Lineare Algebra Beispiele

[\begin{matrix} 5 & 4 & - 3 - 5 & - 3 & - 2 \end{matrix}]

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

Schritt 1

Ermittele die normierte Zeilenstufenform.

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Schritt 1.1

Multiply each element of

R_{1}

$R_{1}$ by

\frac{1}{5}

$\frac{1}{5}$ to make the entry at

1, 1

$1, 1$ a

1

$1$ .

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Schritt 1.1.1

Multiply each element of

R_{1}

$R_{1}$ by

\frac{1}{5}

$\frac{1}{5}$ to make the entry at

1, 1

$1, 1$ a

1

$1$ .

[\begin{matrix} \frac{5}{5} & \frac{4}{5} & \frac{- 3}{5} - 5 & - 3 & - 2 \end{matrix}]

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

Schritt 1.1.2

Vereinfache

R_{1}

$R_{1}$ .

[\begin{matrix} 1 & \frac{4}{5} & - \frac{3}{5} - 5 & - 3 & - 2 \end{matrix}]

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

[\begin{matrix} 1 & \frac{4}{5} & - \frac{3}{5} - 5 & - 3 & - 2 \end{matrix}]

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

Schritt 1.2

Perform the row operation

R_{2} = R_{2} + 5 R_{1}

$R_{2} = R_{2} + 5 R_{1}$ to make the entry at

2, 1

$2, 1$ a

0

$0$ .

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Schritt 1.2.1

Perform the row operation

R_{2} = R_{2} + 5 R_{1}

$R_{2} = R_{2} + 5 R_{1}$ to make the entry at

2, 1

$2, 1$ a

0

$0$ .

⎡ ⎢ ⎣ \begin{matrix} 1 & \frac{4}{5} & - \frac{3}{5} - 5 + 5 \cdot 1 & - 3 + 5 (\frac{4}{5}) & - 2 + 5 (- \frac{3}{5}) \end{matrix} ⎤ ⎥ ⎦

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

Schritt 1.2.2

Vereinfache

R_{2}

$R_{2}$ .

[\begin{matrix} 1 & \frac{4}{5} & - \frac{3}{5} 0 & 1 & - 5 \end{matrix}]

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

[\begin{matrix} 1 & \frac{4}{5} & - \frac{3}{5} 0 & 1 & - 5 \end{matrix}]

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

Schritt 1.3

Perform the row operation

R_{1} = R_{1} - \frac{4}{5} R_{2}

$R_{1} = R_{1} - \frac{4}{5} R_{2}$ to make the entry at

1, 2

$1, 2$ a

0

$0$ .

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Schritt 1.3.1

Perform the row operation

R_{1} = R_{1} - \frac{4}{5} R_{2}

$R_{1} = R_{1} - \frac{4}{5} R_{2}$ to make the entry at

1, 2

$1, 2$ a

0

$0$ .

[\begin{matrix} 1 - \frac{4}{5} \cdot 0 & \frac{4}{5} - \frac{4}{5} \cdot 1 & - \frac{3}{5} - \frac{4}{5} \cdot - 5 0 & 1 & - 5 \end{matrix}]

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

Schritt 1.3.2

Vereinfache

R_{1}

$R_{1}$ .

[\begin{matrix} 1 & 0 & \frac{17}{5} 0 & 1 & - 5 \end{matrix}]

$[\begin{array}{c} 1 & 0 & \frac{17}{5} \\ 0 & 1 & - 5 \end{array}]$

[\begin{matrix} 1 & 0 & \frac{17}{5} 0 & 1 & - 5 \end{matrix}]

$[\begin{array}{c} 1 & 0 & \frac{17}{5} \\ 0 & 1 & - 5 \end{array}]$

[\begin{matrix} 1 & 0 & \frac{17}{5} 0 & 1 & - 5 \end{matrix}]

$[\begin{array}{c} 1 & 0 & \frac{17}{5} \\ 0 & 1 & - 5 \end{array}]$

Schritt 2

The pivot positions are the locations with the leading

1

$1$ in each row. The pivot columns are the columns that have a pivot position.

Pivot Positions:

a_{11}

$a_{11}$ and

a_{22}

$a_{22}$

Pivot Columns:

1

$1$ and

2

$2$