Lineare Algebra Beispiele

[\begin{array}{c} 6 & 6 \sqrt{3} \\ - 6 \sqrt{3} & 6 \end{array}]

$[\begin{array}{c} 6 & 6 \sqrt{3} \\ - 6 \sqrt{3} & 6 \end{array}]$

Schritt 1

Ermittele die normierte Zeilenstufenform.

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Schritt 1.1

Multiply each element of

R_{1}

$R_{1}$ by

\frac{1}{6}

$\frac{1}{6}$ 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}{6}

$\frac{1}{6}$ to make the entry at

1, 1

$1, 1$ a

1

$1$ .

[\begin{array}{c} \frac{6}{6} & \frac{6 \sqrt{3}}{6} \\ - 6 \sqrt{3} & 6 \end{array}]

$[\begin{array}{c} \frac{6}{6} & \frac{6 \sqrt{3}}{6} \\ - 6 \sqrt{3} & 6 \end{array}]$

Schritt 1.1.2

Vereinfache

R_{1}

$R_{1}$ .

[\begin{array}{c} 1 & \sqrt{3} \\ - 6 \sqrt{3} & 6 \end{array}]

$[\begin{array}{c} 1 & \sqrt{3} \\ - 6 \sqrt{3} & 6 \end{array}]$

[\begin{array}{c} 1 & \sqrt{3} \\ - 6 \sqrt{3} & 6 \end{array}]

$[\begin{array}{c} 1 & \sqrt{3} \\ - 6 \sqrt{3} & 6 \end{array}]$

Schritt 1.2

Perform the row operation

R_{2} = R_{2} + 6 \sqrt{3} R_{1}

$R_{2} = R_{2} + 6 \sqrt{3} R_{1}$ 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} + 6 \sqrt{3} R_{1}

$R_{2} = R_{2} + 6 \sqrt{3} R_{1}$ to make the entry at

2, 1

$2, 1$ a

0

$0$ .

[\begin{array}{c} 1 & \sqrt{3} \\ - 6 \sqrt{3} + 6 \sqrt{3} \cdot 1 & 6 + 6 \sqrt{3} \sqrt{3} \end{array}]

$[\begin{array}{c} 1 & \sqrt{3} \\ - 6 \sqrt{3} + 6 \sqrt{3} \cdot 1 & 6 + 6 \sqrt{3} \sqrt{3} \end{array}]$

Schritt 1.2.2

Vereinfache

R_{2}

$R_{2}$ .

[\begin{array}{c} 1 & \sqrt{3} \\ 0 & 24 \end{array}]

$[\begin{array}{c} 1 & \sqrt{3} \\ 0 & 24 \end{array}]$

[\begin{array}{c} 1 & \sqrt{3} \\ 0 & 24 \end{array}]

$[\begin{array}{c} 1 & \sqrt{3} \\ 0 & 24 \end{array}]$

Schritt 1.3

Multiply each element of

R_{2}

$R_{2}$ by

\frac{1}{24}

$\frac{1}{24}$ to make the entry at

2, 2

$2, 2$ a

1

$1$ .

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Schritt 1.3.1

Multiply each element of

R_{2}

$R_{2}$ by

\frac{1}{24}

$\frac{1}{24}$ to make the entry at

2, 2

$2, 2$ a

1

$1$ .

[\begin{array}{c} 1 & \sqrt{3} \\ \frac{0}{24} & \frac{24}{24} \end{array}]

$[\begin{array}{c} 1 & \sqrt{3} \\ \frac{0}{24} & \frac{24}{24} \end{array}]$

Schritt 1.3.2

Vereinfache

R_{2}

$R_{2}$ .

[\begin{array}{c} 1 & \sqrt{3} \\ 0 & 1 \end{array}]

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

[\begin{array}{c} 1 & \sqrt{3} \\ 0 & 1 \end{array}]

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

Schritt 1.4

Perform the row operation

R_{1} = R_{1} - \sqrt{3} R_{2}

$R_{1} = R_{1} - \sqrt{3} R_{2}$ to make the entry at

1, 2

$1, 2$ a

0

$0$ .

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Schritt 1.4.1

Perform the row operation

R_{1} = R_{1} - \sqrt{3} R_{2}

$R_{1} = R_{1} - \sqrt{3} R_{2}$ to make the entry at

1, 2

$1, 2$ a

0

$0$ .

[\begin{array}{c} 1 - \sqrt{3} \cdot 0 & \sqrt{3} - \sqrt{3} \cdot 1 \\ 0 & 1 \end{array}]

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

Schritt 1.4.2

Vereinfache

R_{1}

$R_{1}$ .

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

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

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

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

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

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

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

$2$