Finite Mathematik Beispiele

$x - y - z = 1$ , $- x + 2 y - 3 z = - 7$ , $5 x - 3 y - 13 z = - 7$

Schritt 1

Write the system as a matrix.

$[\begin{array}{c} 1 & - 1 & - 1 & 1 \\ - 1 & 2 & - 3 & - 7 \\ 5 & - 3 & - 13 & - 7 \end{array}]$

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 $0$ .

$[\begin{array}{c} 1 & - 1 & - 1 & 1 \\ - 1 + 1 \cdot 1 & 2 - 1 & - 3 - 1 & - 7 + 1 \cdot 1 \\ 5 & - 3 & - 13 & - 7 \end{array}]$

Schritt 2.1.2

Vereinfache $R_{2}$ .

$[\begin{array}{c} 1 & - 1 & - 1 & 1 \\ 0 & 1 & - 4 & - 6 \\ 5 & - 3 & - 13 & - 7 \end{array}]$

Schritt 2.2

Perform the row operation $R_{3} = R_{3} - 5 R_{1}$ to make the entry at $3, 1$ a $0$ .

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Schritt 2.2.1

Perform the row operation $R_{3} = R_{3} - 5 R_{1}$ to make the entry at $3, 1$ a $0$ .

$[\begin{array}{c} 1 & - 1 & - 1 & 1 \\ 0 & 1 & - 4 & - 6 \\ 5 - 5 \cdot 1 & - 3 - 5 \cdot - 1 & - 13 - 5 \cdot - 1 & - 7 - 5 \cdot 1 \end{array}]$

Schritt 2.2.2

Vereinfache $R_{3}$ .

$[\begin{array}{c} 1 & - 1 & - 1 & 1 \\ 0 & 1 & - 4 & - 6 \\ 0 & 2 & - 8 & - 12 \end{array}]$

Schritt 2.3

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

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Schritt 2.3.1

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

$[\begin{array}{c} 1 & - 1 & - 1 & 1 \\ 0 & 1 & - 4 & - 6 \\ 0 - 2 \cdot 0 & 2 - 2 \cdot 1 & - 8 - 2 \cdot - 4 & - 12 - 2 \cdot - 6 \end{array}]$

Schritt 2.3.2

Vereinfache $R_{3}$ .

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

Schritt 2.4

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

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Schritt 2.4.1

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

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

Schritt 2.4.2

Vereinfache $R_{1}$ .

$[\begin{array}{c} 1 & 0 & - 5 & - 5 \\ 0 & 1 & - 4 & - 6 \\ 0 & 0 & 0 & 0 \end{array}]$

Schritt 3

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

$x - 5 z = - 5$

$y - 4 z = - 6$

$0 = 0$

Schritt 4

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

$(- 5 + 5 z, - 6 + 4 z, z)$