Lineare Algebra Beispiele

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

Schritt 1

Write as an augmented matrix for $A x = 0$ .

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

Schritt 2

Ermittele die normierte Zeilenstufenform.

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Schritt 2.1

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

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Schritt 2.1.1

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

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

Schritt 2.1.2

Vereinfache $R_{1}$ .

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

Schritt 2.2

Perform the row operation $R_{2} = R_{2} - 6 R_{1}$ to make the entry at $2, 1$ a $0$ .

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Schritt 2.2.1

Perform the row operation $R_{2} = R_{2} - 6 R_{1}$ to make the entry at $2, 1$ a $0$ .

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

Schritt 2.2.2

Vereinfache $R_{2}$ .

$[\begin{array}{c} 1 & \frac{6}{7} & \frac{2}{7} & 0 \\ 0 & - \frac{15}{7} & - \frac{5}{7} & 0 \\ 5 & 5 & 6 & 0 \end{array}]$

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 $0$ .

$[\begin{array}{c} 1 & \frac{6}{7} & \frac{2}{7} & 0 \\ 0 & - \frac{15}{7} & - \frac{5}{7} & 0 \\ 5 - 5 \cdot 1 & 5 - 5 (\frac{6}{7}) & 6 - 5 (\frac{2}{7}) & 0 - 5 \cdot 0 \end{array}]$

Schritt 2.3.2

Vereinfache $R_{3}$ .

$[\begin{array}{c} 1 & \frac{6}{7} & \frac{2}{7} & 0 \\ 0 & - \frac{15}{7} & - \frac{5}{7} & 0 \\ 0 & \frac{5}{7} & \frac{32}{7} & 0 \end{array}]$

Schritt 2.4

Multiply each element of $R_{2}$ by $- \frac{7}{15}$ to make the entry at $2, 2$ a $1$ .

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Schritt 2.4.1

Multiply each element of $R_{2}$ by $- \frac{7}{15}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & \frac{6}{7} & \frac{2}{7} & 0 \\ - \frac{7}{15} \cdot 0 & - \frac{7}{15} (- \frac{15}{7}) & - \frac{7}{15} (- \frac{5}{7}) & - \frac{7}{15} \cdot 0 \\ 0 & \frac{5}{7} & \frac{32}{7} & 0 \end{array}]$

Schritt 2.4.2

Vereinfache $R_{2}$ .

$[\begin{array}{c} 1 & \frac{6}{7} & \frac{2}{7} & 0 \\ 0 & 1 & \frac{1}{3} & 0 \\ 0 & \frac{5}{7} & \frac{32}{7} & 0 \end{array}]$

Schritt 2.5

Perform the row operation $R_{3} = R_{3} - \frac{5}{7} R_{2}$ to make the entry at $3, 2$ a $0$ .

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Schritt 2.5.1

Perform the row operation $R_{3} = R_{3} - \frac{5}{7} R_{2}$ to make the entry at $3, 2$ a $0$ .

$[\begin{array}{c} 1 & \frac{6}{7} & \frac{2}{7} & 0 \\ 0 & 1 & \frac{1}{3} & 0 \\ 0 - \frac{5}{7} \cdot 0 & \frac{5}{7} - \frac{5}{7} \cdot 1 & \frac{32}{7} - \frac{5}{7} \cdot \frac{1}{3} & 0 - \frac{5}{7} \cdot 0 \end{array}]$

Schritt 2.5.2

Vereinfache $R_{3}$ .

$[\begin{array}{c} 1 & \frac{6}{7} & \frac{2}{7} & 0 \\ 0 & 1 & \frac{1}{3} & 0 \\ 0 & 0 & \frac{13}{3} & 0 \end{array}]$

Schritt 2.6

Multiply each element of $R_{3}$ by $\frac{3}{13}$ to make the entry at $3, 3$ a $1$ .

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Schritt 2.6.1

Multiply each element of $R_{3}$ by $\frac{3}{13}$ to make the entry at $3, 3$ a $1$ .

$[\begin{array}{c} 1 & \frac{6}{7} & \frac{2}{7} & 0 \\ 0 & 1 & \frac{1}{3} & 0 \\ \frac{3}{13} \cdot 0 & \frac{3}{13} \cdot 0 & \frac{3}{13} \cdot \frac{13}{3} & \frac{3}{13} \cdot 0 \end{array}]$

Schritt 2.6.2

Vereinfache $R_{3}$ .

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

Schritt 2.7

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

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Schritt 2.7.1

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

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

Schritt 2.7.2

Vereinfache $R_{2}$ .

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

Schritt 2.8

Perform the row operation $R_{1} = R_{1} - \frac{2}{7} R_{3}$ to make the entry at $1, 3$ a $0$ .

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Schritt 2.8.1

Perform the row operation $R_{1} = R_{1} - \frac{2}{7} R_{3}$ to make the entry at $1, 3$ a $0$ .

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

Schritt 2.8.2

Vereinfache $R_{1}$ .

$[\begin{array}{c} 1 & \frac{6}{7} & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \end{array}]$

Schritt 2.9

Perform the row operation $R_{1} = R_{1} - \frac{6}{7} R_{2}$ to make the entry at $1, 2$ a $0$ .

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Schritt 2.9.1

Perform the row operation $R_{1} = R_{1} - \frac{6}{7} R_{2}$ to make the entry at $1, 2$ a $0$ .

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

Schritt 2.9.2

Vereinfache $R_{1}$ .

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

Schritt 3

Use the result matrix to declare the final solution to the system of equations.

$x = 0$

$y = 0$

$z = 0$

Schritt 4

Write a solution vector by solving in terms of the free variables in each row.

$[\begin{array}{c} x \\ y \\ z \end{array}] = [\begin{array}{c} 0 \\ 0 \\ 0 \end{array}]$

Schritt 5

Write as a solution set.

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