

Perform the row operation $R_{3} = R_{3} + 4 R_{1}$ to make the entry at $3, 1$ a $0$ .

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Schritt 2.3.1

Perform the row operation $R_{3} = R_{3} + 4 R_{1}$ to make the entry at $3, 1$ a $0$ .

$[\begin{array}{c} 1 & - \frac{1}{5} & - \frac{1}{5} & \frac{11}{5} \\ 0 & \frac{32}{5} & \frac{32}{5} & - \frac{32}{5} \\ - 4 + 4 \cdot 1 & - 9 + 4 (- \frac{1}{5}) & - 9 + 4 (- \frac{1}{5}) & 2 + 4 (\frac{11}{5}) \end{array}]$

Schritt 2.3.2

Vereinfache $R_{3}$ .

$[\begin{array}{c} 1 & - \frac{1}{5} & - \frac{1}{5} & \frac{11}{5} \\ 0 & \frac{32}{5} & \frac{32}{5} & - \frac{32}{5} \\ 0 & - \frac{49}{5} & - \frac{49}{5} & \frac{54}{5} \end{array}]$

Schritt 2.4

Multiply each element of $R_{2}$ by $\frac{5}{32}$ to make the entry at $2, 2$ a $1$ .

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Schritt 2.4.1

Multiply each element of $R_{2}$ by $\frac{5}{32}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & - \frac{1}{5} & - \frac{1}{5} & \frac{11}{5} \\ \frac{5}{32} \cdot 0 & \frac{5}{32} \cdot \frac{32}{5} & \frac{5}{32} \cdot \frac{32}{5} & \frac{5}{32} (- \frac{32}{5}) \\ 0 & - \frac{49}{5} & - \frac{49}{5} & \frac{54}{5} \end{array}]$

Schritt 2.4.2

Vereinfache $R_{2}$ .

$[\begin{array}{c} 1 & - \frac{1}{5} & - \frac{1}{5} & \frac{11}{5} \\ 0 & 1 & 1 & - 1 \\ 0 & - \frac{49}{5} & - \frac{49}{5} & \frac{54}{5} \end{array}]$

Schritt 2.5

Perform the row operation $R_{3} = R_{3} + \frac{49}{5} R_{2}$ to make the entry at $3, 2$ a $0$ .

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Schritt 2.5.1

Perform the row operation $R_{3} = R_{3} + \frac{49}{5} R_{2}$ to make the entry at $3, 2$ a $0$ .

$[\begin{array}{c} 1 & - \frac{1}{5} & - \frac{1}{5} & \frac{11}{5} \\ 0 & 1 & 1 & - 1 \\ 0 + \frac{49}{5} \cdot 0 & - \frac{49}{5} + \frac{49}{5} \cdot 1 & - \frac{49}{5} + \frac{49}{5} \cdot 1 & \frac{54}{5} + \frac{49}{5} \cdot - 1 \end{array}]$

Schritt 2.5.2

Vereinfache $R_{3}$ .

$[\begin{array}{c} 1 & - \frac{1}{5} & - \frac{1}{5} & \frac{11}{5} \\ 0 & 1 & 1 & - 1 \\ 0 & 0 & 0 & 1 \end{array}]$

Schritt 2.6

Perform the row operation $R_{2} = R_{2} + R_{3}$ to make the entry at $2, 4$ a $0$ .

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Schritt 2.6.1

Perform the row operation $R_{2} = R_{2} + R_{3}$ to make the entry at $2, 4$ a $0$ .

$[\begin{array}{c} 1 & - \frac{1}{5} & - \frac{1}{5} & \frac{11}{5} \\ 0 + 0 & 1 + 0 & 1 + 0 & - 1 + 1 \cdot 1 \\ 0 & 0 & 0 & 1 \end{array}]$

Schritt 2.6.2

Vereinfache $R_{2}$ .

$[\begin{array}{c} 1 & - \frac{1}{5} & - \frac{1}{5} & \frac{11}{5} \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}]$

Schritt 2.7

Perform the row operation $R_{1} = R_{1} - \frac{11}{5} R_{3}$ to make the entry at $1, 4$ a $0$ .

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Schritt 2.7.1

Perform the row operation $R_{1} = R_{1} - \frac{11}{5} R_{3}$ to make the entry at $1, 4$ a $0$ .

$[\begin{array}{c} 1 - \frac{11}{5} \cdot 0 & - \frac{1}{5} - \frac{11}{5} \cdot 0 & - \frac{1}{5} - \frac{11}{5} \cdot 0 & \frac{11}{5} - \frac{11}{5} \cdot 1 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}]$

Schritt 2.7.2

Vereinfache $R_{1}$ .

$[\begin{array}{c} 1 & - \frac{1}{5} & - \frac{1}{5} & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}]$

Schritt 2.8

Perform the row operation $R_{1} = R_{1} + \frac{1}{5} R_{2}$ to make the entry at $1, 2$ a $0$ .

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Schritt 2.8.1

Perform the row operation $R_{1} = R_{1} + \frac{1}{5} R_{2}$ to make the entry at $1, 2$ a $0$ .

$[\begin{array}{c} 1 + \frac{1}{5} \cdot 0 & - \frac{1}{5} + \frac{1}{5} \cdot 1 & - \frac{1}{5} + \frac{1}{5} \cdot 1 & 0 + \frac{1}{5} \cdot 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}]$

Schritt 2.8.2

Vereinfache $R_{1}$ .

$[\begin{array}{c} 1 & 0 & 0 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}]$

Schritt 3

Use the result matrix to declare the final solution to the system of equations.

$x = 0$

$y + z = 0$

$0 = 1$

Schritt 4

The system is inconsistent so there is no solution.

$Keine Lösung$