Álgebra linear Exemplos

$[\begin{array}{c} 1 & h & 4 \\ - 5 & 10 & - 24 \end{array}]$

Etapa 1

Encontre a forma escalonada reduzida por linhas.

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Etapa 1.1

Perform the row operation $R_{2} = R_{2} + 5 R_{1}$ to make the entry at $2, 1$ a $0$ .

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Etapa 1.1.1

Perform the row operation $R_{2} = R_{2} + 5 R_{1}$ to make the entry at $2, 1$ a $0$ .

$[\begin{array}{c} 1 & h & 4 \\ - 5 + 5 \cdot 1 & 10 + 5 h & - 24 + 5 \cdot 4 \end{array}]$

Etapa 1.1.2

Simplifique $R_{2}$ .

$[\begin{array}{c} 1 & h & 4 \\ 0 & 10 + 5 h & - 4 \end{array}]$

Etapa 1.2

Multiply each element of $R_{2}$ by $\frac{1}{10 + 5 h}$ to make the entry at $2, 2$ a $1$ .

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Etapa 1.2.1

Multiply each element of $R_{2}$ by $\frac{1}{10 + 5 h}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & h & 4 \\ \frac{0}{10 + 5 h} & \frac{10 + 5 h}{10 + 5 h} & \frac{- 4}{10 + 5 h} \end{array}]$

Etapa 1.2.2

Simplifique $R_{2}$ .

$[\begin{array}{c} 1 & h & 4 \\ 0 & 1 & - \frac{4}{5 (2 + h)} \end{array}]$

Etapa 1.3

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

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Etapa 1.3.1

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

$[\begin{array}{c} 1 - h \cdot 0 & h - h \cdot 1 & 4 - h (- \frac{4}{5 (2 + h)}) \\ 0 & 1 & - \frac{4}{5 (2 + h)} \end{array}]$

Etapa 1.3.2

Simplifique $R_{1}$ .

$[\begin{array}{c} 1 & 0 & 4 + \frac{4 h}{5 (2 + h)} \\ 0 & 1 & - \frac{4}{5 (2 + h)} \end{array}]$

Etapa 2

The pivot positions are the locations with the leading $1$ in each row. The pivot columns are the columns that have a pivot position.

Pivot Positions: $a_{11}$ and $a_{22}$

Pivot Columns: $1$ and $2$

Etapa 3

The rank is the number of pivot columns.

$2$