Finite Mathematik Beispiele

$2 x + 3 y + 54 z = 295$ , $x + y + z = 100$ , $12 x + 16 y + 17 y = 1526$

Schritt 1

Addiere $16 y$ und $17 y$ .

$2 x + 3 y + 54 z = 295$

$x + y + z = 100$

$12 x + 33 y = 1526$

Schritt 2

Write the system as a matrix.

$[\begin{array}{c} 2 & 3 & 54 & 295 \\ 1 & 1 & 1 & 100 \\ 12 & 33 & 0 & 1526 \end{array}]$

Schritt 3

Ermittele die normierte Zeilenstufenform.

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Schritt 3.1

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

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Schritt 3.1.1

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

$[\begin{array}{c} \frac{2}{2} & \frac{3}{2} & \frac{54}{2} & \frac{295}{2} \\ 1 & 1 & 1 & 100 \\ 12 & 33 & 0 & 1526 \end{array}]$

Schritt 3.1.2

Vereinfache $R_{1}$ .

$[\begin{array}{c} 1 & \frac{3}{2} & 27 & \frac{295}{2} \\ 1 & 1 & 1 & 100 \\ 12 & 33 & 0 & 1526 \end{array}]$

Schritt 3.2

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

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Schritt 3.2.1

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

$[\begin{array}{c} 1 & \frac{3}{2} & 27 & \frac{295}{2} \\ 1 - 1 & 1 - \frac{3}{2} & 1 - 27 & 100 - \frac{295}{2} \\ 12 & 33 & 0 & 1526 \end{array}]$

Schritt 3.2.2

Vereinfache $R_{2}$ .

$[\begin{array}{c} 1 & \frac{3}{2} & 27 & \frac{295}{2} \\ 0 & - \frac{1}{2} & - 26 & - \frac{95}{2} \\ 12 & 33 & 0 & 1526 \end{array}]$

Schritt 3.3

Perform the row operation $R_{3} = R_{3} - 12 R_{1}$ to make the entry at $3, 1$ a $0$ .

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Schritt 3.3.1

Perform the row operation $R_{3} = R_{3} - 12 R_{1}$ to make the entry at $3, 1$ a $0$ .

$[\begin{array}{c} 1 & \frac{3}{2} & 27 & \frac{295}{2} \\ 0 & - \frac{1}{2} & - 26 & - \frac{95}{2} \\ 12 - 12 \cdot 1 & 33 - 12 (\frac{3}{2}) & 0 - 12 \cdot 27 & 1526 - 12 (\frac{295}{2}) \end{array}]$

Schritt 3.3.2

Vereinfache $R_{3}$ .

$[\begin{array}{c} 1 & \frac{3}{2} & 27 & \frac{295}{2} \\ 0 & - \frac{1}{2} & - 26 & - \frac{95}{2} \\ 0 & 15 & - 324 & - 244 \end{array}]$

Schritt 3.4

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

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Schritt 3.4.1

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

$[\begin{array}{c} 1 & \frac{3}{2} & 27 & \frac{295}{2} \\ - 2 \cdot 0 & - 2 (- \frac{1}{2}) & - 2 \cdot - 26 & - 2 (- \frac{95}{2}) \\ 0 & 15 & - 324 & - 244 \end{array}]$

Schritt 3.4.2

Vereinfache $R_{2}$ .

$[\begin{array}{c} 1 & \frac{3}{2} & 27 & \frac{295}{2} \\ 0 & 1 & 52 & 95 \\ 0 & 15 & - 324 & - 244 \end{array}]$

Schritt 3.5

Perform the row operation $R_{3} = R_{3} - 15 R_{2}$ to make the entry at $3, 2$ a $0$ .

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Schritt 3.5.1

Perform the row operation $R_{3} = R_{3} - 15 R_{2}$ to make the entry at $3, 2$ a $0$ .

$[\begin{array}{c} 1 & \frac{3}{2} & 27 & \frac{295}{2} \\ 0 & 1 & 52 & 95 \\ 0 - 15 \cdot 0 & 15 - 15 \cdot 1 & - 324 - 15 \cdot 52 & - 244 - 15 \cdot 95 \end{array}]$

Schritt 3.5.2

Vereinfache $R_{3}$ .

$[\begin{array}{c} 1 & \frac{3}{2} & 27 & \frac{295}{2} \\ 0 & 1 & 52 & 95 \\ 0 & 0 & - 1104 & - 1669 \end{array}]$

Schritt 3.6

Multiply each element of $R_{3}$ by $- \frac{1}{1104}$ to make the entry at $3, 3$ a $1$ .

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Schritt 3.6.1

Multiply each element of $R_{3}$ by $- \frac{1}{1104}$ to make the entry at $3, 3$ a $1$ .

$[\begin{array}{c} 1 & \frac{3}{2} & 27 & \frac{295}{2} \\ 0 & 1 & 52 & 95 \\ - \frac{1}{1104} \cdot 0 & - \frac{1}{1104} \cdot 0 & - \frac{1}{1104} \cdot - 1104 & - \frac{1}{1104} \cdot - 1669 \end{array}]$

Schritt 3.6.2

Vereinfache $R_{3}$ .

$[\begin{array}{c} 1 & \frac{3}{2} & 27 & \frac{295}{2} \\ 0 & 1 & 52 & 95 \\ 0 & 0 & 1 & \frac{1669}{1104} \end{array}]$

Schritt 3.7

Perform the row operation $R_{2} = R_{2} - 52 R_{3}$ to make the entry at $2, 3$ a $0$ .

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Schritt 3.7.1

Perform the row operation $R_{2} = R_{2} - 52 R_{3}$ to make the entry at $2, 3$ a $0$ .

$[\begin{array}{c} 1 & \frac{3}{2} & 27 & \frac{295}{2} \\ 0 - 52 \cdot 0 & 1 - 52 \cdot 0 & 52 - 52 \cdot 1 & 95 - 52 (\frac{1669}{1104}) \\ 0 & 0 & 1 & \frac{1669}{1104} \end{array}]$

Schritt 3.7.2

Vereinfache $R_{2}$ .

$[\begin{array}{c} 1 & \frac{3}{2} & 27 & \frac{295}{2} \\ 0 & 1 & 0 & \frac{4523}{276} \\ 0 & 0 & 1 & \frac{1669}{1104} \end{array}]$

Schritt 3.8

Perform the row operation $R_{1} = R_{1} - 27 R_{3}$ to make the entry at $1, 3$ a $0$ .

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Schritt 3.8.1

Perform the row operation $R_{1} = R_{1} - 27 R_{3}$ to make the entry at $1, 3$ a $0$ .

$[\begin{array}{c} 1 - 27 \cdot 0 & \frac{3}{2} - 27 \cdot 0 & 27 - 27 \cdot 1 & \frac{295}{2} - 27 (\frac{1669}{1104}) \\ 0 & 1 & 0 & \frac{4523}{276} \\ 0 & 0 & 1 & \frac{1669}{1104} \end{array}]$

Schritt 3.8.2

Vereinfache $R_{1}$ .

$[\begin{array}{c} 1 & \frac{3}{2} & 0 & \frac{39259}{368} \\ 0 & 1 & 0 & \frac{4523}{276} \\ 0 & 0 & 1 & \frac{1669}{1104} \end{array}]$

Schritt 3.9

Perform the row operation $R_{1} = R_{1} - \frac{3}{2} R_{2}$ to make the entry at $1, 2$ a $0$ .

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Schritt 3.9.1

Perform the row operation $R_{1} = R_{1} - \frac{3}{2} R_{2}$ to make the entry at $1, 2$ a $0$ .

$[\begin{array}{c} 1 - \frac{3}{2} \cdot 0 & \frac{3}{2} - \frac{3}{2} \cdot 1 & 0 - \frac{3}{2} \cdot 0 & \frac{39259}{368} - \frac{3}{2} \cdot \frac{4523}{276} \\ 0 & 1 & 0 & \frac{4523}{276} \\ 0 & 0 & 1 & \frac{1669}{1104} \end{array}]$

Schritt 3.9.2

Vereinfache $R_{1}$ .

$[\begin{array}{c} 1 & 0 & 0 & \frac{30213}{368} \\ 0 & 1 & 0 & \frac{4523}{276} \\ 0 & 0 & 1 & \frac{1669}{1104} \end{array}]$

Schritt 4

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

$x = \frac{30213}{368}$

$y = \frac{4523}{276}$

$z = \frac{1669}{1104}$

Schritt 5

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

$(\frac{30213}{368}, \frac{4523}{276}, \frac{1669}{1104})$