Lineare Algebra Beispiele

$[\begin{array}{c} - 3 & 6 & - 1 & 1 & - 7 \\ 1 & - 2 & 2 & 3 & - 1 \\ 2 & - 4 & 5 & 8 & - 4 \end{array}]$

Schritt 1

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

$[\begin{array}{c} - 3 & 6 & - 1 & 1 & - 7 & 0 \\ 1 & - 2 & 2 & 3 & - 1 & 0 \\ 2 & - 4 & 5 & 8 & - 4 & 0 \end{array}]$

Schritt 2

Ermittele die normierte Zeilenstufenform.

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Schritt 2.1

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

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Schritt 2.1.1

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

$[\begin{array}{c} - \frac{1}{3} \cdot - 3 & - \frac{1}{3} \cdot 6 & - \frac{1}{3} \cdot - 1 & - \frac{1}{3} \cdot 1 & - \frac{1}{3} \cdot - 7 & - \frac{1}{3} \cdot 0 \\ 1 & - 2 & 2 & 3 & - 1 & 0 \\ 2 & - 4 & 5 & 8 & - 4 & 0 \end{array}]$

Schritt 2.1.2

Vereinfache $R_{1}$ .

$[\begin{array}{c} 1 & - 2 & \frac{1}{3} & - \frac{1}{3} & \frac{7}{3} & 0 \\ 1 & - 2 & 2 & 3 & - 1 & 0 \\ 2 & - 4 & 5 & 8 & - 4 & 0 \end{array}]$

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 $0$ .

$[\begin{array}{c} 1 & - 2 & \frac{1}{3} & - \frac{1}{3} & \frac{7}{3} & 0 \\ 1 - 1 & - 2 + 2 & 2 - \frac{1}{3} & 3 + \frac{1}{3} & - 1 - \frac{7}{3} & 0 - 0 \\ 2 & - 4 & 5 & 8 & - 4 & 0 \end{array}]$

Schritt 2.2.2

Vereinfache $R_{2}$ .

$[\begin{array}{c} 1 & - 2 & \frac{1}{3} & - \frac{1}{3} & \frac{7}{3} & 0 \\ 0 & 0 & \frac{5}{3} & \frac{10}{3} & - \frac{10}{3} & 0 \\ 2 & - 4 & 5 & 8 & - 4 & 0 \end{array}]$

Schritt 2.3

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

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Schritt 2.3.1

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

$[\begin{array}{c} 1 & - 2 & \frac{1}{3} & - \frac{1}{3} & \frac{7}{3} & 0 \\ 0 & 0 & \frac{5}{3} & \frac{10}{3} & - \frac{10}{3} & 0 \\ 2 - 2 \cdot 1 & - 4 - 2 \cdot - 2 & 5 - 2 (\frac{1}{3}) & 8 - 2 (- \frac{1}{3}) & - 4 - 2 (\frac{7}{3}) & 0 - 2 \cdot 0 \end{array}]$

Schritt 2.3.2

Vereinfache $R_{3}$ .

$[\begin{array}{c} 1 & - 2 & \frac{1}{3} & - \frac{1}{3} & \frac{7}{3} & 0 \\ 0 & 0 & \frac{5}{3} & \frac{10}{3} & - \frac{10}{3} & 0 \\ 0 & 0 & \frac{13}{3} & \frac{26}{3} & - \frac{26}{3} & 0 \end{array}]$

Schritt 2.4

Multiply each element of $R_{2}$ by $\frac{3}{5}$ to make the entry at $2, 3$ a $1$ .

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Schritt 2.4.1

Multiply each element of $R_{2}$ by $\frac{3}{5}$ to make the entry at $2, 3$ a $1$ .

$[\begin{array}{c} 1 & - 2 & \frac{1}{3} & - \frac{1}{3} & \frac{7}{3} & 0 \\ \frac{3}{5} \cdot 0 & \frac{3}{5} \cdot 0 & \frac{3}{5} \cdot \frac{5}{3} & \frac{3}{5} \cdot \frac{10}{3} & \frac{3}{5} (- \frac{10}{3}) & \frac{3}{5} \cdot 0 \\ 0 & 0 & \frac{13}{3} & \frac{26}{3} & - \frac{26}{3} & 0 \end{array}]$

Schritt 2.4.2

Vereinfache $R_{2}$ .

$[\begin{array}{c} 1 & - 2 & \frac{1}{3} & - \frac{1}{3} & \frac{7}{3} & 0 \\ 0 & 0 & 1 & 2 & - 2 & 0 \\ 0 & 0 & \frac{13}{3} & \frac{26}{3} & - \frac{26}{3} & 0 \end{array}]$

Schritt 2.5

Perform the row operation $R_{3} = R_{3} - \frac{13}{3} R_{2}$ to make the entry at $3, 3$ a $0$ .

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Schritt 2.5.1

Perform the row operation $R_{3} = R_{3} - \frac{13}{3} R_{2}$ to make the entry at $3, 3$ a $0$ .

$[\begin{array}{c} 1 & - 2 & \frac{1}{3} & - \frac{1}{3} & \frac{7}{3} & 0 \\ 0 & 0 & 1 & 2 & - 2 & 0 \\ 0 - \frac{13}{3} \cdot 0 & 0 - \frac{13}{3} \cdot 0 & \frac{13}{3} - \frac{13}{3} \cdot 1 & \frac{26}{3} - \frac{13}{3} \cdot 2 & - \frac{26}{3} - \frac{13}{3} \cdot - 2 & 0 - \frac{13}{3} \cdot 0 \end{array}]$

Schritt 2.5.2

Vereinfache $R_{3}$ .

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

Schritt 2.6

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

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Schritt 2.6.1

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

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

Schritt 2.6.2

Vereinfache $R_{1}$ .

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

Schritt 3

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

$x_{1} - 2 x_{2} - x_{4} + 3 x_{5} = 0$

$x_{3} + 2 x_{4} - 2 x_{5} = 0$

$0 = 0$

Schritt 4

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

$[\begin{array}{c} x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \\ x_{5} \end{array}] = [\begin{array}{c} 2 x_{2} + x_{4} - 3 x_{5} \\ x_{2} \\ - 2 x_{4} + 2 x_{5} \\ x_{4} \\ x_{5} \end{array}]$

Schritt 5

Write the solution as a linear combination of vectors.

$[\begin{array}{c} x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \\ x_{5} \end{array}] = x_{2} [\begin{array}{c} 2 \\ 1 \\ 0 \\ 0 \\ 0 \end{array}] + x_{4} [\begin{array}{c} 1 \\ 0 \\ - 2 \\ 1 \\ 0 \end{array}] + x_{5} [\begin{array}{c} - 3 \\ 0 \\ 2 \\ 0 \\ 1 \end{array}]$

Schritt 6

Write as a solution set.

${x_{2} [\begin{array}{c} 2 \\ 1 \\ 0 \\ 0 \\ 0 \end{array}] + x_{4} [\begin{array}{c} 1 \\ 0 \\ - 2 \\ 1 \\ 0 \end{array}] + x_{5} [\begin{array}{c} - 3 \\ 0 \\ 2 \\ 0 \\ 1 \end{array}] | x_{2}, x_{4}, x_{5} \in R}$