Finite Mathematik Beispiele

$(x) = 42 x$ , $x = 600 + 40 x$

Schritt 1

Move variables to the left and constant terms to the right.

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Schritt 1.1

Subtrahiere $42 x$ von beiden Seiten der Gleichung.

$(x) - 42 x = 0$

$x = 600 + 40 x$

Schritt 1.2

Subtrahiere $42 x$ von $x$ .

$- 41 x = 0$

$x = 600 + 40 x$

Schritt 1.3

Subtrahiere $40 x$ von beiden Seiten der Gleichung.

$- 41 x = 0$

$x - 40 x = 600$

Schritt 1.4

Subtrahiere $40 x$ von $x$ .

$- 41 x = 0$

$- 39 x = 600$

$- 41 x = 0$

$- 39 x = 600$

Schritt 2

Write the system as a matrix.

$[\begin{array}{c} - 41 & 0 \\ - 39 & 600 \end{array}]$

Schritt 3

Ermittele die normierte Zeilenstufenform.

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Schritt 3.1

Multiply each element of $R_{1}$ by $- \frac{1}{41}$ to make the entry at $1, 1$ a $1$ .

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Schritt 3.1.1

Multiply each element of $R_{1}$ by $- \frac{1}{41}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} - \frac{1}{41} \cdot - 41 & - \frac{1}{41} \cdot 0 \\ - 39 & 600 \end{array}]$

Schritt 3.1.2

Vereinfache $R_{1}$ .

$[\begin{array}{c} 1 & 0 \\ - 39 & 600 \end{array}]$

Schritt 3.2

Perform the row operation $R_{2} = R_{2} + 39 R_{1}$ to make the entry at $2, 1$ a $0$ .

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Schritt 3.2.1

Perform the row operation $R_{2} = R_{2} + 39 R_{1}$ to make the entry at $2, 1$ a $0$ .

$[\begin{array}{c} 1 & 0 \\ - 39 + 39 \cdot 1 & 600 + 39 \cdot 0 \end{array}]$

Schritt 3.2.2

Vereinfache $R_{2}$ .

$[\begin{array}{c} 1 & 0 \\ 0 & 600 \end{array}]$

Schritt 3.3

Multiply each element of $R_{2}$ by $\frac{1}{600}$ to make the entry at $2, 2$ a $1$ .

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Schritt 3.3.1

Multiply each element of $R_{2}$ by $\frac{1}{600}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & 0 \\ \frac{0}{600} & \frac{600}{600} \end{array}]$

Schritt 3.3.2

Vereinfache $R_{2}$ .

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

Schritt 4

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

$x = 0$

$0 = 1$

Schritt 5

The system is inconsistent so there is no solution.

$Keine Lösung$