Lineare Algebra Beispiele

$0.15 g - 0.35 h = - 0.5$ , $- 0.012 g + 0.25 h = 0.1$

Schritt 1

Write the system as a matrix.

$[\begin{array}{c} 0.15 & - 0.35 & - 0.5 \\ - 0.012 & 0.25 & 0.1 \end{array}]$

Schritt 2

Ermittele die normierte Zeilenstufenform.

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Schritt 2.1

Multiply each element of $R_{1}$ by $\frac{1}{0.15}$ to make the entry at $1, 1$ a $1$ .

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Schritt 2.1.1

Multiply each element of $R_{1}$ by $\frac{1}{0.15}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} \frac{0.15}{0.15} & \frac{- 0.35}{0.15} & \frac{- 0.5}{0.15} \\ - 0.012 & 0.25 & 0.1 \end{array}]$

Schritt 2.1.2

Vereinfache $R_{1}$ .

$[\begin{array}{c} 1 & - 2.\overline{3} & - 3.\overline{3} \\ - 0.012 & 0.25 & 0.1 \end{array}]$

Schritt 2.2

Perform the row operation $R_{2} = R_{2} + 0.012 R_{1}$ to make the entry at $2, 1$ a $0$ .

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Schritt 2.2.1

Perform the row operation $R_{2} = R_{2} + 0.012 R_{1}$ to make the entry at $2, 1$ a $0$ .

$[\begin{array}{c} 1 & - 2.\overline{3} & - 3.\overline{3} \\ - 0.012 + 0.012 \cdot 1 & 0.25 + 0.012 \cdot - 2.\overline{3} & 0.1 + 0.012 \cdot - 3.\overline{3} \end{array}]$

Schritt 2.2.2

Vereinfache $R_{2}$ .

$[\begin{array}{c} 1 & - 2.\overline{3} & - 3.\overline{3} \\ 0 & 0.222 & 0.06 \end{array}]$

Schritt 2.3

Multiply each element of $R_{2}$ by $\frac{1}{0.222}$ to make the entry at $2, 2$ a $1$ .

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Schritt 2.3.1

Multiply each element of $R_{2}$ by $\frac{1}{0.222}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & - 2.\overline{3} & - 3.\overline{3} \\ \frac{0}{0.222} & \frac{0.222}{0.222} & \frac{0.06}{0.222} \end{array}]$

Schritt 2.3.2

Vereinfache $R_{2}$ .

$[\begin{array}{c} 1 & - 2.\overline{3} & - 3.\overline{3} \\ 0 & 1 & 0.\overline{270} \end{array}]$

Schritt 2.4

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

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Schritt 2.4.1

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

$[\begin{array}{c} 1 + 2.\overline{3} \cdot 0 & - 2.\overline{3} + 2.\overline{3} \cdot 1 & - 3.\overline{3} + 2.\overline{3} \cdot 0.\overline{270} \\ 0 & 1 & 0.\overline{270} \end{array}]$

Schritt 2.4.2

Vereinfache $R_{1}$ .

$[\begin{array}{c} 1 & 0 & - 2.\overline{702} \\ 0 & 1 & 0.\overline{270} \end{array}]$

Schritt 3

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

$g = - 2.\overline{702}$

$h = 0.\overline{270}$

Schritt 4

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

$(- 2.\overline{702}, 0.\overline{270})$