Lineare Algebra Beispiele

$2 a - 2 b = 0$ , $d + 3 g = 2$ , $3 a - 3 b = 0$ , $a - b + 2 d + 6 g = 4$

Schritt 1

Write the system as a matrix.

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

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 $1$ .

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

Schritt 2.1.2

Vereinfache $R_{1}$ .

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

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 $0$ .

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

Schritt 2.2.2

Vereinfache $R_{3}$ .

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

Schritt 2.3

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

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Schritt 2.3.1

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

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

Schritt 2.3.2

Vereinfache $R_{4}$ .

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

Schritt 2.4

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

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Schritt 2.4.1

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

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

Schritt 2.4.2

Vereinfache $R_{4}$ .

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

Schritt 3

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

$a - b = 0$

$d + 3 g = 2$

$0 = 0$

Schritt 4

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

$(b, b, 2 - 3 g, g)$