Lineare Algebra Beispiele

y = 4 x + 3 x - 2

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

y = 6

$y = 6$

Schritt 1

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

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Schritt 1.1

Bringe alle Terme, die Variablen enthalten, auf die linke Seite der Gleichung.

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Schritt 1.1.1

Subtrahiere

4 x

$4 x$ von beiden Seiten der Gleichung.

y - 4 x = 3 x - 2

$y - 4 x = 3 x - 2$

y = 6

$y = 6$

Schritt 1.1.2

Subtrahiere

3 x

$3 x$ von beiden Seiten der Gleichung.

y - 4 x - 3 x = - 2

$y - 4 x - 3 x = - 2$

y = 6

$y = 6$

y - 4 x - 3 x = - 2

$y - 4 x - 3 x = - 2$

y = 6

$y = 6$

Schritt 1.2

Subtrahiere

3 x

$3 x$ von

- 4 x

$- 4 x$ .

y - 7 x = - 2

$y - 7 x = - 2$

y = 6

$y = 6$

Schritt 1.3

Stelle

y

$y$ und

- 7 x

$- 7 x$ um.

- 7 x + y = - 2

$- 7 x + y = - 2$

y = 6

$y = 6$

- 7 x + y = - 2

$- 7 x + y = - 2$

y = 6

$y = 6$

Schritt 2

Write the system as a matrix.

[\begin{array}{c} - 7 & 1 & - 2 \\ 0 & 1 & 6 \end{array}]

$[\begin{array}{c} - 7 & 1 & - 2 \\ 0 & 1 & 6 \end{array}]$

Schritt 3

Ermittele die normierte Zeilenstufenform.

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Schritt 3.1

Multiply each element of

R_{1}

$R_{1}$ by

- \frac{1}{7}

$- \frac{1}{7}$ to make the entry at

1, 1

$1, 1$ a

1

$1$ .

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Schritt 3.1.1

Multiply each element of

R_{1}

$R_{1}$ by

- \frac{1}{7}

$- \frac{1}{7}$ to make the entry at

1, 1

$1, 1$ a

1

$1$ .

[\begin{array}{c} - \frac{1}{7} \cdot - 7 & - \frac{1}{7} \cdot 1 & - \frac{1}{7} \cdot - 2 \\ 0 & 1 & 6 \end{array}]

$[\begin{array}{c} - \frac{1}{7} \cdot - 7 & - \frac{1}{7} \cdot 1 & - \frac{1}{7} \cdot - 2 \\ 0 & 1 & 6 \end{array}]$

Schritt 3.1.2

Vereinfache

R_{1}

$R_{1}$ .

[\begin{array}{c} 1 & - \frac{1}{7} & \frac{2}{7} \\ 0 & 1 & 6 \end{array}]

$[\begin{array}{c} 1 & - \frac{1}{7} & \frac{2}{7} \\ 0 & 1 & 6 \end{array}]$

[\begin{array}{c} 1 & - \frac{1}{7} & \frac{2}{7} \\ 0 & 1 & 6 \end{array}]

$[\begin{array}{c} 1 & - \frac{1}{7} & \frac{2}{7} \\ 0 & 1 & 6 \end{array}]$

Schritt 3.2

Perform the row operation

R_{1} = R_{1} + \frac{1}{7} R_{2}

$R_{1} = R_{1} + \frac{1}{7} R_{2}$ to make the entry at

1, 2

$1, 2$ a

0

$0$ .

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Schritt 3.2.1

Perform the row operation

R_{1} = R_{1} + \frac{1}{7} R_{2}

$R_{1} = R_{1} + \frac{1}{7} R_{2}$ to make the entry at

1, 2

$1, 2$ a

0

$0$ .

[\begin{array}{c} 1 + \frac{1}{7} \cdot 0 & - \frac{1}{7} + \frac{1}{7} \cdot 1 & \frac{2}{7} + \frac{1}{7} \cdot 6 \\ 0 & 1 & 6 \end{array}]

$[\begin{array}{c} 1 + \frac{1}{7} \cdot 0 & - \frac{1}{7} + \frac{1}{7} \cdot 1 & \frac{2}{7} + \frac{1}{7} \cdot 6 \\ 0 & 1 & 6 \end{array}]$

Schritt 3.2.2

Vereinfache

R_{1}

$R_{1}$ .

[\begin{array}{c} 1 & 0 & \frac{8}{7} \\ 0 & 1 & 6 \end{array}]

$[\begin{array}{c} 1 & 0 & \frac{8}{7} \\ 0 & 1 & 6 \end{array}]$

[\begin{array}{c} 1 & 0 & \frac{8}{7} \\ 0 & 1 & 6 \end{array}]

$[\begin{array}{c} 1 & 0 & \frac{8}{7} \\ 0 & 1 & 6 \end{array}]$

[\begin{array}{c} 1 & 0 & \frac{8}{7} \\ 0 & 1 & 6 \end{array}]

$[\begin{array}{c} 1 & 0 & \frac{8}{7} \\ 0 & 1 & 6 \end{array}]$

Schritt 4

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

x = \frac{8}{7}

$x = \frac{8}{7}$

y = 6

$y = 6$

Schritt 5

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

(\frac{8}{7}, 6)

$(\frac{8}{7}, 6)$