Elementarmathematik Beispiele

$x + y - z = 3$ , $x - y + 5 z = - 1$ , $x + 2 y - 3 z = 6$

Schritt 1

Write the system as a matrix.

$[\begin{array}{c} 1 & 1 & - 1 & 3 \\ 1 & - 1 & 5 & - 1 \\ 1 & 2 & - 3 & 6 \end{array}]$

Schritt 2

Ermittele die normierte Zeilenstufenform.

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Schritt 2.1

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

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Schritt 2.1.1

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

$[\begin{array}{c} 1 & 1 & - 1 & 3 \\ 1 - 1 & - 1 - 1 & 5 + 1 & - 1 - 3 \\ 1 & 2 & - 3 & 6 \end{array}]$

Schritt 2.1.2

Vereinfache $R_{2}$ .

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

Schritt 2.2

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

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Schritt 2.2.1

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

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

Schritt 2.2.2

Vereinfache $R_{3}$ .

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

Schritt 2.3

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

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Schritt 2.3.1

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

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

Schritt 2.3.2

Vereinfache $R_{2}$ .

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

Schritt 2.4

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

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Schritt 2.4.1

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

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

Schritt 2.4.2

Vereinfache $R_{3}$ .

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

Schritt 2.5

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

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Schritt 2.5.1

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

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

Schritt 2.5.2

Vereinfache $R_{2}$ .

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

Schritt 2.6

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

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Schritt 2.6.1

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

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

Schritt 2.6.2

Vereinfache $R_{1}$ .

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

Schritt 2.7

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

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Schritt 2.7.1

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

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

Schritt 2.7.2

Vereinfache $R_{1}$ .

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

Schritt 3

Use the result matrix to declare the final solution to the system of equations.

$x = - 1$

$y = 5$

$z = 1$

Schritt 4

The solution is the set of ordered pairs that make the system true.

$(- 1, 5, 1)$