Finite Mathematik Beispiele

$7 x - 2 y + 8 z = - 9$ , $4 x - 3 y + 6 z = 1$ , $y - x = - 7$

Schritt 1

Stelle $y$ und $- x$ um.

$7 x - 2 y + 8 z = - 9$

$4 x - 3 y + 6 z = 1$

$- x + y = - 7$

Schritt 2

Write the system as a matrix.

$[\begin{array}{c} 7 & - 2 & 8 & - 9 \\ 4 & - 3 & 6 & 1 \\ - 1 & 1 & 0 & - 7 \end{array}]$

Schritt 3

Ermittele die normierte Zeilenstufenform.

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Schritt 3.1

Multiply each element of $R_{1}$ by $\frac{1}{7}$ to make the entry at $1, 1$ a $1$ .

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Schritt 3.1.1

Multiply each element of $R_{1}$ by $\frac{1}{7}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} \frac{7}{7} & \frac{- 2}{7} & \frac{8}{7} & \frac{- 9}{7} \\ 4 & - 3 & 6 & 1 \\ - 1 & 1 & 0 & - 7 \end{array}]$

Schritt 3.1.2

Vereinfache $R_{1}$ .

$[\begin{array}{c} 1 & - \frac{2}{7} & \frac{8}{7} & - \frac{9}{7} \\ 4 & - 3 & 6 & 1 \\ - 1 & 1 & 0 & - 7 \end{array}]$

Schritt 3.2

Perform the row operation $R_{2} = R_{2} - 4 R_{1}$ to make the entry at $2, 1$ a $0$ .

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Schritt 3.2.1

Perform the row operation $R_{2} = R_{2} - 4 R_{1}$ to make the entry at $2, 1$ a $0$ .

$[\begin{array}{c} 1 & - \frac{2}{7} & \frac{8}{7} & - \frac{9}{7} \\ 4 - 4 \cdot 1 & - 3 - 4 (- \frac{2}{7}) & 6 - 4 (\frac{8}{7}) & 1 - 4 (- \frac{9}{7}) \\ - 1 & 1 & 0 & - 7 \end{array}]$

Schritt 3.2.2

Vereinfache $R_{2}$ .

$[\begin{array}{c} 1 & - \frac{2}{7} & \frac{8}{7} & - \frac{9}{7} \\ 0 & - \frac{13}{7} & \frac{10}{7} & \frac{43}{7} \\ - 1 & 1 & 0 & - 7 \end{array}]$

Schritt 3.3

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

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Schritt 3.3.1

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

$[\begin{array}{c} 1 & - \frac{2}{7} & \frac{8}{7} & - \frac{9}{7} \\ 0 & - \frac{13}{7} & \frac{10}{7} & \frac{43}{7} \\ - 1 + 1 \cdot 1 & 1 - \frac{2}{7} & 0 + \frac{8}{7} & - 7 - \frac{9}{7} \end{array}]$

Schritt 3.3.2

Vereinfache $R_{3}$ .

$[\begin{array}{c} 1 & - \frac{2}{7} & \frac{8}{7} & - \frac{9}{7} \\ 0 & - \frac{13}{7} & \frac{10}{7} & \frac{43}{7} \\ 0 & \frac{5}{7} & \frac{8}{7} & - \frac{58}{7} \end{array}]$

Schritt 3.4

Multiply each element of $R_{2}$ by $- \frac{7}{13}$ to make the entry at $2, 2$ a $1$ .

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Schritt 3.4.1

Multiply each element of $R_{2}$ by $- \frac{7}{13}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & - \frac{2}{7} & \frac{8}{7} & - \frac{9}{7} \\ - \frac{7}{13} \cdot 0 & - \frac{7}{13} (- \frac{13}{7}) & - \frac{7}{13} \cdot \frac{10}{7} & - \frac{7}{13} \cdot \frac{43}{7} \\ 0 & \frac{5}{7} & \frac{8}{7} & - \frac{58}{7} \end{array}]$

Schritt 3.4.2

Vereinfache $R_{2}$ .

$[\begin{array}{c} 1 & - \frac{2}{7} & \frac{8}{7} & - \frac{9}{7} \\ 0 & 1 & - \frac{10}{13} & - \frac{43}{13} \\ 0 & \frac{5}{7} & \frac{8}{7} & - \frac{58}{7} \end{array}]$

Schritt 3.5

Perform the row operation $R_{3} = R_{3} - \frac{5}{7} R_{2}$ to make the entry at $3, 2$ a $0$ .

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Schritt 3.5.1

Perform the row operation $R_{3} = R_{3} - \frac{5}{7} R_{2}$ to make the entry at $3, 2$ a $0$ .

$[\begin{array}{c} 1 & - \frac{2}{7} & \frac{8}{7} & - \frac{9}{7} \\ 0 & 1 & - \frac{10}{13} & - \frac{43}{13} \\ 0 - \frac{5}{7} \cdot 0 & \frac{5}{7} - \frac{5}{7} \cdot 1 & \frac{8}{7} - \frac{5}{7} (- \frac{10}{13}) & - \frac{58}{7} - \frac{5}{7} (- \frac{43}{13}) \end{array}]$

Schritt 3.5.2

Vereinfache $R_{3}$ .

$[\begin{array}{c} 1 & - \frac{2}{7} & \frac{8}{7} & - \frac{9}{7} \\ 0 & 1 & - \frac{10}{13} & - \frac{43}{13} \\ 0 & 0 & \frac{22}{13} & - \frac{77}{13} \end{array}]$

Schritt 3.6

Multiply each element of $R_{3}$ by $\frac{13}{22}$ to make the entry at $3, 3$ a $1$ .

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Schritt 3.6.1

Multiply each element of $R_{3}$ by $\frac{13}{22}$ to make the entry at $3, 3$ a $1$ .

$[\begin{array}{c} 1 & - \frac{2}{7} & \frac{8}{7} & - \frac{9}{7} \\ 0 & 1 & - \frac{10}{13} & - \frac{43}{13} \\ \frac{13}{22} \cdot 0 & \frac{13}{22} \cdot 0 & \frac{13}{22} \cdot \frac{22}{13} & \frac{13}{22} (- \frac{77}{13}) \end{array}]$

Schritt 3.6.2

Vereinfache $R_{3}$ .

$[\begin{array}{c} 1 & - \frac{2}{7} & \frac{8}{7} & - \frac{9}{7} \\ 0 & 1 & - \frac{10}{13} & - \frac{43}{13} \\ 0 & 0 & 1 & - \frac{7}{2} \end{array}]$

Schritt 3.7

Perform the row operation $R_{2} = R_{2} + \frac{10}{13} R_{3}$ to make the entry at $2, 3$ a $0$ .

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Schritt 3.7.1

Perform the row operation $R_{2} = R_{2} + \frac{10}{13} R_{3}$ to make the entry at $2, 3$ a $0$ .

$[\begin{array}{c} 1 & - \frac{2}{7} & \frac{8}{7} & - \frac{9}{7} \\ 0 + \frac{10}{13} \cdot 0 & 1 + \frac{10}{13} \cdot 0 & - \frac{10}{13} + \frac{10}{13} \cdot 1 & - \frac{43}{13} + \frac{10}{13} (- \frac{7}{2}) \\ 0 & 0 & 1 & - \frac{7}{2} \end{array}]$

Schritt 3.7.2

Vereinfache $R_{2}$ .

$[\begin{array}{c} 1 & - \frac{2}{7} & \frac{8}{7} & - \frac{9}{7} \\ 0 & 1 & 0 & - 6 \\ 0 & 0 & 1 & - \frac{7}{2} \end{array}]$

Schritt 3.8

Perform the row operation $R_{1} = R_{1} - \frac{8}{7} R_{3}$ to make the entry at $1, 3$ a $0$ .

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Schritt 3.8.1

Perform the row operation $R_{1} = R_{1} - \frac{8}{7} R_{3}$ to make the entry at $1, 3$ a $0$ .

$[\begin{array}{c} 1 - \frac{8}{7} \cdot 0 & - \frac{2}{7} - \frac{8}{7} \cdot 0 & \frac{8}{7} - \frac{8}{7} \cdot 1 & - \frac{9}{7} - \frac{8}{7} (- \frac{7}{2}) \\ 0 & 1 & 0 & - 6 \\ 0 & 0 & 1 & - \frac{7}{2} \end{array}]$

Schritt 3.8.2

Vereinfache $R_{1}$ .

$[\begin{array}{c} 1 & - \frac{2}{7} & 0 & \frac{19}{7} \\ 0 & 1 & 0 & - 6 \\ 0 & 0 & 1 & - \frac{7}{2} \end{array}]$

Schritt 3.9

Perform the row operation $R_{1} = R_{1} + \frac{2}{7} R_{2}$ to make the entry at $1, 2$ a $0$ .

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Schritt 3.9.1

Perform the row operation $R_{1} = R_{1} + \frac{2}{7} R_{2}$ to make the entry at $1, 2$ a $0$ .

$[\begin{array}{c} 1 + \frac{2}{7} \cdot 0 & - \frac{2}{7} + \frac{2}{7} \cdot 1 & 0 + \frac{2}{7} \cdot 0 & \frac{19}{7} + \frac{2}{7} \cdot - 6 \\ 0 & 1 & 0 & - 6 \\ 0 & 0 & 1 & - \frac{7}{2} \end{array}]$

Schritt 3.9.2

Vereinfache $R_{1}$ .

$[\begin{array}{c} 1 & 0 & 0 & 1 \\ 0 & 1 & 0 & - 6 \\ 0 & 0 & 1 & - \frac{7}{2} \end{array}]$

Schritt 4

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

$x = 1$

$y = - 6$

$z = - \frac{7}{2}$

Schritt 5

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

$(1, - 6, - \frac{7}{2})$