Lineare Algebra Beispiele

$y = 2 x - 1$ , $y = - 3 + 4$ , $y = - 1 + 2$ , $y = - \frac{1}{3} x + 1$ , $y = \frac{4}{3} x - 4$

Schritt 1

Subtrahiere $2 x$ von beiden Seiten der Gleichung.

$y - 2 x = - 1, y = - 3 + 4, y = - 1 + 2, y = - \frac{1}{3} x + 1, y = \frac{4}{3} x - 4$

Schritt 2

Addiere $- 3$ und $4$ .

$y - 2 x = - 1, y = 1, y = - 1 + 2, y = - \frac{1}{3} x + 1, y = \frac{4}{3} x - 4$

Schritt 3

Addiere $- 1$ und $2$ .

$y - 2 x = - 1, y = 1, y = 1, y = - \frac{1}{3} x + 1, y = \frac{4}{3} x - 4$

Schritt 4

Kombiniere $x$ und $\frac{1}{3}$ .

$y - 2 x = - 1, y = 1, y = 1, y = - \frac{x}{3} + 1, y = \frac{4}{3} x - 4$

Schritt 5

Addiere $\frac{x}{3}$ zu beiden Seiten der Gleichung.

$y - 2 x = - 1, y = 1, y = 1, y + \frac{x}{3} = 1, y = \frac{4}{3} x - 4$

Schritt 6

Kombiniere $\frac{4}{3}$ und $x$ .

$y - 2 x = - 1, y = 1, y = 1, y + \frac{x}{3} = 1, y = \frac{4 x}{3} - 4$

Schritt 7

Subtrahiere $\frac{4 x}{3}$ von beiden Seiten der Gleichung.

$y - 2 x = - 1, y = 1, y = 1, y + \frac{x}{3} = 1, y - \frac{4 x}{3} = - 4$

Schritt 8

Schreibe das Gleichungssystem in Matrixform.

$[\begin{array}{c} - 2 & 1 & - 1 \\ 0 & 1 & 1 \\ 0 & 1 & 1 \\ \frac{1}{3} & 1 & 1 \\ - \frac{4}{3} & 1 & - 4 \end{array}]$

Schritt 9

Ermittele die normierte Zeilenstufenform.

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Schritt 9.1

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

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Schritt 9.1.1

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

$[\begin{array}{c} - \frac{1}{2} \cdot - 2 & - \frac{1}{2} \cdot 1 & - \frac{1}{2} \cdot - 1 \\ 0 & 1 & 1 \\ 0 & 1 & 1 \\ \frac{1}{3} & 1 & 1 \\ - \frac{4}{3} & 1 & - 4 \end{array}]$

Schritt 9.1.2

Vereinfache $R_{1}$ .

$[\begin{array}{c} 1 & - \frac{1}{2} & \frac{1}{2} \\ 0 & 1 & 1 \\ 0 & 1 & 1 \\ \frac{1}{3} & 1 & 1 \\ - \frac{4}{3} & 1 & - 4 \end{array}]$

Schritt 9.2

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

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Schritt 9.2.1

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

$[\begin{array}{c} 1 & - \frac{1}{2} & \frac{1}{2} \\ 0 & 1 & 1 \\ 0 & 1 & 1 \\ \frac{1}{3} - \frac{1}{3} \cdot 1 & 1 - \frac{1}{3} (- \frac{1}{2}) & 1 - \frac{1}{3} \cdot \frac{1}{2} \\ - \frac{4}{3} & 1 & - 4 \end{array}]$

Schritt 9.2.2

Vereinfache $R_{4}$ .

$[\begin{array}{c} 1 & - \frac{1}{2} & \frac{1}{2} \\ 0 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & \frac{7}{6} & \frac{5}{6} \\ - \frac{4}{3} & 1 & - 4 \end{array}]$

Schritt 9.3

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

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Schritt 9.3.1

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

$[\begin{array}{c} 1 & - \frac{1}{2} & \frac{1}{2} \\ 0 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & \frac{7}{6} & \frac{5}{6} \\ - \frac{4}{3} + \frac{4}{3} \cdot 1 & 1 + \frac{4}{3} (- \frac{1}{2}) & - 4 + \frac{4}{3} \cdot \frac{1}{2} \end{array}]$

Schritt 9.3.2

Vereinfache $R_{5}$ .

$[\begin{array}{c} 1 & - \frac{1}{2} & \frac{1}{2} \\ 0 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & \frac{7}{6} & \frac{5}{6} \\ 0 & \frac{1}{3} & - \frac{10}{3} \end{array}]$

Schritt 9.4

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

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Schritt 9.4.1

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

$[\begin{array}{c} 1 & - \frac{1}{2} & \frac{1}{2} \\ 0 & 1 & 1 \\ 0 - 0 & 1 - 1 & 1 - 1 \\ 0 & \frac{7}{6} & \frac{5}{6} \\ 0 & \frac{1}{3} & - \frac{10}{3} \end{array}]$

Schritt 9.4.2

Vereinfache $R_{3}$ .

$[\begin{array}{c} 1 & - \frac{1}{2} & \frac{1}{2} \\ 0 & 1 & 1 \\ 0 & 0 & 0 \\ 0 & \frac{7}{6} & \frac{5}{6} \\ 0 & \frac{1}{3} & - \frac{10}{3} \end{array}]$

Schritt 9.5

Perform the row operation $R_{4} = R_{4} - \frac{7}{6} R_{2}$ to make the entry at $4, 2$ a $0$ .

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Schritt 9.5.1

Perform the row operation $R_{4} = R_{4} - \frac{7}{6} R_{2}$ to make the entry at $4, 2$ a $0$ .

$[\begin{array}{c} 1 & - \frac{1}{2} & \frac{1}{2} \\ 0 & 1 & 1 \\ 0 & 0 & 0 \\ 0 - \frac{7}{6} \cdot 0 & \frac{7}{6} - \frac{7}{6} \cdot 1 & \frac{5}{6} - \frac{7}{6} \cdot 1 \\ 0 & \frac{1}{3} & - \frac{10}{3} \end{array}]$

Schritt 9.5.2

Vereinfache $R_{4}$ .

$[\begin{array}{c} 1 & - \frac{1}{2} & \frac{1}{2} \\ 0 & 1 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & - \frac{1}{3} \\ 0 & \frac{1}{3} & - \frac{10}{3} \end{array}]$

Schritt 9.6

Perform the row operation $R_{5} = R_{5} - \frac{1}{3} R_{2}$ to make the entry at $5, 2$ a $0$ .

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Schritt 9.6.1

Perform the row operation $R_{5} = R_{5} - \frac{1}{3} R_{2}$ to make the entry at $5, 2$ a $0$ .

$[\begin{array}{c} 1 & - \frac{1}{2} & \frac{1}{2} \\ 0 & 1 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & - \frac{1}{3} \\ 0 - \frac{1}{3} \cdot 0 & \frac{1}{3} - \frac{1}{3} \cdot 1 & - \frac{10}{3} - \frac{1}{3} \cdot 1 \end{array}]$

Schritt 9.6.2

Vereinfache $R_{5}$ .

$[\begin{array}{c} 1 & - \frac{1}{2} & \frac{1}{2} \\ 0 & 1 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & - \frac{1}{3} \\ 0 & 0 & - \frac{11}{3} \end{array}]$

Schritt 9.7

Swap $R_{4}$ with $R_{3}$ to put a nonzero entry at $3, 3$ .

$[\begin{array}{c} 1 & - \frac{1}{2} & \frac{1}{2} \\ 0 & 1 & 1 \\ 0 & 0 & - \frac{1}{3} \\ 0 & 0 & 0 \\ 0 & 0 & - \frac{11}{3} \end{array}]$

Schritt 9.8

Multiply each element of $R_{3}$ by $- 3$ to make the entry at $3, 3$ a $1$ .

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Schritt 9.8.1

Multiply each element of $R_{3}$ by $- 3$ to make the entry at $3, 3$ a $1$ .

$[\begin{array}{c} 1 & - \frac{1}{2} & \frac{1}{2} \\ 0 & 1 & 1 \\ - 3 \cdot 0 & - 3 \cdot 0 & - 3 (- \frac{1}{3}) \\ 0 & 0 & 0 \\ 0 & 0 & - \frac{11}{3} \end{array}]$

Schritt 9.8.2

Vereinfache $R_{3}$ .

$[\begin{array}{c} 1 & - \frac{1}{2} & \frac{1}{2} \\ 0 & 1 & 1 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & - \frac{11}{3} \end{array}]$

Schritt 9.9

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

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Schritt 9.9.1

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

$[\begin{array}{c} 1 & - \frac{1}{2} & \frac{1}{2} \\ 0 & 1 & 1 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 + \frac{11}{3} \cdot 0 & 0 + \frac{11}{3} \cdot 0 & - \frac{11}{3} + \frac{11}{3} \cdot 1 \end{array}]$

Schritt 9.9.2

Vereinfache $R_{5}$ .

$[\begin{array}{c} 1 & - \frac{1}{2} & \frac{1}{2} \\ 0 & 1 & 1 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}]$

Schritt 9.10

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

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Schritt 9.10.1

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

$[\begin{array}{c} 1 & - \frac{1}{2} & \frac{1}{2} \\ 0 - 0 & 1 - 0 & 1 - 1 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}]$

Schritt 9.10.2

Vereinfache $R_{2}$ .

$[\begin{array}{c} 1 & - \frac{1}{2} & \frac{1}{2} \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}]$

Schritt 9.11

Perform the row operation $R_{1} = R_{1} - \frac{1}{2} R_{3}$ to make the entry at $1, 3$ a $0$ .

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Schritt 9.11.1

Perform the row operation $R_{1} = R_{1} - \frac{1}{2} R_{3}$ to make the entry at $1, 3$ a $0$ .

$[\begin{array}{c} 1 - \frac{1}{2} \cdot 0 & - \frac{1}{2} - \frac{1}{2} \cdot 0 & \frac{1}{2} - \frac{1}{2} \cdot 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}]$

Schritt 9.11.2

Vereinfache $R_{1}$ .

$[\begin{array}{c} 1 & - \frac{1}{2} & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}]$

Schritt 9.12

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

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Schritt 9.12.1

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

$[\begin{array}{c} 1 + \frac{1}{2} \cdot 0 & - \frac{1}{2} + \frac{1}{2} \cdot 1 & 0 + \frac{1}{2} \cdot 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}]$

Schritt 9.12.2

Vereinfache $R_{1}$ .

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

Schritt 10

Verwende die Ergebnismatrix, um die endgültigen Lösungen für das Gleichungssystem anzugeben.

$x = 0$

$y = 0$

$0 = 1$

Schritt 11

Da $0 \neq 1$ , gibt es keine Lösungen.

Keine Lösung