Lineare Algebra Beispiele

$[\begin{array}{c} 1 & h & 4 \\ 3 & 6 & 8 \end{array}]$

Schritt 1

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

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Schritt 1.1

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

$[\begin{array}{c} 1 & h & 4 \\ 3 - 3 \cdot 1 & 6 - 3 h & 8 - 3 \cdot 4 \end{array}]$

Schritt 1.2

Vereinfache $R_{2}$ .

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

Schritt 2

Multiply each element of $R_{2}$ by $\frac{1}{6 - 3 h}$ to make the entry at $2, 2$ a $1$ .

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Schritt 2.1

Multiply each element of $R_{2}$ by $\frac{1}{6 - 3 h}$ to make the entry at $2, 2$ a $1$ .

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

Schritt 2.2

Vereinfache $R_{2}$ .

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

Schritt 3

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

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Schritt 3.1

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

$[\begin{array}{c} 1 - h \cdot 0 & h - h \cdot 1 & 4 - h (- \frac{4}{3 (2 - h)}) \\ 0 & 1 & - \frac{4}{3 (2 - h)} \end{array}]$

Schritt 3.2

Vereinfache $R_{1}$ .

$[\begin{array}{c} 1 & 0 & 4 + \frac{4 h}{3 (2 - h)} \\ 0 & 1 & - \frac{4}{3 (2 - h)} \end{array}]$