Finite Mathematik Beispiele

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

Schritt 1

Bringe alle Terme, die keine Variable enthalten, auf die rechte Seite der Gleichung.

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Schritt 1.1

Subtrahiere $3$ von beiden Seiten der Gleichung.

$3 x + 2 y - z = 2$

$4 x + z = 5 - 3$

Schritt 1.2

Subtrahiere $3$ von $5$ .

$3 x + 2 y - z = 2$

$4 x + z = 2$

$3 x + 2 y - z = 2$

$4 x + z = 2$

Schritt 2

Write the system as a matrix.

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

Schritt 3

Ermittele die normierte Zeilenstufenform.

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Schritt 3.1

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

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Schritt 3.1.1

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

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

Schritt 3.1.2

Vereinfache $R_{1}$ .

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

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 $0$ .

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

Schritt 3.2.2

Vereinfache $R_{2}$ .

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

Schritt 3.3

Multiply each element of $R_{2}$ by $- \frac{3}{8}$ to make the entry at $2, 2$ a $1$ .

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Schritt 3.3.1

Multiply each element of $R_{2}$ by $- \frac{3}{8}$ to make the entry at $2, 2$ a $1$ .

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

Schritt 3.3.2

Vereinfache $R_{2}$ .

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

Schritt 3.4

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

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Schritt 3.4.1

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

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

Schritt 3.4.2

Vereinfache $R_{1}$ .

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

Schritt 4

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

$x + \frac{1}{4} z = \frac{1}{2}$

$y - \frac{7}{8} z = \frac{1}{4}$

Schritt 5

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

$(\frac{1}{2} - \frac{z}{4}, \frac{1}{4} + \frac{7 z}{8}, z)$