Lineare Algebra Beispiele

$x + 3 y - 3 z - w = 8$ , $4 x + y + 4 z + 3 w = 1$ , $- 3 x - y + z - 5 w = - 24$ , $x - y - 3 z - 5 w = - 12$

Schritt 1

Schreibe das Gleichungssystem in Matrixform.

$[\begin{array}{c} - 1 & 1 & 3 & - 3 & 8 \\ 3 & 4 & 1 & 4 & 1 \\ - 5 & - 3 & - 1 & 1 & - 24 \\ - 5 & 1 & - 1 & - 3 & - 12 \end{array}]$

Schritt 2

Ermittele die normierte Zeilenstufenform.

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Schritt 2.1

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

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Schritt 2.1.1

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

$[\begin{array}{c} - - 1 & - 1 \cdot 1 & - 1 \cdot 3 & - - 3 & - 1 \cdot 8 \\ 3 & 4 & 1 & 4 & 1 \\ - 5 & - 3 & - 1 & 1 & - 24 \\ - 5 & 1 & - 1 & - 3 & - 12 \end{array}]$

Schritt 2.1.2

Vereinfache $R_{1}$ .

$[\begin{array}{c} 1 & - 1 & - 3 & 3 & - 8 \\ 3 & 4 & 1 & 4 & 1 \\ - 5 & - 3 & - 1 & 1 & - 24 \\ - 5 & 1 & - 1 & - 3 & - 12 \end{array}]$

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 $0$ .

$[\begin{array}{c} 1 & - 1 & - 3 & 3 & - 8 \\ 3 - 3 \cdot 1 & 4 - 3 \cdot - 1 & 1 - 3 \cdot - 3 & 4 - 3 \cdot 3 & 1 - 3 \cdot - 8 \\ - 5 & - 3 & - 1 & 1 & - 24 \\ - 5 & 1 & - 1 & - 3 & - 12 \end{array}]$

Schritt 2.2.2

Vereinfache $R_{2}$ .

$[\begin{array}{c} 1 & - 1 & - 3 & 3 & - 8 \\ 0 & 7 & 10 & - 5 & 25 \\ - 5 & - 3 & - 1 & 1 & - 24 \\ - 5 & 1 & - 1 & - 3 & - 12 \end{array}]$

Schritt 2.3

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

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Schritt 2.3.1

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

$[\begin{array}{c} 1 & - 1 & - 3 & 3 & - 8 \\ 0 & 7 & 10 & - 5 & 25 \\ - 5 + 5 \cdot 1 & - 3 + 5 \cdot - 1 & - 1 + 5 \cdot - 3 & 1 + 5 \cdot 3 & - 24 + 5 \cdot - 8 \\ - 5 & 1 & - 1 & - 3 & - 12 \end{array}]$

Schritt 2.3.2

Vereinfache $R_{3}$ .

$[\begin{array}{c} 1 & - 1 & - 3 & 3 & - 8 \\ 0 & 7 & 10 & - 5 & 25 \\ 0 & - 8 & - 16 & 16 & - 64 \\ - 5 & 1 & - 1 & - 3 & - 12 \end{array}]$

Schritt 2.4

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

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Schritt 2.4.1

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

$[\begin{array}{c} 1 & - 1 & - 3 & 3 & - 8 \\ 0 & 7 & 10 & - 5 & 25 \\ 0 & - 8 & - 16 & 16 & - 64 \\ - 5 + 5 \cdot 1 & 1 + 5 \cdot - 1 & - 1 + 5 \cdot - 3 & - 3 + 5 \cdot 3 & - 12 + 5 \cdot - 8 \end{array}]$

Schritt 2.4.2

Vereinfache $R_{4}$ .

$[\begin{array}{c} 1 & - 1 & - 3 & 3 & - 8 \\ 0 & 7 & 10 & - 5 & 25 \\ 0 & - 8 & - 16 & 16 & - 64 \\ 0 & - 4 & - 16 & 12 & - 52 \end{array}]$

Schritt 2.5

Multiply each element of $R_{2}$ by $\frac{1}{7}$ to make the entry at $2, 2$ a $1$ .

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Schritt 2.5.1

Multiply each element of $R_{2}$ by $\frac{1}{7}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & - 1 & - 3 & 3 & - 8 \\ \frac{0}{7} & \frac{7}{7} & \frac{10}{7} & \frac{- 5}{7} & \frac{25}{7} \\ 0 & - 8 & - 16 & 16 & - 64 \\ 0 & - 4 & - 16 & 12 & - 52 \end{array}]$

Schritt 2.5.2

Vereinfache $R_{2}$ .

$[\begin{array}{c} 1 & - 1 & - 3 & 3 & - 8 \\ 0 & 1 & \frac{10}{7} & - \frac{5}{7} & \frac{25}{7} \\ 0 & - 8 & - 16 & 16 & - 64 \\ 0 & - 4 & - 16 & 12 & - 52 \end{array}]$

Schritt 2.6

Perform the row operation $R_{3} = R_{3} + 8 R_{2}$ to make the entry at $3, 2$ a $0$ .

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Schritt 2.6.1

Perform the row operation $R_{3} = R_{3} + 8 R_{2}$ to make the entry at $3, 2$ a $0$ .

$[\begin{array}{c} 1 & - 1 & - 3 & 3 & - 8 \\ 0 & 1 & \frac{10}{7} & - \frac{5}{7} & \frac{25}{7} \\ 0 + 8 \cdot 0 & - 8 + 8 \cdot 1 & - 16 + 8 (\frac{10}{7}) & 16 + 8 (- \frac{5}{7}) & - 64 + 8 (\frac{25}{7}) \\ 0 & - 4 & - 16 & 12 & - 52 \end{array}]$

Schritt 2.6.2

Vereinfache $R_{3}$ .

$[\begin{array}{c} 1 & - 1 & - 3 & 3 & - 8 \\ 0 & 1 & \frac{10}{7} & - \frac{5}{7} & \frac{25}{7} \\ 0 & 0 & - \frac{32}{7} & \frac{72}{7} & - \frac{248}{7} \\ 0 & - 4 & - 16 & 12 & - 52 \end{array}]$

Schritt 2.7

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

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Schritt 2.7.1

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

$[\begin{array}{c} 1 & - 1 & - 3 & 3 & - 8 \\ 0 & 1 & \frac{10}{7} & - \frac{5}{7} & \frac{25}{7} \\ 0 & 0 & - \frac{32}{7} & \frac{72}{7} & - \frac{248}{7} \\ 0 + 4 \cdot 0 & - 4 + 4 \cdot 1 & - 16 + 4 (\frac{10}{7}) & 12 + 4 (- \frac{5}{7}) & - 52 + 4 (\frac{25}{7}) \end{array}]$

Schritt 2.7.2

Vereinfache $R_{4}$ .

$[\begin{array}{c} 1 & - 1 & - 3 & 3 & - 8 \\ 0 & 1 & \frac{10}{7} & - \frac{5}{7} & \frac{25}{7} \\ 0 & 0 & - \frac{32}{7} & \frac{72}{7} & - \frac{248}{7} \\ 0 & 0 & - \frac{72}{7} & \frac{64}{7} & - \frac{264}{7} \end{array}]$

Schritt 2.8

Multiply each element of $R_{3}$ by $- \frac{7}{32}$ to make the entry at $3, 3$ a $1$ .

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Schritt 2.8.1

Multiply each element of $R_{3}$ by $- \frac{7}{32}$ to make the entry at $3, 3$ a $1$ .

$[\begin{array}{c} 1 & - 1 & - 3 & 3 & - 8 \\ 0 & 1 & \frac{10}{7} & - \frac{5}{7} & \frac{25}{7} \\ - \frac{7}{32} \cdot 0 & - \frac{7}{32} \cdot 0 & - \frac{7}{32} (- \frac{32}{7}) & - \frac{7}{32} \cdot \frac{72}{7} & - \frac{7}{32} (- \frac{248}{7}) \\ 0 & 0 & - \frac{72}{7} & \frac{64}{7} & - \frac{264}{7} \end{array}]$

Schritt 2.8.2

Vereinfache $R_{3}$ .

$[\begin{array}{c} 1 & - 1 & - 3 & 3 & - 8 \\ 0 & 1 & \frac{10}{7} & - \frac{5}{7} & \frac{25}{7} \\ 0 & 0 & 1 & - \frac{9}{4} & \frac{31}{4} \\ 0 & 0 & - \frac{72}{7} & \frac{64}{7} & - \frac{264}{7} \end{array}]$

Schritt 2.9

Perform the row operation $R_{4} = R_{4} + \frac{72}{7} R_{3}$ to make the entry at $4, 3$ a $0$ .

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Schritt 2.9.1

Perform the row operation $R_{4} = R_{4} + \frac{72}{7} R_{3}$ to make the entry at $4, 3$ a $0$ .

$[\begin{array}{c} 1 & - 1 & - 3 & 3 & - 8 \\ 0 & 1 & \frac{10}{7} & - \frac{5}{7} & \frac{25}{7} \\ 0 & 0 & 1 & - \frac{9}{4} & \frac{31}{4} \\ 0 + \frac{72}{7} \cdot 0 & 0 + \frac{72}{7} \cdot 0 & - \frac{72}{7} + \frac{72}{7} \cdot 1 & \frac{64}{7} + \frac{72}{7} (- \frac{9}{4}) & - \frac{264}{7} + \frac{72}{7} \cdot \frac{31}{4} \end{array}]$

Schritt 2.9.2

Vereinfache $R_{4}$ .

$[\begin{array}{c} 1 & - 1 & - 3 & 3 & - 8 \\ 0 & 1 & \frac{10}{7} & - \frac{5}{7} & \frac{25}{7} \\ 0 & 0 & 1 & - \frac{9}{4} & \frac{31}{4} \\ 0 & 0 & 0 & - 14 & 42 \end{array}]$

Schritt 2.10

Multiply each element of $R_{4}$ by $- \frac{1}{14}$ to make the entry at $4, 4$ a $1$ .

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Schritt 2.10.1

Multiply each element of $R_{4}$ by $- \frac{1}{14}$ to make the entry at $4, 4$ a $1$ .

$[\begin{array}{c} 1 & - 1 & - 3 & 3 & - 8 \\ 0 & 1 & \frac{10}{7} & - \frac{5}{7} & \frac{25}{7} \\ 0 & 0 & 1 & - \frac{9}{4} & \frac{31}{4} \\ - \frac{1}{14} \cdot 0 & - \frac{1}{14} \cdot 0 & - \frac{1}{14} \cdot 0 & - \frac{1}{14} \cdot - 14 & - \frac{1}{14} \cdot 42 \end{array}]$

Schritt 2.10.2

Vereinfache $R_{4}$ .

$[\begin{array}{c} 1 & - 1 & - 3 & 3 & - 8 \\ 0 & 1 & \frac{10}{7} & - \frac{5}{7} & \frac{25}{7} \\ 0 & 0 & 1 & - \frac{9}{4} & \frac{31}{4} \\ 0 & 0 & 0 & 1 & - 3 \end{array}]$

Schritt 2.11

Perform the row operation $R_{3} = R_{3} + \frac{9}{4} R_{4}$ to make the entry at $3, 4$ a $0$ .

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Schritt 2.11.1

Perform the row operation $R_{3} = R_{3} + \frac{9}{4} R_{4}$ to make the entry at $3, 4$ a $0$ .

$[\begin{array}{c} 1 & - 1 & - 3 & 3 & - 8 \\ 0 & 1 & \frac{10}{7} & - \frac{5}{7} & \frac{25}{7} \\ 0 + \frac{9}{4} \cdot 0 & 0 + \frac{9}{4} \cdot 0 & 1 + \frac{9}{4} \cdot 0 & - \frac{9}{4} + \frac{9}{4} \cdot 1 & \frac{31}{4} + \frac{9}{4} \cdot - 3 \\ 0 & 0 & 0 & 1 & - 3 \end{array}]$

Schritt 2.11.2

Vereinfache $R_{3}$ .

$[\begin{array}{c} 1 & - 1 & - 3 & 3 & - 8 \\ 0 & 1 & \frac{10}{7} & - \frac{5}{7} & \frac{25}{7} \\ 0 & 0 & 1 & 0 & 1 \\ 0 & 0 & 0 & 1 & - 3 \end{array}]$

Schritt 2.12

Perform the row operation $R_{2} = R_{2} + \frac{5}{7} R_{4}$ to make the entry at $2, 4$ a $0$ .

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Schritt 2.12.1

Perform the row operation $R_{2} = R_{2} + \frac{5}{7} R_{4}$ to make the entry at $2, 4$ a $0$ .

$[\begin{array}{c} 1 & - 1 & - 3 & 3 & - 8 \\ 0 + \frac{5}{7} \cdot 0 & 1 + \frac{5}{7} \cdot 0 & \frac{10}{7} + \frac{5}{7} \cdot 0 & - \frac{5}{7} + \frac{5}{7} \cdot 1 & \frac{25}{7} + \frac{5}{7} \cdot - 3 \\ 0 & 0 & 1 & 0 & 1 \\ 0 & 0 & 0 & 1 & - 3 \end{array}]$

Schritt 2.12.2

Vereinfache $R_{2}$ .

$[\begin{array}{c} 1 & - 1 & - 3 & 3 & - 8 \\ 0 & 1 & \frac{10}{7} & 0 & \frac{10}{7} \\ 0 & 0 & 1 & 0 & 1 \\ 0 & 0 & 0 & 1 & - 3 \end{array}]$

Schritt 2.13

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

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Schritt 2.13.1

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

$[\begin{array}{c} 1 - 3 \cdot 0 & - 1 - 3 \cdot 0 & - 3 - 3 \cdot 0 & 3 - 3 \cdot 1 & - 8 - 3 \cdot - 3 \\ 0 & 1 & \frac{10}{7} & 0 & \frac{10}{7} \\ 0 & 0 & 1 & 0 & 1 \\ 0 & 0 & 0 & 1 & - 3 \end{array}]$

Schritt 2.13.2

Vereinfache $R_{1}$ .

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

Schritt 2.14

Perform the row operation $R_{2} = R_{2} - \frac{10}{7} R_{3}$ to make the entry at $2, 3$ a $0$ .

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Schritt 2.14.1

Perform the row operation $R_{2} = R_{2} - \frac{10}{7} R_{3}$ to make the entry at $2, 3$ a $0$ .

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

Schritt 2.14.2

Vereinfache $R_{2}$ .

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

Schritt 2.15

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

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Schritt 2.15.1

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

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

Schritt 2.15.2

Vereinfache $R_{1}$ .

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

Schritt 2.16

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

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Schritt 2.16.1

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

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

Schritt 2.16.2

Vereinfache $R_{1}$ .

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

Schritt 3

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

$w = 4$

$x = 0$

$y = 1$

$z = - 3$

Schritt 4

Die Lösung ist die Menge geordneter Paare, die das System erfüllen.

$(4, 0, 1, - 3)$

Schritt 5

Das Zerlegen eines Lösungsvektors durch Umstellen jeder Gleichung, die in der reduzierten Zeilenstufenform der erweiterten Matrix wiedergegeben ist, durch Auflösen nach der abhängigen Variablen in jeder Zeile, ergibt die Vektorgleichung.

$X = [\begin{array}{c} w \\ x \\ y \\ z \end{array}] = [\begin{array}{c} 4 \\ 0 \\ 1 \\ - 3 \end{array}]$