Lineare Algebra Beispiele

$5 x + 3 y + z = - 21$ , $x - 3 y + 2 z = - 15$ , $14 x - 2 y + 3 z = 60$

Schritt 1

Schreibe das Gleichungssystem in Matrixform.

$[\begin{array}{c} 5 & 3 & 1 & - 21 \\ 1 & - 3 & 2 & - 15 \\ 14 & - 2 & 3 & 60 \end{array}]$

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 $1$ .

$[\begin{array}{c} \frac{5}{5} & \frac{3}{5} & \frac{1}{5} & \frac{- 21}{5} \\ 1 & - 3 & 2 & - 15 \\ 14 & - 2 & 3 & 60 \end{array}]$

Schritt 2.1.2

Vereinfache $R_{1}$ .

$[\begin{array}{c} 1 & \frac{3}{5} & \frac{1}{5} & - \frac{21}{5} \\ 1 & - 3 & 2 & - 15 \\ 14 & - 2 & 3 & 60 \end{array}]$

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 $0$ .

$[\begin{array}{c} 1 & \frac{3}{5} & \frac{1}{5} & - \frac{21}{5} \\ 1 - 1 & - 3 - \frac{3}{5} & 2 - \frac{1}{5} & - 15 + \frac{21}{5} \\ 14 & - 2 & 3 & 60 \end{array}]$

Schritt 2.2.2

Vereinfache $R_{2}$ .

$[\begin{array}{c} 1 & \frac{3}{5} & \frac{1}{5} & - \frac{21}{5} \\ 0 & - \frac{18}{5} & \frac{9}{5} & - \frac{54}{5} \\ 14 & - 2 & 3 & 60 \end{array}]$

Schritt 2.3

Perform the row operation $R_{3} = R_{3} - 14 R_{1}$ to make the entry at $3, 1$ a $0$ .

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Schritt 2.3.1

Perform the row operation $R_{3} = R_{3} - 14 R_{1}$ to make the entry at $3, 1$ a $0$ .

$[\begin{array}{c} 1 & \frac{3}{5} & \frac{1}{5} & - \frac{21}{5} \\ 0 & - \frac{18}{5} & \frac{9}{5} & - \frac{54}{5} \\ 14 - 14 \cdot 1 & - 2 - 14 (\frac{3}{5}) & 3 - 14 (\frac{1}{5}) & 60 - 14 (- \frac{21}{5}) \end{array}]$

Schritt 2.3.2

Vereinfache $R_{3}$ .

$[\begin{array}{c} 1 & \frac{3}{5} & \frac{1}{5} & - \frac{21}{5} \\ 0 & - \frac{18}{5} & \frac{9}{5} & - \frac{54}{5} \\ 0 & - \frac{52}{5} & \frac{1}{5} & \frac{594}{5} \end{array}]$

Schritt 2.4

Multiply each element of $R_{2}$ by $- \frac{5}{18}$ to make the entry at $2, 2$ a $1$ .

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Schritt 2.4.1

Multiply each element of $R_{2}$ by $- \frac{5}{18}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & \frac{3}{5} & \frac{1}{5} & - \frac{21}{5} \\ - \frac{5}{18} \cdot 0 & - \frac{5}{18} (- \frac{18}{5}) & - \frac{5}{18} \cdot \frac{9}{5} & - \frac{5}{18} (- \frac{54}{5}) \\ 0 & - \frac{52}{5} & \frac{1}{5} & \frac{594}{5} \end{array}]$

Schritt 2.4.2

Vereinfache $R_{2}$ .

$[\begin{array}{c} 1 & \frac{3}{5} & \frac{1}{5} & - \frac{21}{5} \\ 0 & 1 & - \frac{1}{2} & 3 \\ 0 & - \frac{52}{5} & \frac{1}{5} & \frac{594}{5} \end{array}]$

Schritt 2.5

Perform the row operation $R_{3} = R_{3} + \frac{52}{5} R_{2}$ to make the entry at $3, 2$ a $0$ .

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Schritt 2.5.1

Perform the row operation $R_{3} = R_{3} + \frac{52}{5} R_{2}$ to make the entry at $3, 2$ a $0$ .

$[\begin{array}{c} 1 & \frac{3}{5} & \frac{1}{5} & - \frac{21}{5} \\ 0 & 1 & - \frac{1}{2} & 3 \\ 0 + \frac{52}{5} \cdot 0 & - \frac{52}{5} + \frac{52}{5} \cdot 1 & \frac{1}{5} + \frac{52}{5} (- \frac{1}{2}) & \frac{594}{5} + \frac{52}{5} \cdot 3 \end{array}]$

Schritt 2.5.2

Vereinfache $R_{3}$ .

$[\begin{array}{c} 1 & \frac{3}{5} & \frac{1}{5} & - \frac{21}{5} \\ 0 & 1 & - \frac{1}{2} & 3 \\ 0 & 0 & - 5 & 150 \end{array}]$

Schritt 2.6

Multiply each element of $R_{3}$ by $- \frac{1}{5}$ to make the entry at $3, 3$ a $1$ .

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Schritt 2.6.1

Multiply each element of $R_{3}$ by $- \frac{1}{5}$ to make the entry at $3, 3$ a $1$ .

$[\begin{array}{c} 1 & \frac{3}{5} & \frac{1}{5} & - \frac{21}{5} \\ 0 & 1 & - \frac{1}{2} & 3 \\ - \frac{1}{5} \cdot 0 & - \frac{1}{5} \cdot 0 & - \frac{1}{5} \cdot - 5 & - \frac{1}{5} \cdot 150 \end{array}]$

Schritt 2.6.2

Vereinfache $R_{3}$ .

$[\begin{array}{c} 1 & \frac{3}{5} & \frac{1}{5} & - \frac{21}{5} \\ 0 & 1 & - \frac{1}{2} & 3 \\ 0 & 0 & 1 & - 30 \end{array}]$

Schritt 2.7

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

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Schritt 2.7.1

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

$[\begin{array}{c} 1 & \frac{3}{5} & \frac{1}{5} & - \frac{21}{5} \\ 0 + \frac{1}{2} \cdot 0 & 1 + \frac{1}{2} \cdot 0 & - \frac{1}{2} + \frac{1}{2} \cdot 1 & 3 + \frac{1}{2} \cdot - 30 \\ 0 & 0 & 1 & - 30 \end{array}]$

Schritt 2.7.2

Vereinfache $R_{2}$ .

$[\begin{array}{c} 1 & \frac{3}{5} & \frac{1}{5} & - \frac{21}{5} \\ 0 & 1 & 0 & - 12 \\ 0 & 0 & 1 & - 30 \end{array}]$

Schritt 2.8

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

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Schritt 2.8.1

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

$[\begin{array}{c} 1 - \frac{1}{5} \cdot 0 & \frac{3}{5} - \frac{1}{5} \cdot 0 & \frac{1}{5} - \frac{1}{5} \cdot 1 & - \frac{21}{5} - \frac{1}{5} \cdot - 30 \\ 0 & 1 & 0 & - 12 \\ 0 & 0 & 1 & - 30 \end{array}]$

Schritt 2.8.2

Vereinfache $R_{1}$ .

$[\begin{array}{c} 1 & \frac{3}{5} & 0 & \frac{9}{5} \\ 0 & 1 & 0 & - 12 \\ 0 & 0 & 1 & - 30 \end{array}]$

Schritt 2.9

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

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Schritt 2.9.1

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

$[\begin{array}{c} 1 - \frac{3}{5} \cdot 0 & \frac{3}{5} - \frac{3}{5} \cdot 1 & 0 - \frac{3}{5} \cdot 0 & \frac{9}{5} - \frac{3}{5} \cdot - 12 \\ 0 & 1 & 0 & - 12 \\ 0 & 0 & 1 & - 30 \end{array}]$

Schritt 2.9.2

Vereinfache $R_{1}$ .

$[\begin{array}{c} 1 & 0 & 0 & 9 \\ 0 & 1 & 0 & - 12 \\ 0 & 0 & 1 & - 30 \end{array}]$

Schritt 3

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

$x = 9$

$y = - 12$

$z = - 30$

Schritt 4

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

$(9, - 12, - 30)$

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.

$X = [\begin{array}{c} x \\ y \\ z \end{array}] = [\begin{array}{c} 9 \\ - 12 \\ - 30 \end{array}]$