Lineare Algebra Beispiele

$[\begin{array}{c} 0 & 4 & 10 & 1 \\ 4 & 8 & 18 & 7 \\ 10 & 18 & 40 & 17 \\ 1 & 7 & 17 & 3 \end{array}]$

Schritt 1

Ermittele die normierte Zeilenstufenform.

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Schritt 1.1

Swap $R_{2}$ with $R_{1}$ to put a nonzero entry at $1, 1$ .

$[\begin{array}{c} 4 & 8 & 18 & 7 \\ 0 & 4 & 10 & 1 \\ 10 & 18 & 40 & 17 \\ 1 & 7 & 17 & 3 \end{array}]$

Schritt 1.2

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

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Schritt 1.2.1

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

$[\begin{array}{c} \frac{4}{4} & \frac{8}{4} & \frac{18}{4} & \frac{7}{4} \\ 0 & 4 & 10 & 1 \\ 10 & 18 & 40 & 17 \\ 1 & 7 & 17 & 3 \end{array}]$

Schritt 1.2.2

Vereinfache $R_{1}$ .

$[\begin{array}{c} 1 & 2 & \frac{9}{2} & \frac{7}{4} \\ 0 & 4 & 10 & 1 \\ 10 & 18 & 40 & 17 \\ 1 & 7 & 17 & 3 \end{array}]$

Schritt 1.3

Perform the row operation $R_{3} = R_{3} - 10 R_{1}$ to make the entry at $3, 1$ a $0$ .

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Schritt 1.3.1

Perform the row operation $R_{3} = R_{3} - 10 R_{1}$ to make the entry at $3, 1$ a $0$ .

$[\begin{array}{c} 1 & 2 & \frac{9}{2} & \frac{7}{4} \\ 0 & 4 & 10 & 1 \\ 10 - 10 \cdot 1 & 18 - 10 \cdot 2 & 40 - 10 (\frac{9}{2}) & 17 - 10 (\frac{7}{4}) \\ 1 & 7 & 17 & 3 \end{array}]$

Schritt 1.3.2

Vereinfache $R_{3}$ .

$[\begin{array}{c} 1 & 2 & \frac{9}{2} & \frac{7}{4} \\ 0 & 4 & 10 & 1 \\ 0 & - 2 & - 5 & - \frac{1}{2} \\ 1 & 7 & 17 & 3 \end{array}]$

Schritt 1.4

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

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Schritt 1.4.1

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

$[\begin{array}{c} 1 & 2 & \frac{9}{2} & \frac{7}{4} \\ 0 & 4 & 10 & 1 \\ 0 & - 2 & - 5 & - \frac{1}{2} \\ 1 - 1 & 7 - 2 & 17 - \frac{9}{2} & 3 - \frac{7}{4} \end{array}]$

Schritt 1.4.2

Vereinfache $R_{4}$ .

$[\begin{array}{c} 1 & 2 & \frac{9}{2} & \frac{7}{4} \\ 0 & 4 & 10 & 1 \\ 0 & - 2 & - 5 & - \frac{1}{2} \\ 0 & 5 & \frac{25}{2} & \frac{5}{4} \end{array}]$

Schritt 1.5

Multiply each element of $R_{2}$ by $\frac{1}{4}$ to make the entry at $2, 2$ a $1$ .

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Schritt 1.5.1

Multiply each element of $R_{2}$ by $\frac{1}{4}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & 2 & \frac{9}{2} & \frac{7}{4} \\ \frac{0}{4} & \frac{4}{4} & \frac{10}{4} & \frac{1}{4} \\ 0 & - 2 & - 5 & - \frac{1}{2} \\ 0 & 5 & \frac{25}{2} & \frac{5}{4} \end{array}]$

Schritt 1.5.2

Vereinfache $R_{2}$ .

$[\begin{array}{c} 1 & 2 & \frac{9}{2} & \frac{7}{4} \\ 0 & 1 & \frac{5}{2} & \frac{1}{4} \\ 0 & - 2 & - 5 & - \frac{1}{2} \\ 0 & 5 & \frac{25}{2} & \frac{5}{4} \end{array}]$

Schritt 1.6

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

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Schritt 1.6.1

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

$[\begin{array}{c} 1 & 2 & \frac{9}{2} & \frac{7}{4} \\ 0 & 1 & \frac{5}{2} & \frac{1}{4} \\ 0 + 2 \cdot 0 & - 2 + 2 \cdot 1 & - 5 + 2 (\frac{5}{2}) & - \frac{1}{2} + 2 (\frac{1}{4}) \\ 0 & 5 & \frac{25}{2} & \frac{5}{4} \end{array}]$

Schritt 1.6.2

Vereinfache $R_{3}$ .

$[\begin{array}{c} 1 & 2 & \frac{9}{2} & \frac{7}{4} \\ 0 & 1 & \frac{5}{2} & \frac{1}{4} \\ 0 & 0 & 0 & 0 \\ 0 & 5 & \frac{25}{2} & \frac{5}{4} \end{array}]$

Schritt 1.7

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

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Schritt 1.7.1

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

$[\begin{array}{c} 1 & 2 & \frac{9}{2} & \frac{7}{4} \\ 0 & 1 & \frac{5}{2} & \frac{1}{4} \\ 0 & 0 & 0 & 0 \\ 0 - 5 \cdot 0 & 5 - 5 \cdot 1 & \frac{25}{2} - 5 (\frac{5}{2}) & \frac{5}{4} - 5 (\frac{1}{4}) \end{array}]$

Schritt 1.7.2

Vereinfache $R_{4}$ .

$[\begin{array}{c} 1 & 2 & \frac{9}{2} & \frac{7}{4} \\ 0 & 1 & \frac{5}{2} & \frac{1}{4} \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array}]$

Schritt 1.8

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

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Schritt 1.8.1

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

$[\begin{array}{c} 1 - 2 \cdot 0 & 2 - 2 \cdot 1 & \frac{9}{2} - 2 (\frac{5}{2}) & \frac{7}{4} - 2 (\frac{1}{4}) \\ 0 & 1 & \frac{5}{2} & \frac{1}{4} \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array}]$

Schritt 1.8.2

Vereinfache $R_{1}$ .

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

Schritt 2

The pivot positions are the locations with the leading $1$ in each row. The pivot columns are the columns that have a pivot position.

Pivot Positions: $a_{11}$ and $a_{22}$

Pivot Columns: $1$ and $2$

Schritt 3

The rank is the number of pivot columns.

$2$