Lineare Algebra Beispiele

x + 2 y - z = 4

$x + 2 y - z = 4$ ,

2 x + y + z = - 2

$2 x + y + z = - 2$ ,

x + 2 y + z = 2

$x + 2 y + z = 2$

Schritt 1

Ermittle

A X = B

$A X = B$ aus dem Gleichungssystem.

[\begin{array}{c} 1 & 2 & - 1 \\ 2 & 1 & 1 \\ 1 & 2 & 1 \end{array}] \cdot [\begin{array}{c} x \\ y \\ z \end{array}] = [\begin{array}{c} 4 \\ - 2 \\ 2 \end{array}]

$[\begin{array}{c} 1 & 2 & - 1 \\ 2 & 1 & 1 \\ 1 & 2 & 1 \end{array}] \cdot [\begin{array}{c} x \\ y \\ z \end{array}] = [\begin{array}{c} 4 \\ - 2 \\ 2 \end{array}]$

Schritt 2

Finde die Inverse der Koeffizientenmatrix.

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Schritt 2.1

Find the determinant.

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Schritt 2.1.1

Choose the row or column with the most

0

$0$ elements. If there are no

0

$0$ elements choose any row or column. Multiply every element in row

1

$1$ by its cofactor and add.

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Schritt 2.1.1.1

Consider the corresponding sign chart.

| \begin{array}{c} + & - & + \\ - & + & - \\ + & - & + \end{array} |

$| \begin{array}{c} + & - & + \\ - & + & - \\ + & - & + \end{array} |$

Schritt 2.1.1.2

The cofactor is the minor with the sign changed if the indices match a

-

$-$ position on the sign chart.

Schritt 2.1.1.3

The minor for

a_{11}

$a_{11}$ is the determinant with row

1

$1$ and column

1

$1$ deleted.

| \begin{array}{c} 1 & 1 \\ 2 & 1 \end{array} |

$| \begin{array}{c} 1 & 1 \\ 2 & 1 \end{array} |$

Schritt 2.1.1.4

Multiply element

a_{11}

$a_{11}$ by its cofactor.

1 | \begin{array}{c} 1 & 1 \\ 2 & 1 \end{array} |

$1 | \begin{array}{c} 1 & 1 \\ 2 & 1 \end{array} |$

Schritt 2.1.1.5

The minor for

a_{12}

$a_{12}$ is the determinant with row

1

$1$ and column

2

$2$ deleted.

| \begin{array}{c} 2 & 1 \\ 1 & 1 \end{array} |

$| \begin{array}{c} 2 & 1 \\ 1 & 1 \end{array} |$

Schritt 2.1.1.6

Multiply element

a_{12}

$a_{12}$ by its cofactor.

- 2 | \begin{array}{c} 2 & 1 \\ 1 & 1 \end{array} |

$- 2 | \begin{array}{c} 2 & 1 \\ 1 & 1 \end{array} |$

Schritt 2.1.1.7

The minor for

a_{13}

$a_{13}$ is the determinant with row

1

$1$ and column

3

$3$ deleted.

| \begin{array}{c} 2 & 1 \\ 1 & 2 \end{array} |

$| \begin{array}{c} 2 & 1 \\ 1 & 2 \end{array} |$

Schritt 2.1.1.8

Multiply element

a_{13}

$a_{13}$ by its cofactor.

- 1 | \begin{array}{c} 2 & 1 \\ 1 & 2 \end{array} |

$- 1 | \begin{array}{c} 2 & 1 \\ 1 & 2 \end{array} |$

Schritt 2.1.1.9

Add the terms together.

1 | \begin{array}{c} 1 & 1 \\ 2 & 1 \end{array} | - 2 | \begin{array}{c} 2 & 1 \\ 1 & 1 \end{array} | - 1 | \begin{array}{c} 2 & 1 \\ 1 & 2 \end{array} |

$1 | \begin{array}{c} 1 & 1 \\ 2 & 1 \end{array} | - 2 | \begin{array}{c} 2 & 1 \\ 1 & 1 \end{array} | - 1 | \begin{array}{c} 2 & 1 \\ 1 & 2 \end{array} |$

1 | \begin{array}{c} 1 & 1 \\ 2 & 1 \end{array} | - 2 | \begin{array}{c} 2 & 1 \\ 1 & 1 \end{array} | - 1 | \begin{array}{c} 2 & 1 \\ 1 & 2 \end{array} |

$1 | \begin{array}{c} 1 & 1 \\ 2 & 1 \end{array} | - 2 | \begin{array}{c} 2 & 1 \\ 1 & 1 \end{array} | - 1 | \begin{array}{c} 2 & 1 \\ 1 & 2 \end{array} |$

Schritt 2.1.2

Berechne

| \begin{array}{c} 1 & 1 \\ 2 & 1 \end{array} |

$| \begin{array}{c} 1 & 1 \\ 2 & 1 \end{array} |$ .

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Schritt 2.1.2.1

Die Determinante einer

2 \times 2

$2 \times 2$ -Matrix kann mithilfe der Formel

| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b

$| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ bestimmt werden.

1 (1 \cdot 1 - 2 \cdot 1) - 2 | \begin{array}{c} 2 & 1 \\ 1 & 1 \end{array} | - 1 | \begin{array}{c} 2 & 1 \\ 1 & 2 \end{array} |

$1 (1 \cdot 1 - 2 \cdot 1) - 2 | \begin{array}{c} 2 & 1 \\ 1 & 1 \end{array} | - 1 | \begin{array}{c} 2 & 1 \\ 1 & 2 \end{array} |$

Schritt 2.1.2.2

Vereinfache die Determinante.

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Schritt 2.1.2.2.1

Vereinfache jeden Term.

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Schritt 2.1.2.2.1.1

Mutltipliziere

1

$1$ mit

1

$1$ .

1 (1 - 2 \cdot 1) - 2 | \begin{array}{c} 2 & 1 \\ 1 & 1 \end{array} | - 1 | \begin{array}{c} 2 & 1 \\ 1 & 2 \end{array} |

$1 (1 - 2 \cdot 1) - 2 | \begin{array}{c} 2 & 1 \\ 1 & 1 \end{array} | - 1 | \begin{array}{c} 2 & 1 \\ 1 & 2 \end{array} |$

Schritt 2.1.2.2.1.2

Mutltipliziere

- 2

$- 2$ mit

1

$1$ .

1 (1 - 2) - 2 | \begin{array}{c} 2 & 1 \\ 1 & 1 \end{array} | - 1 | \begin{array}{c} 2 & 1 \\ 1 & 2 \end{array} |

$1 (1 - 2) - 2 | \begin{array}{c} 2 & 1 \\ 1 & 1 \end{array} | - 1 | \begin{array}{c} 2 & 1 \\ 1 & 2 \end{array} |$

1 (1 - 2) - 2 | \begin{array}{c} 2 & 1 \\ 1 & 1 \end{array} | - 1 | \begin{array}{c} 2 & 1 \\ 1 & 2 \end{array} |

$1 (1 - 2) - 2 | \begin{array}{c} 2 & 1 \\ 1 & 1 \end{array} | - 1 | \begin{array}{c} 2 & 1 \\ 1 & 2 \end{array} |$

Schritt 2.1.2.2.2

Subtrahiere

2

$2$ von

1

$1$ .

1 \cdot - 1 - 2 | \begin{array}{c} 2 & 1 \\ 1 & 1 \end{array} | - 1 | \begin{array}{c} 2 & 1 \\ 1 & 2 \end{array} |

$1 \cdot - 1 - 2 | \begin{array}{c} 2 & 1 \\ 1 & 1 \end{array} | - 1 | \begin{array}{c} 2 & 1 \\ 1 & 2 \end{array} |$

1 \cdot - 1 - 2 | \begin{array}{c} 2 & 1 \\ 1 & 1 \end{array} | - 1 | \begin{array}{c} 2 & 1 \\ 1 & 2 \end{array} |

$1 \cdot - 1 - 2 | \begin{array}{c} 2 & 1 \\ 1 & 1 \end{array} | - 1 | \begin{array}{c} 2 & 1 \\ 1 & 2 \end{array} |$

1 \cdot - 1 - 2 | \begin{array}{c} 2 & 1 \\ 1 & 1 \end{array} | - 1 | \begin{array}{c} 2 & 1 \\ 1 & 2 \end{array} |

$1 \cdot - 1 - 2 | \begin{array}{c} 2 & 1 \\ 1 & 1 \end{array} | - 1 | \begin{array}{c} 2 & 1 \\ 1 & 2 \end{array} |$

Schritt 2.1.3

Berechne

| \begin{array}{c} 2 & 1 \\ 1 & 1 \end{array} |

$| \begin{array}{c} 2 & 1 \\ 1 & 1 \end{array} |$ .

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Schritt 2.1.3.1

Die Determinante einer

2 \times 2

$2 \times 2$ -Matrix kann mithilfe der Formel

| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b

$| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ bestimmt werden.

1 \cdot - 1 - 2 (2 \cdot 1 - 1 \cdot 1) - 1 | \begin{array}{c} 2 & 1 \\ 1 & 2 \end{array} |

$1 \cdot - 1 - 2 (2 \cdot 1 - 1 \cdot 1) - 1 | \begin{array}{c} 2 & 1 \\ 1 & 2 \end{array} |$

Schritt 2.1.3.2

Vereinfache die Determinante.

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Schritt 2.1.3.2.1

Vereinfache jeden Term.

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Schritt 2.1.3.2.1.1

Mutltipliziere

2

$2$ mit

1

$1$ .

1 \cdot - 1 - 2 (2 - 1 \cdot 1) - 1 | \begin{array}{c} 2 & 1 \\ 1 & 2 \end{array} |

$1 \cdot - 1 - 2 (2 - 1 \cdot 1) - 1 | \begin{array}{c} 2 & 1 \\ 1 & 2 \end{array} |$

Schritt 2.1.3.2.1.2

Mutltipliziere

- 1

$- 1$ mit

1

$1$ .

1 \cdot - 1 - 2 (2 - 1) - 1 | \begin{array}{c} 2 & 1 \\ 1 & 2 \end{array} |

$1 \cdot - 1 - 2 (2 - 1) - 1 | \begin{array}{c} 2 & 1 \\ 1 & 2 \end{array} |$

1 \cdot - 1 - 2 (2 - 1) - 1 | \begin{array}{c} 2 & 1 \\ 1 & 2 \end{array} |

$1 \cdot - 1 - 2 (2 - 1) - 1 | \begin{array}{c} 2 & 1 \\ 1 & 2 \end{array} |$

Schritt 2.1.3.2.2

Subtrahiere

1

$1$ von

2

$2$ .

1 \cdot - 1 - 2 \cdot 1 - 1 | \begin{array}{c} 2 & 1 \\ 1 & 2 \end{array} |

$1 \cdot - 1 - 2 \cdot 1 - 1 | \begin{array}{c} 2 & 1 \\ 1 & 2 \end{array} |$

1 \cdot - 1 - 2 \cdot 1 - 1 | \begin{array}{c} 2 & 1 \\ 1 & 2 \end{array} |

$1 \cdot - 1 - 2 \cdot 1 - 1 | \begin{array}{c} 2 & 1 \\ 1 & 2 \end{array} |$

1 \cdot - 1 - 2 \cdot 1 - 1 | \begin{array}{c} 2 & 1 \\ 1 & 2 \end{array} |

$1 \cdot - 1 - 2 \cdot 1 - 1 | \begin{array}{c} 2 & 1 \\ 1 & 2 \end{array} |$

Schritt 2.1.4

Berechne

| \begin{array}{c} 2 & 1 \\ 1 & 2 \end{array} |

$| \begin{array}{c} 2 & 1 \\ 1 & 2 \end{array} |$ .

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Schritt 2.1.4.1

Die Determinante einer

2 \times 2

$2 \times 2$ -Matrix kann mithilfe der Formel

| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b

$| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ bestimmt werden.

1 \cdot - 1 - 2 \cdot 1 - 1 (2 \cdot 2 - 1 \cdot 1)

$1 \cdot - 1 - 2 \cdot 1 - 1 (2 \cdot 2 - 1 \cdot 1)$

Schritt 2.1.4.2

Vereinfache die Determinante.

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Schritt 2.1.4.2.1

Vereinfache jeden Term.

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Schritt 2.1.4.2.1.1

Mutltipliziere

2

$2$ mit

2

$2$ .

1 \cdot - 1 - 2 \cdot 1 - 1 (4 - 1 \cdot 1)

$1 \cdot - 1 - 2 \cdot 1 - 1 (4 - 1 \cdot 1)$

Schritt 2.1.4.2.1.2

Mutltipliziere

- 1

$- 1$ mit

1

$1$ .

1 \cdot - 1 - 2 \cdot 1 - 1 (4 - 1)

$1 \cdot - 1 - 2 \cdot 1 - 1 (4 - 1)$

1 \cdot - 1 - 2 \cdot 1 - 1 (4 - 1)

$1 \cdot - 1 - 2 \cdot 1 - 1 (4 - 1)$

Schritt 2.1.4.2.2

Subtrahiere

1

$1$ von

4

$4$ .

1 \cdot - 1 - 2 \cdot 1 - 1 \cdot 3

$1 \cdot - 1 - 2 \cdot 1 - 1 \cdot 3$

1 \cdot - 1 - 2 \cdot 1 - 1 \cdot 3

$1 \cdot - 1 - 2 \cdot 1 - 1 \cdot 3$

1 \cdot - 1 - 2 \cdot 1 - 1 \cdot 3

$1 \cdot - 1 - 2 \cdot 1 - 1 \cdot 3$

Schritt 2.1.5

Vereinfache die Determinante.

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Schritt 2.1.5.1

Vereinfache jeden Term.

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Schritt 2.1.5.1.1

Mutltipliziere

- 1

$- 1$ mit

1

$1$ .

- 1 - 2 \cdot 1 - 1 \cdot 3

$- 1 - 2 \cdot 1 - 1 \cdot 3$

Schritt 2.1.5.1.2

Mutltipliziere

- 2

$- 2$ mit

1

$1$ .

- 1 - 2 - 1 \cdot 3

$- 1 - 2 - 1 \cdot 3$

Schritt 2.1.5.1.3

Mutltipliziere

- 1

$- 1$ mit

3

$3$ .

- 1 - 2 - 3

$- 1 - 2 - 3$

- 1 - 2 - 3

$- 1 - 2 - 3$

Schritt 2.1.5.2

Subtrahiere

2

$2$ von

- 1

$- 1$ .

- 3 - 3

$- 3 - 3$

Schritt 2.1.5.3

Subtrahiere

3

$3$ von

- 3

$- 3$ .

- 6

$- 6$

- 6

$- 6$

- 6

$- 6$

Schritt 2.2

Since the determinant is non-zero, the inverse exists.

Schritt 2.3

Set up a

3 \times 6

$3 \times 6$ matrix where the left half is the original matrix and the right half is its identity matrix.

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

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

Schritt 2.4

Ermittele die normierte Zeilenstufenform.

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Schritt 2.4.1

Perform the row operation

R_{2} = R_{2} - 2 R_{1}

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

2, 1

$2, 1$ a

0

$0$ .

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Schritt 2.4.1.1

Perform the row operation

R_{2} = R_{2} - 2 R_{1}

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

2, 1

$2, 1$ a

0

$0$ .

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

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

Schritt 2.4.1.2

Vereinfache

R_{2}

$R_{2}$ .

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

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

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

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

Schritt 2.4.2

Perform the row operation

R_{3} = R_{3} - R_{1}

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

3, 1

$3, 1$ a

0

$0$ .

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Schritt 2.4.2.1

Perform the row operation

R_{3} = R_{3} - R_{1}

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

3, 1

$3, 1$ a

0

$0$ .

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

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

Schritt 2.4.2.2

Vereinfache

R_{3}

$R_{3}$ .

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

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

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

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

Schritt 2.4.3

Multiply each element of

R_{2}

$R_{2}$ by

- \frac{1}{3}

$- \frac{1}{3}$ to make the entry at

2, 2

$2, 2$ a

1

$1$ .

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Schritt 2.4.3.1

Multiply each element of

R_{2}

$R_{2}$ by

- \frac{1}{3}

$- \frac{1}{3}$ to make the entry at

2, 2

$2, 2$ a

1

$1$ .

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

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

Schritt 2.4.3.2

Vereinfache

R_{2}

$R_{2}$ .

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

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

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

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

Schritt 2.4.4

Multiply each element of

R_{3}

$R_{3}$ by

\frac{1}{2}

$\frac{1}{2}$ to make the entry at

3, 3

$3, 3$ a

1

$1$ .

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Schritt 2.4.4.1

Multiply each element of

R_{3}

$R_{3}$ by

\frac{1}{2}

$\frac{1}{2}$ to make the entry at

3, 3

$3, 3$ a

1

$1$ .

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

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

Schritt 2.4.4.2

Vereinfache

R_{3}

$R_{3}$ .

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

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

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

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

Schritt 2.4.5

Perform the row operation

R_{2} = R_{2} + R_{3}

$R_{2} = R_{2} + R_{3}$ to make the entry at

2, 3

$2, 3$ a

0

$0$ .

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Schritt 2.4.5.1

Perform the row operation

R_{2} = R_{2} + R_{3}

$R_{2} = R_{2} + R_{3}$ to make the entry at

2, 3

$2, 3$ a

0

$0$ .

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

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

Schritt 2.4.5.2

Vereinfache

R_{2}

$R_{2}$ .

[\begin{array}{c} 1 & 2 & - 1 & 1 & 0 & 0 \\ 0 & 1 & 0 & \frac{1}{6} & - \frac{1}{3} & \frac{1}{2} \\ 0 & 0 & 1 & - \frac{1}{2} & 0 & \frac{1}{2} \end{array}]

$[\begin{array}{c} 1 & 2 & - 1 & 1 & 0 & 0 \\ 0 & 1 & 0 & \frac{1}{6} & - \frac{1}{3} & \frac{1}{2} \\ 0 & 0 & 1 & - \frac{1}{2} & 0 & \frac{1}{2} \end{array}]$

[\begin{array}{c} 1 & 2 & - 1 & 1 & 0 & 0 \\ 0 & 1 & 0 & \frac{1}{6} & - \frac{1}{3} & \frac{1}{2} \\ 0 & 0 & 1 & - \frac{1}{2} & 0 & \frac{1}{2} \end{array}]

$[\begin{array}{c} 1 & 2 & - 1 & 1 & 0 & 0 \\ 0 & 1 & 0 & \frac{1}{6} & - \frac{1}{3} & \frac{1}{2} \\ 0 & 0 & 1 & - \frac{1}{2} & 0 & \frac{1}{2} \end{array}]$

Schritt 2.4.6

Perform the row operation

R_{1} = R_{1} + R_{3}

$R_{1} = R_{1} + R_{3}$ to make the entry at

1, 3

$1, 3$ a

0

$0$ .

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Schritt 2.4.6.1

Perform the row operation

R_{1} = R_{1} + R_{3}

$R_{1} = R_{1} + R_{3}$ to make the entry at

1, 3

$1, 3$ a

0

$0$ .

[\begin{array}{c} 1 + 0 & 2 + 0 & - 1 + 1 \cdot 1 & 1 - \frac{1}{2} & 0 + 0 & 0 + \frac{1}{2} \\ 0 & 1 & 0 & \frac{1}{6} & - \frac{1}{3} & \frac{1}{2} \\ 0 & 0 & 1 & - \frac{1}{2} & 0 & \frac{1}{2} \end{array}]

$[\begin{array}{c} 1 + 0 & 2 + 0 & - 1 + 1 \cdot 1 & 1 - \frac{1}{2} & 0 + 0 & 0 + \frac{1}{2} \\ 0 & 1 & 0 & \frac{1}{6} & - \frac{1}{3} & \frac{1}{2} \\ 0 & 0 & 1 & - \frac{1}{2} & 0 & \frac{1}{2} \end{array}]$

Schritt 2.4.6.2

Vereinfache

R_{1}

$R_{1}$ .

[\begin{array}{c} 1 & 2 & 0 & \frac{1}{2} & 0 & \frac{1}{2} \\ 0 & 1 & 0 & \frac{1}{6} & - \frac{1}{3} & \frac{1}{2} \\ 0 & 0 & 1 & - \frac{1}{2} & 0 & \frac{1}{2} \end{array}]

$[\begin{array}{c} 1 & 2 & 0 & \frac{1}{2} & 0 & \frac{1}{2} \\ 0 & 1 & 0 & \frac{1}{6} & - \frac{1}{3} & \frac{1}{2} \\ 0 & 0 & 1 & - \frac{1}{2} & 0 & \frac{1}{2} \end{array}]$

[\begin{array}{c} 1 & 2 & 0 & \frac{1}{2} & 0 & \frac{1}{2} \\ 0 & 1 & 0 & \frac{1}{6} & - \frac{1}{3} & \frac{1}{2} \\ 0 & 0 & 1 & - \frac{1}{2} & 0 & \frac{1}{2} \end{array}]

$[\begin{array}{c} 1 & 2 & 0 & \frac{1}{2} & 0 & \frac{1}{2} \\ 0 & 1 & 0 & \frac{1}{6} & - \frac{1}{3} & \frac{1}{2} \\ 0 & 0 & 1 & - \frac{1}{2} & 0 & \frac{1}{2} \end{array}]$

Schritt 2.4.7

Perform the row operation

R_{1} = R_{1} - 2 R_{2}

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

1, 2

$1, 2$ a

0

$0$ .

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Schritt 2.4.7.1

Perform the row operation

R_{1} = R_{1} - 2 R_{2}

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

1, 2

$1, 2$ a

0

$0$ .

[\begin{array}{c} 1 - 2 \cdot 0 & 2 - 2 \cdot 1 & 0 - 2 \cdot 0 & \frac{1}{2} - 2 (\frac{1}{6}) & 0 - 2 (- \frac{1}{3}) & \frac{1}{2} - 2 (\frac{1}{2}) \\ 0 & 1 & 0 & \frac{1}{6} & - \frac{1}{3} & \frac{1}{2} \\ 0 & 0 & 1 & - \frac{1}{2} & 0 & \frac{1}{2} \end{array}]

$[\begin{array}{c} 1 - 2 \cdot 0 & 2 - 2 \cdot 1 & 0 - 2 \cdot 0 & \frac{1}{2} - 2 (\frac{1}{6}) & 0 - 2 (- \frac{1}{3}) & \frac{1}{2} - 2 (\frac{1}{2}) \\ 0 & 1 & 0 & \frac{1}{6} & - \frac{1}{3} & \frac{1}{2} \\ 0 & 0 & 1 & - \frac{1}{2} & 0 & \frac{1}{2} \end{array}]$

Schritt 2.4.7.2

Vereinfache

R_{1}

$R_{1}$ .

[\begin{array}{c} 1 & 0 & 0 & \frac{1}{6} & \frac{2}{3} & - \frac{1}{2} \\ 0 & 1 & 0 & \frac{1}{6} & - \frac{1}{3} & \frac{1}{2} \\ 0 & 0 & 1 & - \frac{1}{2} & 0 & \frac{1}{2} \end{array}]

$[\begin{array}{c} 1 & 0 & 0 & \frac{1}{6} & \frac{2}{3} & - \frac{1}{2} \\ 0 & 1 & 0 & \frac{1}{6} & - \frac{1}{3} & \frac{1}{2} \\ 0 & 0 & 1 & - \frac{1}{2} & 0 & \frac{1}{2} \end{array}]$

[\begin{array}{c} 1 & 0 & 0 & \frac{1}{6} & \frac{2}{3} & - \frac{1}{2} \\ 0 & 1 & 0 & \frac{1}{6} & - \frac{1}{3} & \frac{1}{2} \\ 0 & 0 & 1 & - \frac{1}{2} & 0 & \frac{1}{2} \end{array}]

$[\begin{array}{c} 1 & 0 & 0 & \frac{1}{6} & \frac{2}{3} & - \frac{1}{2} \\ 0 & 1 & 0 & \frac{1}{6} & - \frac{1}{3} & \frac{1}{2} \\ 0 & 0 & 1 & - \frac{1}{2} & 0 & \frac{1}{2} \end{array}]$

[\begin{array}{c} 1 & 0 & 0 & \frac{1}{6} & \frac{2}{3} & - \frac{1}{2} \\ 0 & 1 & 0 & \frac{1}{6} & - \frac{1}{3} & \frac{1}{2} \\ 0 & 0 & 1 & - \frac{1}{2} & 0 & \frac{1}{2} \end{array}]

$[\begin{array}{c} 1 & 0 & 0 & \frac{1}{6} & \frac{2}{3} & - \frac{1}{2} \\ 0 & 1 & 0 & \frac{1}{6} & - \frac{1}{3} & \frac{1}{2} \\ 0 & 0 & 1 & - \frac{1}{2} & 0 & \frac{1}{2} \end{array}]$

Schritt 2.5

The right half of the reduced row echelon form is the inverse.

[\begin{array}{c} \frac{1}{6} & \frac{2}{3} & - \frac{1}{2} \\ \frac{1}{6} & - \frac{1}{3} & \frac{1}{2} \\ - \frac{1}{2} & 0 & \frac{1}{2} \end{array}]

$[\begin{array}{c} \frac{1}{6} & \frac{2}{3} & - \frac{1}{2} \\ \frac{1}{6} & - \frac{1}{3} & \frac{1}{2} \\ - \frac{1}{2} & 0 & \frac{1}{2} \end{array}]$

[\begin{array}{c} \frac{1}{6} & \frac{2}{3} & - \frac{1}{2} \\ \frac{1}{6} & - \frac{1}{3} & \frac{1}{2} \\ - \frac{1}{2} & 0 & \frac{1}{2} \end{array}]

$[\begin{array}{c} \frac{1}{6} & \frac{2}{3} & - \frac{1}{2} \\ \frac{1}{6} & - \frac{1}{3} & \frac{1}{2} \\ - \frac{1}{2} & 0 & \frac{1}{2} \end{array}]$

Schritt 3

Multipliziere beide Seiten der Matrizengleichung von links mit der inversen Matrix.

([\begin{array}{c} \frac{1}{6} & \frac{2}{3} & - \frac{1}{2} \\ \frac{1}{6} & - \frac{1}{3} & \frac{1}{2} \\ - \frac{1}{2} & 0 & \frac{1}{2} \end{array}] \cdot [\begin{array}{c} 1 & 2 & - 1 \\ 2 & 1 & 1 \\ 1 & 2 & 1 \end{array}]) \cdot [\begin{array}{c} x \\ y \\ z \end{array}] = [\begin{array}{c} \frac{1}{6} & \frac{2}{3} & - \frac{1}{2} \\ \frac{1}{6} & - \frac{1}{3} & \frac{1}{2} \\ - \frac{1}{2} & 0 & \frac{1}{2} \end{array}] \cdot [\begin{array}{c} 4 \\ - 2 \\ 2 \end{array}]

$([\begin{array}{c} \frac{1}{6} & \frac{2}{3} & - \frac{1}{2} \\ \frac{1}{6} & - \frac{1}{3} & \frac{1}{2} \\ - \frac{1}{2} & 0 & \frac{1}{2} \end{array}] \cdot [\begin{array}{c} 1 & 2 & - 1 \\ 2 & 1 & 1 \\ 1 & 2 & 1 \end{array}]) \cdot [\begin{array}{c} x \\ y \\ z \end{array}] = [\begin{array}{c} \frac{1}{6} & \frac{2}{3} & - \frac{1}{2} \\ \frac{1}{6} & - \frac{1}{3} & \frac{1}{2} \\ - \frac{1}{2} & 0 & \frac{1}{2} \end{array}] \cdot [\begin{array}{c} 4 \\ - 2 \\ 2 \end{array}]$

Schritt 4

Jede Matrix multipliziert mit ihrer Inversen ist immer gleich

1

$1$ .

A \cdot A^{- 1} = 1

$A \cdot A^{- 1} = 1$ .

[\begin{array}{c} x \\ y \\ z \end{array}] = [\begin{array}{c} \frac{1}{6} & \frac{2}{3} & - \frac{1}{2} \\ \frac{1}{6} & - \frac{1}{3} & \frac{1}{2} \\ - \frac{1}{2} & 0 & \frac{1}{2} \end{array}] \cdot [\begin{array}{c} 4 \\ - 2 \\ 2 \end{array}]

$[\begin{array}{c} x \\ y \\ z \end{array}] = [\begin{array}{c} \frac{1}{6} & \frac{2}{3} & - \frac{1}{2} \\ \frac{1}{6} & - \frac{1}{3} & \frac{1}{2} \\ - \frac{1}{2} & 0 & \frac{1}{2} \end{array}] \cdot [\begin{array}{c} 4 \\ - 2 \\ 2 \end{array}]$

Schritt 5

Multipliziere

[\begin{array}{c} \frac{1}{6} & \frac{2}{3} & - \frac{1}{2} \\ \frac{1}{6} & - \frac{1}{3} & \frac{1}{2} \\ - \frac{1}{2} & 0 & \frac{1}{2} \end{array}] [\begin{array}{c} 4 \\ - 2 \\ 2 \end{array}]

$[\begin{array}{c} \frac{1}{6} & \frac{2}{3} & - \frac{1}{2} \\ \frac{1}{6} & - \frac{1}{3} & \frac{1}{2} \\ - \frac{1}{2} & 0 & \frac{1}{2} \end{array}] [\begin{array}{c} 4 \\ - 2 \\ 2 \end{array}]$ .

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Schritt 5.1

Two matrices can be multiplied if and only if the number of columns in the first matrix is equal to the number of rows in the second matrix. In this case, the first matrix is

3 \times 3

$3 \times 3$ and the second matrix is

3 \times 1

$3 \times 1$ .

Schritt 5.2

Multipliziere jede Zeile in der ersten Matrix mit jeder Spalte in der zweiten Matrix.

[\begin{array}{c} \frac{1}{6} \cdot 4 + \frac{2}{3} \cdot - 2 - \frac{1}{2} \cdot 2 \\ \frac{1}{6} \cdot 4 - \frac{1}{3} \cdot - 2 + \frac{1}{2} \cdot 2 \\ - \frac{1}{2} \cdot 4 + 0 \cdot - 2 + \frac{1}{2} \cdot 2 \end{array}]

$[\begin{array}{c} \frac{1}{6} \cdot 4 + \frac{2}{3} \cdot - 2 - \frac{1}{2} \cdot 2 \\ \frac{1}{6} \cdot 4 - \frac{1}{3} \cdot - 2 + \frac{1}{2} \cdot 2 \\ - \frac{1}{2} \cdot 4 + 0 \cdot - 2 + \frac{1}{2} \cdot 2 \end{array}]$

Schritt 5.3

Vereinfache jedes Element der Matrix durch Ausmultiplizieren aller Ausdrücke.

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

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

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

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

Schritt 6

Vereinfache die linke und rechte Seite.

[\begin{array}{c} x \\ y \\ z \end{array}] = [\begin{array}{c} - \frac{5}{3} \\ \frac{7}{3} \\ - 1 \end{array}]

$[\begin{array}{c} x \\ y \\ z \end{array}] = [\begin{array}{c} - \frac{5}{3} \\ \frac{7}{3} \\ - 1 \end{array}]$

Schritt 7

Ermittle die Lösung.

x = - \frac{5}{3}

$x = - \frac{5}{3}$

y = \frac{7}{3}

$y = \frac{7}{3}$

z = - 1

$z = - 1$