Lineare Algebra Beispiele

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

Schritt 1

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

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

Schritt 2

Ermittele die normierte Zeilenstufenform.

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Schritt 2.1

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

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Schritt 2.1.1

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

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

Schritt 2.1.2

Vereinfache $R_{1}$ .

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

Schritt 3

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

$x - \frac{9}{2} y + 3 z = 0$

Schritt 4

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

$[\begin{array}{c} x \\ y \\ z \end{array}] = [\begin{array}{c} \frac{9 y}{2} - 3 z \\ y \\ z \end{array}]$

Schritt 5

Write the solution as a linear combination of vectors.

$[\begin{array}{c} x \\ y \\ z \end{array}] = y [\begin{array}{c} \frac{9}{2} \\ 1 \\ 0 \end{array}] + z [\begin{array}{c} - 3 \\ 0 \\ 1 \end{array}]$

Schritt 6

Write as a solution set.

${y [\begin{array}{c} \frac{9}{2} \\ 1 \\ 0 \end{array}] + z [\begin{array}{c} - 3 \\ 0 \\ 1 \end{array}] | y, z \in R}$