Lineare Algebra Beispiele

$x - 4 y + 2 z = 8$ , $- 2 x + y - 7 z = 0$ , $- 7 y - 3 z = 16$

Schritt 1

Write the system as a matrix.

$[\begin{array}{c} 1 & - 4 & 2 & 8 \\ - 2 & 1 & - 7 & 0 \\ 0 & - 7 & - 3 & 16 \end{array}]$

Schritt 2

Ermittele die normierte Zeilenstufenform.

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Schritt 2.1

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

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Schritt 2.1.1

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

$[\begin{array}{c} 1 & - 4 & 2 & 8 \\ - 2 + 2 \cdot 1 & 1 + 2 \cdot - 4 & - 7 + 2 \cdot 2 & 0 + 2 \cdot 8 \\ 0 & - 7 & - 3 & 16 \end{array}]$

Schritt 2.1.2

Vereinfache $R_{2}$ .

$[\begin{array}{c} 1 & - 4 & 2 & 8 \\ 0 & - 7 & - 3 & 16 \\ 0 & - 7 & - 3 & 16 \end{array}]$

Schritt 2.2

Multiply each element of $R_{2}$ by $- \frac{1}{7}$ to make the entry at $2, 2$ a $1$ .

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Schritt 2.2.1

Multiply each element of $R_{2}$ by $- \frac{1}{7}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & - 4 & 2 & 8 \\ - \frac{1}{7} \cdot 0 & - \frac{1}{7} \cdot - 7 & - \frac{1}{7} \cdot - 3 & - \frac{1}{7} \cdot 16 \\ 0 & - 7 & - 3 & 16 \end{array}]$

Schritt 2.2.2

Vereinfache $R_{2}$ .

$[\begin{array}{c} 1 & - 4 & 2 & 8 \\ 0 & 1 & \frac{3}{7} & - \frac{16}{7} \\ 0 & - 7 & - 3 & 16 \end{array}]$

Schritt 2.3

Perform the row operation $R_{3} = R_{3} + 7 R_{2}$ to make the entry at $3, 2$ a $0$ .

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Schritt 2.3.1

Perform the row operation $R_{3} = R_{3} + 7 R_{2}$ to make the entry at $3, 2$ a $0$ .

$[\begin{array}{c} 1 & - 4 & 2 & 8 \\ 0 & 1 & \frac{3}{7} & - \frac{16}{7} \\ 0 + 7 \cdot 0 & - 7 + 7 \cdot 1 & - 3 + 7 (\frac{3}{7}) & 16 + 7 (- \frac{16}{7}) \end{array}]$

Schritt 2.3.2

Vereinfache $R_{3}$ .

$[\begin{array}{c} 1 & - 4 & 2 & 8 \\ 0 & 1 & \frac{3}{7} & - \frac{16}{7} \\ 0 & 0 & 0 & 0 \end{array}]$

Schritt 2.4

Perform the row operation $R_{1} = R_{1} + 4 R_{2}$ to make the entry at $1, 2$ a $0$ .

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Schritt 2.4.1

Perform the row operation $R_{1} = R_{1} + 4 R_{2}$ to make the entry at $1, 2$ a $0$ .

$[\begin{array}{c} 1 + 4 \cdot 0 & - 4 + 4 \cdot 1 & 2 + 4 (\frac{3}{7}) & 8 + 4 (- \frac{16}{7}) \\ 0 & 1 & \frac{3}{7} & - \frac{16}{7} \\ 0 & 0 & 0 & 0 \end{array}]$

Schritt 2.4.2

Vereinfache $R_{1}$ .

$[\begin{array}{c} 1 & 0 & \frac{26}{7} & - \frac{8}{7} \\ 0 & 1 & \frac{3}{7} & - \frac{16}{7} \\ 0 & 0 & 0 & 0 \end{array}]$

Schritt 3

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

$x + \frac{26}{7} z = - \frac{8}{7}$

$y + \frac{3}{7} z = - \frac{16}{7}$

$0 = 0$

Schritt 4

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

$(- \frac{8}{7} - \frac{26 z}{7}, - \frac{16}{7} - \frac{3 z}{7}, z)$