Finite Mathematik Beispiele

$5 x - 2 y = 1$ , $3 x + y = - 4$

Schritt 1

Write the system as a matrix.

$[\begin{array}{c} 5 & - 2 & 1 \\ 3 & 1 & - 4 \end{array}]$

Schritt 2

Ermittele die normierte Zeilenstufenform.

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Schritt 2.1

Multiply each element of $R_{1}$ by $\frac{1}{5}$ to make the entry at $1, 1$ a $1$ .

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Schritt 2.1.1

Multiply each element of $R_{1}$ by $\frac{1}{5}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} \frac{5}{5} & \frac{- 2}{5} & \frac{1}{5} \\ 3 & 1 & - 4 \end{array}]$

Schritt 2.1.2

Vereinfache $R_{1}$ .

$[\begin{array}{c} 1 & - \frac{2}{5} & \frac{1}{5} \\ 3 & 1 & - 4 \end{array}]$

Schritt 2.2

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

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Schritt 2.2.1

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

$[\begin{array}{c} 1 & - \frac{2}{5} & \frac{1}{5} \\ 3 - 3 \cdot 1 & 1 - 3 (- \frac{2}{5}) & - 4 - 3 (\frac{1}{5}) \end{array}]$

Schritt 2.2.2

Vereinfache $R_{2}$ .

$[\begin{array}{c} 1 & - \frac{2}{5} & \frac{1}{5} \\ 0 & \frac{11}{5} & - \frac{23}{5} \end{array}]$

Schritt 2.3

Multiply each element of $R_{2}$ by $\frac{5}{11}$ to make the entry at $2, 2$ a $1$ .

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Schritt 2.3.1

Multiply each element of $R_{2}$ by $\frac{5}{11}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & - \frac{2}{5} & \frac{1}{5} \\ \frac{5}{11} \cdot 0 & \frac{5}{11} \cdot \frac{11}{5} & \frac{5}{11} (- \frac{23}{5}) \end{array}]$

Schritt 2.3.2

Vereinfache $R_{2}$ .

$[\begin{array}{c} 1 & - \frac{2}{5} & \frac{1}{5} \\ 0 & 1 & - \frac{23}{11} \end{array}]$

Schritt 2.4

Perform the row operation $R_{1} = R_{1} + \frac{2}{5} R_{2}$ to make the entry at $1, 2$ a $0$ .

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Schritt 2.4.1

Perform the row operation $R_{1} = R_{1} + \frac{2}{5} R_{2}$ to make the entry at $1, 2$ a $0$ .

$[\begin{array}{c} 1 + \frac{2}{5} \cdot 0 & - \frac{2}{5} + \frac{2}{5} \cdot 1 & \frac{1}{5} + \frac{2}{5} (- \frac{23}{11}) \\ 0 & 1 & - \frac{23}{11} \end{array}]$

Schritt 2.4.2

Vereinfache $R_{1}$ .

$[\begin{array}{c} 1 & 0 & - \frac{7}{11} \\ 0 & 1 & - \frac{23}{11} \end{array}]$

Schritt 3

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

$x = - \frac{7}{11}$

$y = - \frac{23}{11}$

Schritt 4

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

$(- \frac{7}{11}, - \frac{23}{11})$