Finite Mathematik Beispiele

$x - y + z = 9$ , $6 x + 3 y - 32 = 0$

Schritt 1

Addiere $32$ zu beiden Seiten der Gleichung.

$x - y + z = 9$

$6 x + 3 y = 32$

Schritt 2

Write the system as a matrix.

$[\begin{array}{c} 1 & - 1 & 1 & 9 \\ 6 & 3 & 0 & 32 \end{array}]$

Schritt 3

Ermittele die normierte Zeilenstufenform.

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Schritt 3.1

Perform the row operation $R_{2} = R_{2} - 6 R_{1}$ to make the entry at $2, 1$ a $0$ .

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Schritt 3.1.1

Perform the row operation $R_{2} = R_{2} - 6 R_{1}$ to make the entry at $2, 1$ a $0$ .

$[\begin{array}{c} 1 & - 1 & 1 & 9 \\ 6 - 6 \cdot 1 & 3 - 6 \cdot - 1 & 0 - 6 \cdot 1 & 32 - 6 \cdot 9 \end{array}]$

Schritt 3.1.2

Vereinfache $R_{2}$ .

$[\begin{array}{c} 1 & - 1 & 1 & 9 \\ 0 & 9 & - 6 & - 22 \end{array}]$

Schritt 3.2

Multiply each element of $R_{2}$ by $\frac{1}{9}$ to make the entry at $2, 2$ a $1$ .

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Schritt 3.2.1

Multiply each element of $R_{2}$ by $\frac{1}{9}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & - 1 & 1 & 9 \\ \frac{0}{9} & \frac{9}{9} & \frac{- 6}{9} & \frac{- 22}{9} \end{array}]$

Schritt 3.2.2

Vereinfache $R_{2}$ .

$[\begin{array}{c} 1 & - 1 & 1 & 9 \\ 0 & 1 & - \frac{2}{3} & - \frac{22}{9} \end{array}]$

Schritt 3.3

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

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Schritt 3.3.1

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

$[\begin{array}{c} 1 + 0 & - 1 + 1 \cdot 1 & 1 - \frac{2}{3} & 9 - \frac{22}{9} \\ 0 & 1 & - \frac{2}{3} & - \frac{22}{9} \end{array}]$

Schritt 3.3.2

Vereinfache $R_{1}$ .

$[\begin{array}{c} 1 & 0 & \frac{1}{3} & \frac{59}{9} \\ 0 & 1 & - \frac{2}{3} & - \frac{22}{9} \end{array}]$

Schritt 4

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

$x + \frac{1}{3} z = \frac{59}{9}$

$y - \frac{2}{3} z = - \frac{22}{9}$

Schritt 5

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

$(\frac{59}{9} - \frac{z}{3}, - \frac{22}{9} + \frac{2 z}{3}, z)$