Lineare Algebra Beispiele

$2 w + 3 x - y - z = 9$ , $w + x - y - z = 0$ , $6 w + x - 2 y - z = 4$ , $- w + 2 x + 3 y - z = 5$

Schritt 1

Schreibe das Gleichungssystem in Matrixform.

$[\begin{array}{c} 2 & 3 & - 1 & - 1 & 9 \\ 1 & 1 & - 1 & - 1 & 0 \\ 6 & 1 & - 2 & - 1 & 4 \\ - 1 & 2 & 3 & - 1 & 5 \end{array}]$

Schritt 2

Ermittele die normierte Zeilenstufenform.

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Schritt 2.1

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

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Schritt 2.1.1

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

$[\begin{array}{c} \frac{2}{2} & \frac{3}{2} & \frac{- 1}{2} & \frac{- 1}{2} & \frac{9}{2} \\ 1 & 1 & - 1 & - 1 & 0 \\ 6 & 1 & - 2 & - 1 & 4 \\ - 1 & 2 & 3 & - 1 & 5 \end{array}]$

Schritt 2.1.2

Vereinfache $R_{1}$ .

$[\begin{array}{c} 1 & \frac{3}{2} & - \frac{1}{2} & - \frac{1}{2} & \frac{9}{2} \\ 1 & 1 & - 1 & - 1 & 0 \\ 6 & 1 & - 2 & - 1 & 4 \\ - 1 & 2 & 3 & - 1 & 5 \end{array}]$

Schritt 2.2

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

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Schritt 2.2.1

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

$[\begin{array}{c} 1 & \frac{3}{2} & - \frac{1}{2} & - \frac{1}{2} & \frac{9}{2} \\ 1 - 1 & 1 - \frac{3}{2} & - 1 + \frac{1}{2} & - 1 + \frac{1}{2} & 0 - \frac{9}{2} \\ 6 & 1 & - 2 & - 1 & 4 \\ - 1 & 2 & 3 & - 1 & 5 \end{array}]$

Schritt 2.2.2

Vereinfache $R_{2}$ .

$[\begin{array}{c} 1 & \frac{3}{2} & - \frac{1}{2} & - \frac{1}{2} & \frac{9}{2} \\ 0 & - \frac{1}{2} & - \frac{1}{2} & - \frac{1}{2} & - \frac{9}{2} \\ 6 & 1 & - 2 & - 1 & 4 \\ - 1 & 2 & 3 & - 1 & 5 \end{array}]$

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 $0$ .

$[\begin{array}{c} 1 & \frac{3}{2} & - \frac{1}{2} & - \frac{1}{2} & \frac{9}{2} \\ 0 & - \frac{1}{2} & - \frac{1}{2} & - \frac{1}{2} & - \frac{9}{2} \\ 6 - 6 \cdot 1 & 1 - 6 (\frac{3}{2}) & - 2 - 6 (- \frac{1}{2}) & - 1 - 6 (- \frac{1}{2}) & 4 - 6 (\frac{9}{2}) \\ - 1 & 2 & 3 & - 1 & 5 \end{array}]$

Schritt 2.3.2

Vereinfache $R_{3}$ .

$[\begin{array}{c} 1 & \frac{3}{2} & - \frac{1}{2} & - \frac{1}{2} & \frac{9}{2} \\ 0 & - \frac{1}{2} & - \frac{1}{2} & - \frac{1}{2} & - \frac{9}{2} \\ 0 & - 8 & 1 & 2 & - 23 \\ - 1 & 2 & 3 & - 1 & 5 \end{array}]$

Schritt 2.4

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

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Schritt 2.4.1

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

$[\begin{array}{c} 1 & \frac{3}{2} & - \frac{1}{2} & - \frac{1}{2} & \frac{9}{2} \\ 0 & - \frac{1}{2} & - \frac{1}{2} & - \frac{1}{2} & - \frac{9}{2} \\ 0 & - 8 & 1 & 2 & - 23 \\ - 1 + 1 \cdot 1 & 2 + \frac{3}{2} & 3 - \frac{1}{2} & - 1 - \frac{1}{2} & 5 + \frac{9}{2} \end{array}]$

Schritt 2.4.2

Vereinfache $R_{4}$ .

$[\begin{array}{c} 1 & \frac{3}{2} & - \frac{1}{2} & - \frac{1}{2} & \frac{9}{2} \\ 0 & - \frac{1}{2} & - \frac{1}{2} & - \frac{1}{2} & - \frac{9}{2} \\ 0 & - 8 & 1 & 2 & - 23 \\ 0 & \frac{7}{2} & \frac{5}{2} & - \frac{3}{2} & \frac{19}{2} \end{array}]$

Schritt 2.5

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

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Schritt 2.5.1

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

$[\begin{array}{c} 1 & \frac{3}{2} & - \frac{1}{2} & - \frac{1}{2} & \frac{9}{2} \\ - 2 \cdot 0 & - 2 (- \frac{1}{2}) & - 2 (- \frac{1}{2}) & - 2 (- \frac{1}{2}) & - 2 (- \frac{9}{2}) \\ 0 & - 8 & 1 & 2 & - 23 \\ 0 & \frac{7}{2} & \frac{5}{2} & - \frac{3}{2} & \frac{19}{2} \end{array}]$

Schritt 2.5.2

Vereinfache $R_{2}$ .

$[\begin{array}{c} 1 & \frac{3}{2} & - \frac{1}{2} & - \frac{1}{2} & \frac{9}{2} \\ 0 & 1 & 1 & 1 & 9 \\ 0 & - 8 & 1 & 2 & - 23 \\ 0 & \frac{7}{2} & \frac{5}{2} & - \frac{3}{2} & \frac{19}{2} \end{array}]$

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 $0$ .

$[\begin{array}{c} 1 & \frac{3}{2} & - \frac{1}{2} & - \frac{1}{2} & \frac{9}{2} \\ 0 & 1 & 1 & 1 & 9 \\ 0 + 8 \cdot 0 & - 8 + 8 \cdot 1 & 1 + 8 \cdot 1 & 2 + 8 \cdot 1 & - 23 + 8 \cdot 9 \\ 0 & \frac{7}{2} & \frac{5}{2} & - \frac{3}{2} & \frac{19}{2} \end{array}]$

Schritt 2.6.2

Vereinfache $R_{3}$ .

$[\begin{array}{c} 1 & \frac{3}{2} & - \frac{1}{2} & - \frac{1}{2} & \frac{9}{2} \\ 0 & 1 & 1 & 1 & 9 \\ 0 & 0 & 9 & 10 & 49 \\ 0 & \frac{7}{2} & \frac{5}{2} & - \frac{3}{2} & \frac{19}{2} \end{array}]$

Schritt 2.7

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

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Schritt 2.7.1

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

$[\begin{array}{c} 1 & \frac{3}{2} & - \frac{1}{2} & - \frac{1}{2} & \frac{9}{2} \\ 0 & 1 & 1 & 1 & 9 \\ 0 & 0 & 9 & 10 & 49 \\ 0 - \frac{7}{2} \cdot 0 & \frac{7}{2} - \frac{7}{2} \cdot 1 & \frac{5}{2} - \frac{7}{2} \cdot 1 & - \frac{3}{2} - \frac{7}{2} \cdot 1 & \frac{19}{2} - \frac{7}{2} \cdot 9 \end{array}]$

Schritt 2.7.2

Vereinfache $R_{4}$ .

$[\begin{array}{c} 1 & \frac{3}{2} & - \frac{1}{2} & - \frac{1}{2} & \frac{9}{2} \\ 0 & 1 & 1 & 1 & 9 \\ 0 & 0 & 9 & 10 & 49 \\ 0 & 0 & - 1 & - 5 & - 22 \end{array}]$

Schritt 2.8

Multiply each element of $R_{3}$ by $\frac{1}{9}$ to make the entry at $3, 3$ a $1$ .

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Schritt 2.8.1

Multiply each element of $R_{3}$ by $\frac{1}{9}$ to make the entry at $3, 3$ a $1$ .

$[\begin{array}{c} 1 & \frac{3}{2} & - \frac{1}{2} & - \frac{1}{2} & \frac{9}{2} \\ 0 & 1 & 1 & 1 & 9 \\ \frac{0}{9} & \frac{0}{9} & \frac{9}{9} & \frac{10}{9} & \frac{49}{9} \\ 0 & 0 & - 1 & - 5 & - 22 \end{array}]$

Schritt 2.8.2

Vereinfache $R_{3}$ .

$[\begin{array}{c} 1 & \frac{3}{2} & - \frac{1}{2} & - \frac{1}{2} & \frac{9}{2} \\ 0 & 1 & 1 & 1 & 9 \\ 0 & 0 & 1 & \frac{10}{9} & \frac{49}{9} \\ 0 & 0 & - 1 & - 5 & - 22 \end{array}]$

Schritt 2.9

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

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Schritt 2.9.1

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

$[\begin{array}{c} 1 & \frac{3}{2} & - \frac{1}{2} & - \frac{1}{2} & \frac{9}{2} \\ 0 & 1 & 1 & 1 & 9 \\ 0 & 0 & 1 & \frac{10}{9} & \frac{49}{9} \\ 0 + 0 & 0 + 0 & - 1 + 1 \cdot 1 & - 5 + \frac{10}{9} & - 22 + \frac{49}{9} \end{array}]$

Schritt 2.9.2

Vereinfache $R_{4}$ .

$[\begin{array}{c} 1 & \frac{3}{2} & - \frac{1}{2} & - \frac{1}{2} & \frac{9}{2} \\ 0 & 1 & 1 & 1 & 9 \\ 0 & 0 & 1 & \frac{10}{9} & \frac{49}{9} \\ 0 & 0 & 0 & - \frac{35}{9} & - \frac{149}{9} \end{array}]$

Schritt 2.10

Multiply each element of $R_{4}$ by $- \frac{9}{35}$ to make the entry at $4, 4$ a $1$ .

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Schritt 2.10.1

Multiply each element of $R_{4}$ by $- \frac{9}{35}$ to make the entry at $4, 4$ a $1$ .

$[\begin{array}{c} 1 & \frac{3}{2} & - \frac{1}{2} & - \frac{1}{2} & \frac{9}{2} \\ 0 & 1 & 1 & 1 & 9 \\ 0 & 0 & 1 & \frac{10}{9} & \frac{49}{9} \\ - \frac{9}{35} \cdot 0 & - \frac{9}{35} \cdot 0 & - \frac{9}{35} \cdot 0 & - \frac{9}{35} (- \frac{35}{9}) & - \frac{9}{35} (- \frac{149}{9}) \end{array}]$

Schritt 2.10.2

Vereinfache $R_{4}$ .

$[\begin{array}{c} 1 & \frac{3}{2} & - \frac{1}{2} & - \frac{1}{2} & \frac{9}{2} \\ 0 & 1 & 1 & 1 & 9 \\ 0 & 0 & 1 & \frac{10}{9} & \frac{49}{9} \\ 0 & 0 & 0 & 1 & \frac{149}{35} \end{array}]$

Schritt 2.11

Perform the row operation $R_{3} = R_{3} - \frac{10}{9} R_{4}$ to make the entry at $3, 4$ a $0$ .

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Schritt 2.11.1

Perform the row operation $R_{3} = R_{3} - \frac{10}{9} R_{4}$ to make the entry at $3, 4$ a $0$ .

$[\begin{array}{c} 1 & \frac{3}{2} & - \frac{1}{2} & - \frac{1}{2} & \frac{9}{2} \\ 0 & 1 & 1 & 1 & 9 \\ 0 - \frac{10}{9} \cdot 0 & 0 - \frac{10}{9} \cdot 0 & 1 - \frac{10}{9} \cdot 0 & \frac{10}{9} - \frac{10}{9} \cdot 1 & \frac{49}{9} - \frac{10}{9} \cdot \frac{149}{35} \\ 0 & 0 & 0 & 1 & \frac{149}{35} \end{array}]$

Schritt 2.11.2

Vereinfache $R_{3}$ .

$[\begin{array}{c} 1 & \frac{3}{2} & - \frac{1}{2} & - \frac{1}{2} & \frac{9}{2} \\ 0 & 1 & 1 & 1 & 9 \\ 0 & 0 & 1 & 0 & \frac{5}{7} \\ 0 & 0 & 0 & 1 & \frac{149}{35} \end{array}]$

Schritt 2.12

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

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Schritt 2.12.1

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

$[\begin{array}{c} 1 & \frac{3}{2} & - \frac{1}{2} & - \frac{1}{2} & \frac{9}{2} \\ 0 - 0 & 1 - 0 & 1 - 0 & 1 - 1 & 9 - \frac{149}{35} \\ 0 & 0 & 1 & 0 & \frac{5}{7} \\ 0 & 0 & 0 & 1 & \frac{149}{35} \end{array}]$

Schritt 2.12.2

Vereinfache $R_{2}$ .

$[\begin{array}{c} 1 & \frac{3}{2} & - \frac{1}{2} & - \frac{1}{2} & \frac{9}{2} \\ 0 & 1 & 1 & 0 & \frac{166}{35} \\ 0 & 0 & 1 & 0 & \frac{5}{7} \\ 0 & 0 & 0 & 1 & \frac{149}{35} \end{array}]$

Schritt 2.13

Perform the row operation $R_{1} = R_{1} + \frac{1}{2} R_{4}$ to make the entry at $1, 4$ a $0$ .

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Schritt 2.13.1

Perform the row operation $R_{1} = R_{1} + \frac{1}{2} R_{4}$ to make the entry at $1, 4$ a $0$ .

$[\begin{array}{c} 1 + \frac{1}{2} \cdot 0 & \frac{3}{2} + \frac{1}{2} \cdot 0 & - \frac{1}{2} + \frac{1}{2} \cdot 0 & - \frac{1}{2} + \frac{1}{2} \cdot 1 & \frac{9}{2} + \frac{1}{2} \cdot \frac{149}{35} \\ 0 & 1 & 1 & 0 & \frac{166}{35} \\ 0 & 0 & 1 & 0 & \frac{5}{7} \\ 0 & 0 & 0 & 1 & \frac{149}{35} \end{array}]$

Schritt 2.13.2

Vereinfache $R_{1}$ .

$[\begin{array}{c} 1 & \frac{3}{2} & - \frac{1}{2} & 0 & \frac{232}{35} \\ 0 & 1 & 1 & 0 & \frac{166}{35} \\ 0 & 0 & 1 & 0 & \frac{5}{7} \\ 0 & 0 & 0 & 1 & \frac{149}{35} \end{array}]$

Schritt 2.14

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

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Schritt 2.14.1

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

$[\begin{array}{c} 1 & \frac{3}{2} & - \frac{1}{2} & 0 & \frac{232}{35} \\ 0 - 0 & 1 - 0 & 1 - 1 & 0 - 0 & \frac{166}{35} - \frac{5}{7} \\ 0 & 0 & 1 & 0 & \frac{5}{7} \\ 0 & 0 & 0 & 1 & \frac{149}{35} \end{array}]$

Schritt 2.14.2

Vereinfache $R_{2}$ .

$[\begin{array}{c} 1 & \frac{3}{2} & - \frac{1}{2} & 0 & \frac{232}{35} \\ 0 & 1 & 0 & 0 & \frac{141}{35} \\ 0 & 0 & 1 & 0 & \frac{5}{7} \\ 0 & 0 & 0 & 1 & \frac{149}{35} \end{array}]$

Schritt 2.15

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 2.15.1

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

$[\begin{array}{c} 1 + \frac{1}{2} \cdot 0 & \frac{3}{2} + \frac{1}{2} \cdot 0 & - \frac{1}{2} + \frac{1}{2} \cdot 1 & 0 + \frac{1}{2} \cdot 0 & \frac{232}{35} + \frac{1}{2} \cdot \frac{5}{7} \\ 0 & 1 & 0 & 0 & \frac{141}{35} \\ 0 & 0 & 1 & 0 & \frac{5}{7} \\ 0 & 0 & 0 & 1 & \frac{149}{35} \end{array}]$

Schritt 2.15.2

Vereinfache $R_{1}$ .

$[\begin{array}{c} 1 & \frac{3}{2} & 0 & 0 & \frac{489}{70} \\ 0 & 1 & 0 & 0 & \frac{141}{35} \\ 0 & 0 & 1 & 0 & \frac{5}{7} \\ 0 & 0 & 0 & 1 & \frac{149}{35} \end{array}]$

Schritt 2.16

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

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Schritt 2.16.1

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

$[\begin{array}{c} 1 - \frac{3}{2} \cdot 0 & \frac{3}{2} - \frac{3}{2} \cdot 1 & 0 - \frac{3}{2} \cdot 0 & 0 - \frac{3}{2} \cdot 0 & \frac{489}{70} - \frac{3}{2} \cdot \frac{141}{35} \\ 0 & 1 & 0 & 0 & \frac{141}{35} \\ 0 & 0 & 1 & 0 & \frac{5}{7} \\ 0 & 0 & 0 & 1 & \frac{149}{35} \end{array}]$

Schritt 2.16.2

Vereinfache $R_{1}$ .

$[\begin{array}{c} 1 & 0 & 0 & 0 & \frac{33}{35} \\ 0 & 1 & 0 & 0 & \frac{141}{35} \\ 0 & 0 & 1 & 0 & \frac{5}{7} \\ 0 & 0 & 0 & 1 & \frac{149}{35} \end{array}]$

Schritt 3

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

$w = \frac{33}{35}$

$x = \frac{141}{35}$

$y = \frac{5}{7}$

$z = \frac{149}{35}$

Schritt 4

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

$(\frac{33}{35}, \frac{141}{35}, \frac{5}{7}, \frac{149}{35})$

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.

$X = [\begin{array}{c} w \\ x \\ y \\ z \end{array}] = [\begin{array}{c} \frac{33}{35} \\ \frac{141}{35} \\ \frac{5}{7} \\ \frac{149}{35} \end{array}]$