Lineare Algebra Beispiele

$x - 3 y + 3 z = - 6$ , $4 x + y - z = - 11$ , $3 x + 4 y - 5 z = - 8$

Schritt 1

Write the system as a matrix.

$[\begin{array}{c} 1 & - 3 & 3 & - 6 \\ 4 & 1 & - 1 & - 11 \\ 3 & 4 & - 5 & - 8 \end{array}]$

Schritt 2

Ermittele die normierte Zeilenstufenform.

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Schritt 2.1

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

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Schritt 2.1.1

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

$[\begin{array}{c} 1 & - 3 & 3 & - 6 \\ 4 - 4 \cdot 1 & 1 - 4 \cdot - 3 & - 1 - 4 \cdot 3 & - 11 - 4 \cdot - 6 \\ 3 & 4 & - 5 & - 8 \end{array}]$

Schritt 2.1.2

Vereinfache $R_{2}$ .

$[\begin{array}{c} 1 & - 3 & 3 & - 6 \\ 0 & 13 & - 13 & 13 \\ 3 & 4 & - 5 & - 8 \end{array}]$

Schritt 2.2

Perform the row operation $R_{3} = R_{3} - 3 R_{1}$ to make the entry at $3, 1$ a $0$ .

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Schritt 2.2.1

Perform the row operation $R_{3} = R_{3} - 3 R_{1}$ to make the entry at $3, 1$ a $0$ .

$[\begin{array}{c} 1 & - 3 & 3 & - 6 \\ 0 & 13 & - 13 & 13 \\ 3 - 3 \cdot 1 & 4 - 3 \cdot - 3 & - 5 - 3 \cdot 3 & - 8 - 3 \cdot - 6 \end{array}]$

Schritt 2.2.2

Vereinfache $R_{3}$ .

$[\begin{array}{c} 1 & - 3 & 3 & - 6 \\ 0 & 13 & - 13 & 13 \\ 0 & 13 & - 14 & 10 \end{array}]$

Schritt 2.3

Multiply each element of $R_{2}$ by $\frac{1}{13}$ to make the entry at $2, 2$ a $1$ .

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Schritt 2.3.1

Multiply each element of $R_{2}$ by $\frac{1}{13}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & - 3 & 3 & - 6 \\ \frac{0}{13} & \frac{13}{13} & \frac{- 13}{13} & \frac{13}{13} \\ 0 & 13 & - 14 & 10 \end{array}]$

Schritt 2.3.2

Vereinfache $R_{2}$ .

$[\begin{array}{c} 1 & - 3 & 3 & - 6 \\ 0 & 1 & - 1 & 1 \\ 0 & 13 & - 14 & 10 \end{array}]$

Schritt 2.4

Perform the row operation $R_{3} = R_{3} - 13 R_{2}$ to make the entry at $3, 2$ a $0$ .

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Schritt 2.4.1

Perform the row operation $R_{3} = R_{3} - 13 R_{2}$ to make the entry at $3, 2$ a $0$ .

$[\begin{array}{c} 1 & - 3 & 3 & - 6 \\ 0 & 1 & - 1 & 1 \\ 0 - 13 \cdot 0 & 13 - 13 \cdot 1 & - 14 - 13 \cdot - 1 & 10 - 13 \cdot 1 \end{array}]$

Schritt 2.4.2

Vereinfache $R_{3}$ .

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

Schritt 2.5

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

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Schritt 2.5.1

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

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

Schritt 2.5.2

Vereinfache $R_{3}$ .

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

Schritt 2.6

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

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Schritt 2.6.1

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

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

Schritt 2.6.2

Vereinfache $R_{2}$ .

$[\begin{array}{c} 1 & - 3 & 3 & - 6 \\ 0 & 1 & 0 & 4 \\ 0 & 0 & 1 & 3 \end{array}]$

Schritt 2.7

Perform the row operation $R_{1} = R_{1} - 3 R_{3}$ to make the entry at $1, 3$ a $0$ .

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Schritt 2.7.1

Perform the row operation $R_{1} = R_{1} - 3 R_{3}$ to make the entry at $1, 3$ a $0$ .

$[\begin{array}{c} 1 - 3 \cdot 0 & - 3 - 3 \cdot 0 & 3 - 3 \cdot 1 & - 6 - 3 \cdot 3 \\ 0 & 1 & 0 & 4 \\ 0 & 0 & 1 & 3 \end{array}]$

Schritt 2.7.2

Vereinfache $R_{1}$ .

$[\begin{array}{c} 1 & - 3 & 0 & - 15 \\ 0 & 1 & 0 & 4 \\ 0 & 0 & 1 & 3 \end{array}]$

Schritt 2.8

Perform the row operation $R_{1} = R_{1} + 3 R_{2}$ to make the entry at $1, 2$ a $0$ .

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Schritt 2.8.1

Perform the row operation $R_{1} = R_{1} + 3 R_{2}$ to make the entry at $1, 2$ a $0$ .

$[\begin{array}{c} 1 + 3 \cdot 0 & - 3 + 3 \cdot 1 & 0 + 3 \cdot 0 & - 15 + 3 \cdot 4 \\ 0 & 1 & 0 & 4 \\ 0 & 0 & 1 & 3 \end{array}]$

Schritt 2.8.2

Vereinfache $R_{1}$ .

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

Schritt 3

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

$x = - 3$

$y = 4$

$z = 3$

Schritt 4

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

$(- 3, 4, 3)$