Finite Mathematik Beispiele

$x - 2 y + x = 0$ , $y - 3 z = - 1$ , $2 y + 5 z = - 2$

Schritt 1

Addiere $x$ und $x$ .

$2 x - 2 y = 0$

$y - 3 z = - 1$

$2 y + 5 z = - 2$

Schritt 2

Write the system as a matrix.

$[\begin{array}{c} 2 & - 2 & 0 & 0 \\ 0 & 1 & - 3 & - 1 \\ 0 & 2 & 5 & - 2 \end{array}]$

Schritt 3

Ermittele die normierte Zeilenstufenform.

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Schritt 3.1

Multiply each element of $R_{1}$ by $\frac{1}{2}$ 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 $1$ .

$[\begin{array}{c} \frac{2}{2} & \frac{- 2}{2} & \frac{0}{2} & \frac{0}{2} \\ 0 & 1 & - 3 & - 1 \\ 0 & 2 & 5 & - 2 \end{array}]$

Schritt 3.1.2

Vereinfache $R_{1}$ .

$[\begin{array}{c} 1 & - 1 & 0 & 0 \\ 0 & 1 & - 3 & - 1 \\ 0 & 2 & 5 & - 2 \end{array}]$

Schritt 3.2

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

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Schritt 3.2.1

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

$[\begin{array}{c} 1 & - 1 & 0 & 0 \\ 0 & 1 & - 3 & - 1 \\ 0 - 2 \cdot 0 & 2 - 2 \cdot 1 & 5 - 2 \cdot - 3 & - 2 - 2 \cdot - 1 \end{array}]$

Schritt 3.2.2

Vereinfache $R_{3}$ .

$[\begin{array}{c} 1 & - 1 & 0 & 0 \\ 0 & 1 & - 3 & - 1 \\ 0 & 0 & 11 & 0 \end{array}]$

Schritt 3.3

Multiply each element of $R_{3}$ by $\frac{1}{11}$ to make the entry at $3, 3$ a $1$ .

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Schritt 3.3.1

Multiply each element of $R_{3}$ by $\frac{1}{11}$ to make the entry at $3, 3$ a $1$ .

$[\begin{array}{c} 1 & - 1 & 0 & 0 \\ 0 & 1 & - 3 & - 1 \\ \frac{0}{11} & \frac{0}{11} & \frac{11}{11} & \frac{0}{11} \end{array}]$

Schritt 3.3.2

Vereinfache $R_{3}$ .

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

Schritt 3.4

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

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Schritt 3.4.1

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

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

Schritt 3.4.2

Vereinfache $R_{2}$ .

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

Schritt 3.5

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

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Schritt 3.5.1

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

$[\begin{array}{c} 1 + 0 & - 1 + 1 \cdot 1 & 0 + 0 & 0 - 1 \\ 0 & 1 & 0 & - 1 \\ 0 & 0 & 1 & 0 \end{array}]$

Schritt 3.5.2

Vereinfache $R_{1}$ .

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

Schritt 4

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

$x = - 1$

$y = - 1$

$z = 0$

Schritt 5

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

$(- 1, - 1, 0)$