Lineare Algebra Beispiele

$- 6 x - 2 y + 2 z = - 8$ , $3 x - 2 y - 4 z = 8$ , $6 x - 2 y - 6 z = - 18$

Schritt 1

Schreibe das Gleichungssystem in Matrixform.

$[\begin{array}{c} - 6 & - 2 & 2 & - 8 \\ 3 & - 2 & - 4 & 8 \\ 6 & - 2 & - 6 & - 18 \end{array}]$

Schritt 2

Ermittele die normierte Zeilenstufenform.

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Schritt 2.1

Multiply each element of $R_{1}$ by $- \frac{1}{6}$ to make the entry at $1, 1$ a $1$ .

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Schritt 2.1.1

Multiply each element of $R_{1}$ by $- \frac{1}{6}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} - \frac{1}{6} \cdot - 6 & - \frac{1}{6} \cdot - 2 & - \frac{1}{6} \cdot 2 & - \frac{1}{6} \cdot - 8 \\ 3 & - 2 & - 4 & 8 \\ 6 & - 2 & - 6 & - 18 \end{array}]$

Schritt 2.1.2

Vereinfache $R_{1}$ .

$[\begin{array}{c} 1 & \frac{1}{3} & - \frac{1}{3} & \frac{4}{3} \\ 3 & - 2 & - 4 & 8 \\ 6 & - 2 & - 6 & - 18 \end{array}]$

Schritt 2.2

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

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Schritt 2.2.1

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

$[\begin{array}{c} 1 & \frac{1}{3} & - \frac{1}{3} & \frac{4}{3} \\ 3 - 3 \cdot 1 & - 2 - 3 (\frac{1}{3}) & - 4 - 3 (- \frac{1}{3}) & 8 - 3 (\frac{4}{3}) \\ 6 & - 2 & - 6 & - 18 \end{array}]$

Schritt 2.2.2

Vereinfache $R_{2}$ .

$[\begin{array}{c} 1 & \frac{1}{3} & - \frac{1}{3} & \frac{4}{3} \\ 0 & - 3 & - 3 & 4 \\ 6 & - 2 & - 6 & - 18 \end{array}]$

Schritt 2.3

Perform the row operation $R_{3} = R_{3} - 6 R_{1}$ to make the entry at $3, 1$ a $0$ .

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Schritt 2.3.1

Perform the row operation $R_{3} = R_{3} - 6 R_{1}$ to make the entry at $3, 1$ a $0$ .

$[\begin{array}{c} 1 & \frac{1}{3} & - \frac{1}{3} & \frac{4}{3} \\ 0 & - 3 & - 3 & 4 \\ 6 - 6 \cdot 1 & - 2 - 6 (\frac{1}{3}) & - 6 - 6 (- \frac{1}{3}) & - 18 - 6 (\frac{4}{3}) \end{array}]$

Schritt 2.3.2

Vereinfache $R_{3}$ .

$[\begin{array}{c} 1 & \frac{1}{3} & - \frac{1}{3} & \frac{4}{3} \\ 0 & - 3 & - 3 & 4 \\ 0 & - 4 & - 4 & - 26 \end{array}]$

Schritt 2.4

Multiply each element of $R_{2}$ by $- \frac{1}{3}$ to make the entry at $2, 2$ a $1$ .

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Schritt 2.4.1

Multiply each element of $R_{2}$ by $- \frac{1}{3}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & \frac{1}{3} & - \frac{1}{3} & \frac{4}{3} \\ - \frac{1}{3} \cdot 0 & - \frac{1}{3} \cdot - 3 & - \frac{1}{3} \cdot - 3 & - \frac{1}{3} \cdot 4 \\ 0 & - 4 & - 4 & - 26 \end{array}]$

Schritt 2.4.2

Vereinfache $R_{2}$ .

$[\begin{array}{c} 1 & \frac{1}{3} & - \frac{1}{3} & \frac{4}{3} \\ 0 & 1 & 1 & - \frac{4}{3} \\ 0 & - 4 & - 4 & - 26 \end{array}]$

Schritt 2.5

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

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Schritt 2.5.1

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

$[\begin{array}{c} 1 & \frac{1}{3} & - \frac{1}{3} & \frac{4}{3} \\ 0 & 1 & 1 & - \frac{4}{3} \\ 0 + 4 \cdot 0 & - 4 + 4 \cdot 1 & - 4 + 4 \cdot 1 & - 26 + 4 (- \frac{4}{3}) \end{array}]$

Schritt 2.5.2

Vereinfache $R_{3}$ .

$[\begin{array}{c} 1 & \frac{1}{3} & - \frac{1}{3} & \frac{4}{3} \\ 0 & 1 & 1 & - \frac{4}{3} \\ 0 & 0 & 0 & - \frac{94}{3} \end{array}]$

Schritt 2.6

Multiply each element of $R_{3}$ by $- \frac{3}{94}$ to make the entry at $3, 4$ a $1$ .

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Schritt 2.6.1

Multiply each element of $R_{3}$ by $- \frac{3}{94}$ to make the entry at $3, 4$ a $1$ .

$[\begin{array}{c} 1 & \frac{1}{3} & - \frac{1}{3} & \frac{4}{3} \\ 0 & 1 & 1 & - \frac{4}{3} \\ - \frac{3}{94} \cdot 0 & - \frac{3}{94} \cdot 0 & - \frac{3}{94} \cdot 0 & - \frac{3}{94} (- \frac{94}{3}) \end{array}]$

Schritt 2.6.2

Vereinfache $R_{3}$ .

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

Schritt 2.7

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

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Schritt 2.7.1

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

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

Schritt 2.7.2

Vereinfache $R_{2}$ .

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

Schritt 2.8

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

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Schritt 2.8.1

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

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

Schritt 2.8.2

Vereinfache $R_{1}$ .

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

Schritt 2.9

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

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Schritt 2.9.1

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

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

Schritt 2.9.2

Vereinfache $R_{1}$ .

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

Schritt 3

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

$x - \frac{2 z}{3} = 0$

$y + z = 0$

$0 = 1$

Schritt 4

Addiere $\frac{2 z}{3}$ zu beiden Seiten der Gleichung.

$x = \frac{2 z}{3}$

$y + z = 0$

$0 = 1$

Schritt 5

Subtrahiere $z$ von beiden Seiten der Gleichung.

$y = - z$

$x = \frac{2 z}{3}$

$0 = 1$

Schritt 6

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

Keine Lösung