Algebra Beispiele

3 x + 3 y = 3

$3 x + 3 y = 3$ ,

x < y

$x < y$

Schritt 1

Führe die Schlupfvariablen

u

$u$ und

v

$v$ ein, um die Ungleichungen durch Gleichungen zu ersetzen.

x + Z = y

$x + Z = y$

3 x + 3 y - 3 = 0

$3 x + 3 y - 3 = 0$

Schritt 2

Subtrahiere

y

$y$ von beiden Seiten der Gleichung.

x + Z - y = 0, 3 x + 3 y - 3 = 0

$x + Z - y = 0, 3 x + 3 y - 3 = 0$

Schritt 3

Addiere

3

$3$ zu beiden Seiten der Gleichung.

x + Z - y = 0, 3 x + 3 y = 3

$x + Z - y = 0, 3 x + 3 y = 3$

Schritt 4

Schreibe das Gleichungssystem in Matrixform.

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

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

Schritt 5

Ermittele die normierte Zeilenstufenform.

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Schritt 5.1

Perform the row operation

R_{2} = R_{2} - 3 R_{1}

$R_{2} = R_{2} - 3 R_{1}$ to make the entry at

2, 1

$2, 1$ a

0

$0$ .

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Schritt 5.1.1

Perform the row operation

R_{2} = R_{2} - 3 R_{1}

$R_{2} = R_{2} - 3 R_{1}$ to make the entry at

2, 1

$2, 1$ a

0

$0$ .

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

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

Schritt 5.1.2

Vereinfache

R_{2}

$R_{2}$ .

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

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

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

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

Schritt 5.2

Multiply each element of

R_{2}

$R_{2}$ by

\frac{1}{6}

$\frac{1}{6}$ to make the entry at

2, 2

$2, 2$ a

1

$1$ .

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Schritt 5.2.1

Multiply each element of

R_{2}

$R_{2}$ by

\frac{1}{6}

$\frac{1}{6}$ to make the entry at

2, 2

$2, 2$ a

1

$1$ .

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

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

Schritt 5.2.2

Vereinfache

R_{2}

$R_{2}$ .

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

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

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

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

Schritt 5.3

Perform the row operation

R_{1} = R_{1} + R_{2}

$R_{1} = R_{1} + R_{2}$ to make the entry at

1, 2

$1, 2$ a

0

$0$ .

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Schritt 5.3.1

Perform the row operation

R_{1} = R_{1} + R_{2}

$R_{1} = R_{1} + R_{2}$ to make the entry at

1, 2

$1, 2$ a

0

$0$ .

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

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

Schritt 5.3.2

Vereinfache

R_{1}

$R_{1}$ .

[\begin{array}{c} 1 & 0 & 0 & \frac{1}{2} \\ 0 & 1 & 0 & \frac{1}{2} \end{array}]

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

[\begin{array}{c} 1 & 0 & 0 & \frac{1}{2} \\ 0 & 1 & 0 & \frac{1}{2} \end{array}]

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

[\begin{array}{c} 1 & 0 & 0 & \frac{1}{2} \\ 0 & 1 & 0 & \frac{1}{2} \end{array}]

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

Schritt 6

Verwende die Ergebnismatrix, um die endgültigen Lösungen für das Gleichungssystem anzugeben.

x = \frac{1}{2}

$x = \frac{1}{2}$

y = \frac{1}{2}

$y = \frac{1}{2}$

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