Lineare Algebra Beispiele

$[\begin{array}{c} 1 & - 2 & 2 & - 3 \\ 2 & - 4 & 9 & - 2 \\ - 3 & 6 & - 6 & 9 \end{array}]$

Schritt 1

Write as an augmented matrix for $A x = 0$ .

$[\begin{array}{c} 1 & - 2 & 2 & - 3 & 0 \\ 2 & - 4 & 9 & - 2 & 0 \\ - 3 & 6 & - 6 & 9 & 0 \end{array}]$

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 $0$ .

$[\begin{array}{c} 1 & - 2 & 2 & - 3 & 0 \\ 2 - 2 \cdot 1 & - 4 - 2 \cdot - 2 & 9 - 2 \cdot 2 & - 2 - 2 \cdot - 3 & 0 - 2 \cdot 0 \\ - 3 & 6 & - 6 & 9 & 0 \end{array}]$

Schritt 2.1.2

Vereinfache $R_{2}$ .

$[\begin{array}{c} 1 & - 2 & 2 & - 3 & 0 \\ 0 & 0 & 5 & 4 & 0 \\ - 3 & 6 & - 6 & 9 & 0 \end{array}]$

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 $0$ .

$[\begin{array}{c} 1 & - 2 & 2 & - 3 & 0 \\ 0 & 0 & 5 & 4 & 0 \\ - 3 + 3 \cdot 1 & 6 + 3 \cdot - 2 & - 6 + 3 \cdot 2 & 9 + 3 \cdot - 3 & 0 + 3 \cdot 0 \end{array}]$

Schritt 2.2.2

Vereinfache $R_{3}$ .

$[\begin{array}{c} 1 & - 2 & 2 & - 3 & 0 \\ 0 & 0 & 5 & 4 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{array}]$

Schritt 2.3

Multiply each element of $R_{2}$ by $\frac{1}{5}$ to make the entry at $2, 3$ a $1$ .

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Schritt 2.3.1

Multiply each element of $R_{2}$ by $\frac{1}{5}$ to make the entry at $2, 3$ a $1$ .

$[\begin{array}{c} 1 & - 2 & 2 & - 3 & 0 \\ \frac{0}{5} & \frac{0}{5} & \frac{5}{5} & \frac{4}{5} & \frac{0}{5} \\ 0 & 0 & 0 & 0 & 0 \end{array}]$

Schritt 2.3.2

Vereinfache $R_{2}$ .

$[\begin{array}{c} 1 & - 2 & 2 & - 3 & 0 \\ 0 & 0 & 1 & \frac{4}{5} & 0 \\ 0 & 0 & 0 & 0 & 0 \end{array}]$

Schritt 2.4

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

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Schritt 2.4.1

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

$[\begin{array}{c} 1 - 2 \cdot 0 & - 2 - 2 \cdot 0 & 2 - 2 \cdot 1 & - 3 - 2 (\frac{4}{5}) & 0 - 2 \cdot 0 \\ 0 & 0 & 1 & \frac{4}{5} & 0 \\ 0 & 0 & 0 & 0 & 0 \end{array}]$

Schritt 2.4.2

Vereinfache $R_{1}$ .

$[\begin{array}{c} 1 & - 2 & 0 & - \frac{23}{5} & 0 \\ 0 & 0 & 1 & \frac{4}{5} & 0 \\ 0 & 0 & 0 & 0 & 0 \end{array}]$

Schritt 3

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

$x_{1} - 2 x_{2} - \frac{23}{5} x_{4} = 0$

$x_{3} + \frac{4}{5} x_{4} = 0$

$0 = 0$

Schritt 4

Write a solution vector by solving in terms of the free variables in each row.

$[\begin{array}{c} x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \end{array}] = [\begin{array}{c} 2 x_{2} + \frac{23 x_{4}}{5} \\ x_{2} \\ - \frac{4 x_{4}}{5} \\ x_{4} \end{array}]$

Schritt 5

Write the solution as a linear combination of vectors.

$[\begin{array}{c} x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \end{array}] = x_{2} [\begin{array}{c} 2 \\ 1 \\ 0 \\ 0 \end{array}] + x_{4} [\begin{array}{c} \frac{23}{5} \\ 0 \\ - \frac{4}{5} \\ 1 \end{array}]$

Schritt 6

Write as a solution set.

${x_{2} [\begin{array}{c} 2 \\ 1 \\ 0 \\ 0 \end{array}] + x_{4} [\begin{array}{c} \frac{23}{5} \\ 0 \\ - \frac{4}{5} \\ 1 \end{array}] | x_{2}, x_{4} \in R}$