Lineare Algebra Beispiele

$[\begin{array}{c} 4 & - 3 & 1 & - 7 \\ 0 & \frac{7}{4} & \frac{15}{4} & \frac{23}{4} \\ 0 & 0 & \frac{27}{7} & \frac{54}{7} \end{array}]$

Schritt 1

Multiply each element of $R_{1}$ by $\frac{1}{4}$ to make the entry at $1, 1$ a $1$ .

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Schritt 1.1

Multiply each element of $R_{1}$ by $\frac{1}{4}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} \frac{4}{4} & \frac{- 3}{4} & \frac{1}{4} & \frac{- 7}{4} \\ 0 & \frac{7}{4} & \frac{15}{4} & \frac{23}{4} \\ 0 & 0 & \frac{27}{7} & \frac{54}{7} \end{array}]$

Schritt 1.2

Vereinfache $R_{1}$ .

$[\begin{array}{c} 1 & - \frac{3}{4} & \frac{1}{4} & - \frac{7}{4} \\ 0 & \frac{7}{4} & \frac{15}{4} & \frac{23}{4} \\ 0 & 0 & \frac{27}{7} & \frac{54}{7} \end{array}]$

Schritt 2

Multiply each element of $R_{2}$ by $\frac{4}{7}$ to make the entry at $2, 2$ a $1$ .

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Schritt 2.1

Multiply each element of $R_{2}$ by $\frac{4}{7}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & - \frac{3}{4} & \frac{1}{4} & - \frac{7}{4} \\ \frac{4}{7} \cdot 0 & \frac{4}{7} \cdot \frac{7}{4} & \frac{4}{7} \cdot \frac{15}{4} & \frac{4}{7} \cdot \frac{23}{4} \\ 0 & 0 & \frac{27}{7} & \frac{54}{7} \end{array}]$

Schritt 2.2

Vereinfache $R_{2}$ .

$[\begin{array}{c} 1 & - \frac{3}{4} & \frac{1}{4} & - \frac{7}{4} \\ 0 & 1 & \frac{15}{7} & \frac{23}{7} \\ 0 & 0 & \frac{27}{7} & \frac{54}{7} \end{array}]$

Schritt 3

Multiply each element of $R_{3}$ by $\frac{7}{27}$ to make the entry at $3, 3$ a $1$ .

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Schritt 3.1

Multiply each element of $R_{3}$ by $\frac{7}{27}$ to make the entry at $3, 3$ a $1$ .

$[\begin{array}{c} 1 & - \frac{3}{4} & \frac{1}{4} & - \frac{7}{4} \\ 0 & 1 & \frac{15}{7} & \frac{23}{7} \\ \frac{7}{27} \cdot 0 & \frac{7}{27} \cdot 0 & \frac{7}{27} \cdot \frac{27}{7} & \frac{7}{27} \cdot \frac{54}{7} \end{array}]$

Schritt 3.2

Vereinfache $R_{3}$ .

$[\begin{array}{c} 1 & - \frac{3}{4} & \frac{1}{4} & - \frac{7}{4} \\ 0 & 1 & \frac{15}{7} & \frac{23}{7} \\ 0 & 0 & 1 & 2 \end{array}]$

Schritt 4

Perform the row operation $R_{2} = R_{2} - \frac{15}{7} R_{3}$ to make the entry at $2, 3$ a $0$ .

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Schritt 4.1

Perform the row operation $R_{2} = R_{2} - \frac{15}{7} R_{3}$ to make the entry at $2, 3$ a $0$ .

$[\begin{array}{c} 1 & - \frac{3}{4} & \frac{1}{4} & - \frac{7}{4} \\ 0 - \frac{15}{7} \cdot 0 & 1 - \frac{15}{7} \cdot 0 & \frac{15}{7} - \frac{15}{7} \cdot 1 & \frac{23}{7} - \frac{15}{7} \cdot 2 \\ 0 & 0 & 1 & 2 \end{array}]$

Schritt 4.2

Vereinfache $R_{2}$ .

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

Schritt 5

Perform the row operation $R_{1} = R_{1} - \frac{1}{4} R_{3}$ to make the entry at $1, 3$ a $0$ .

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Schritt 5.1

Perform the row operation $R_{1} = R_{1} - \frac{1}{4} R_{3}$ to make the entry at $1, 3$ a $0$ .

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

Schritt 5.2

Vereinfache $R_{1}$ .

$[\begin{array}{c} 1 & - \frac{3}{4} & 0 & - \frac{9}{4} \\ 0 & 1 & 0 & - 1 \\ 0 & 0 & 1 & 2 \end{array}]$

Schritt 6

Perform the row operation $R_{1} = R_{1} + \frac{3}{4} R_{2}$ to make the entry at $1, 2$ a $0$ .

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Schritt 6.1

Perform the row operation $R_{1} = R_{1} + \frac{3}{4} R_{2}$ to make the entry at $1, 2$ a $0$ .

$[\begin{array}{c} 1 + \frac{3}{4} \cdot 0 & - \frac{3}{4} + \frac{3}{4} \cdot 1 & 0 + \frac{3}{4} \cdot 0 & - \frac{9}{4} + \frac{3}{4} \cdot - 1 \\ 0 & 1 & 0 & - 1 \\ 0 & 0 & 1 & 2 \end{array}]$

Schritt 6.2

Vereinfache $R_{1}$ .

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