Lineare Algebra Beispiele

$[\begin{array}{c} 0 & 0 \\ - 5 & - 8 \end{array}]$

Schritt 1

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

$[\begin{array}{c} 0 & 0 & 0 \\ - 5 & - 8 & 0 \end{array}]$

Schritt 2

Ermittele die normierte Zeilenstufenform.

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Schritt 2.1

Swap $R_{2}$ with $R_{1}$ to put a nonzero entry at $1, 1$ .

$[\begin{array}{c} - 5 & - 8 & 0 \\ 0 & 0 & 0 \end{array}]$

Schritt 2.2

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

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Schritt 2.2.1

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

$[\begin{array}{c} - \frac{1}{5} \cdot - 5 & - \frac{1}{5} \cdot - 8 & - \frac{1}{5} \cdot 0 \\ 0 & 0 & 0 \end{array}]$

Schritt 2.2.2

Vereinfache $R_{1}$ .

$[\begin{array}{c} 1 & \frac{8}{5} & 0 \\ 0 & 0 & 0 \end{array}]$

Schritt 3

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

$x + \frac{8}{5} y = 0$

$0 = 0$

Schritt 4

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

$[\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} - \frac{8 y}{5} \\ y \end{array}]$

Schritt 5

Write the solution as a linear combination of vectors.

$[\begin{array}{c} x \\ y \end{array}] = y [\begin{array}{c} - \frac{8}{5} \\ 1 \end{array}]$

Schritt 6

Write as a solution set.

${y [\begin{array}{c} - \frac{8}{5} \\ 1 \end{array}] | y \in R}$