Álgebra linear Exemplos

$[\begin{array}{c} 10 & 12 \\ 15 & 18 \\ 20 & 24 \end{array}]$

Etapa 1

Encontre a forma escalonada reduzida por linhas.

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Etapa 1.1

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

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Etapa 1.1.1

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

$[\begin{array}{c} \frac{10}{10} & \frac{12}{10} \\ 15 & 18 \\ 20 & 24 \end{array}]$

Etapa 1.1.2

Simplifique $R_{1}$ .

$[\begin{array}{c} 1 & \frac{6}{5} \\ 15 & 18 \\ 20 & 24 \end{array}]$

Etapa 1.2

Perform the row operation $R_{2} = R_{2} - 15 R_{1}$ to make the entry at $2, 1$ a $0$ .

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Etapa 1.2.1

Perform the row operation $R_{2} = R_{2} - 15 R_{1}$ to make the entry at $2, 1$ a $0$ .

$[\begin{array}{c} 1 & \frac{6}{5} \\ 15 - 15 \cdot 1 & 18 - 15 (\frac{6}{5}) \\ 20 & 24 \end{array}]$

Etapa 1.2.2

Simplifique $R_{2}$ .

$[\begin{array}{c} 1 & \frac{6}{5} \\ 0 & 0 \\ 20 & 24 \end{array}]$

Etapa 1.3

Perform the row operation $R_{3} = R_{3} - 20 R_{1}$ to make the entry at $3, 1$ a $0$ .

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Etapa 1.3.1

Perform the row operation $R_{3} = R_{3} - 20 R_{1}$ to make the entry at $3, 1$ a $0$ .

$[\begin{array}{c} 1 & \frac{6}{5} \\ 0 & 0 \\ 20 - 20 \cdot 1 & 24 - 20 (\frac{6}{5}) \end{array}]$

Etapa 1.3.2

Simplifique $R_{3}$ .

$[\begin{array}{c} 1 & \frac{6}{5} \\ 0 & 0 \\ 0 & 0 \end{array}]$

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: $a_{11}$

Pivot Columns: $1$