Lineare Algebra Beispiele

$[\begin{array}{c} 3 & - 7 & 5 & 0 \\ 6 & - 14 & 10 & 0 \\ 12 & - 28 & 20 & 0 \end{array}]$

Schritt 1

Ermittele die normierte Zeilenstufenform.

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Schritt 1.1

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

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Schritt 1.1.1

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

$[\begin{array}{c} \frac{3}{3} & \frac{- 7}{3} & \frac{5}{3} & \frac{0}{3} \\ 6 & - 14 & 10 & 0 \\ 12 & - 28 & 20 & 0 \end{array}]$

Schritt 1.1.2

Vereinfache $R_{1}$ .

$[\begin{array}{c} 1 & - \frac{7}{3} & \frac{5}{3} & 0 \\ 6 & - 14 & 10 & 0 \\ 12 & - 28 & 20 & 0 \end{array}]$

Schritt 1.2

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 1.2.1

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

$[\begin{array}{c} 1 & - \frac{7}{3} & \frac{5}{3} & 0 \\ 6 - 6 \cdot 1 & - 14 - 6 (- \frac{7}{3}) & 10 - 6 (\frac{5}{3}) & 0 - 6 \cdot 0 \\ 12 & - 28 & 20 & 0 \end{array}]$

Schritt 1.2.2

Vereinfache $R_{2}$ .

$[\begin{array}{c} 1 & - \frac{7}{3} & \frac{5}{3} & 0 \\ 0 & 0 & 0 & 0 \\ 12 & - 28 & 20 & 0 \end{array}]$

Schritt 1.3

Perform the row operation $R_{3} = R_{3} - 12 R_{1}$ to make the entry at $3, 1$ a $0$ .

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Schritt 1.3.1

Perform the row operation $R_{3} = R_{3} - 12 R_{1}$ to make the entry at $3, 1$ a $0$ .

$[\begin{array}{c} 1 & - \frac{7}{3} & \frac{5}{3} & 0 \\ 0 & 0 & 0 & 0 \\ 12 - 12 \cdot 1 & - 28 - 12 (- \frac{7}{3}) & 20 - 12 (\frac{5}{3}) & 0 - 12 \cdot 0 \end{array}]$

Schritt 1.3.2

Vereinfache $R_{3}$ .

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

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: $a_{11}$

Pivot Columns: $1$