Álgebra linear Exemplos

$10 x - 2 y + 5 x = - 1$ , $6 x - y + 4 z = - 2$ , $x + z = 3$

Etapa 1

Some $10 x$ e $5 x$ .

$15 x - 2 y = - 1, 6 x - y + 4 z = - 2, x + z = 3$

Etapa 2

Escreva o sistema de equações em formato de matriz.

$[\begin{array}{c} 15 & - 2 & 0 & - 1 \\ 6 & - 1 & 4 & - 2 \\ 1 & 0 & 1 & 3 \end{array}]$

Etapa 3

Encontre a forma escalonada reduzida por linhas.

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Etapa 3.1

Multiply each element of $R_{1}$ by $\frac{1}{15}$ to make the entry at $1, 1$ a $1$ .

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Etapa 3.1.1

Multiply each element of $R_{1}$ by $\frac{1}{15}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} \frac{15}{15} & \frac{- 2}{15} & \frac{0}{15} & \frac{- 1}{15} \\ 6 & - 1 & 4 & - 2 \\ 1 & 0 & 1 & 3 \end{array}]$

Etapa 3.1.2

Simplifique $R_{1}$ .

$[\begin{array}{c} 1 & - \frac{2}{15} & 0 & - \frac{1}{15} \\ 6 & - 1 & 4 & - 2 \\ 1 & 0 & 1 & 3 \end{array}]$

Etapa 3.2

Perform the row operation $R_{2} = R_{2} - 6 R_{1}$ to make the entry at $2, 1$ a $0$ .

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Etapa 3.2.1

Perform the row operation $R_{2} = R_{2} - 6 R_{1}$ to make the entry at $2, 1$ a $0$ .

$[\begin{array}{c} 1 & - \frac{2}{15} & 0 & - \frac{1}{15} \\ 6 - 6 \cdot 1 & - 1 - 6 (- \frac{2}{15}) & 4 - 6 \cdot 0 & - 2 - 6 (- \frac{1}{15}) \\ 1 & 0 & 1 & 3 \end{array}]$

Etapa 3.2.2

Simplifique $R_{2}$ .

$[\begin{array}{c} 1 & - \frac{2}{15} & 0 & - \frac{1}{15} \\ 0 & - \frac{1}{5} & 4 & - \frac{8}{5} \\ 1 & 0 & 1 & 3 \end{array}]$

Etapa 3.3

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

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Etapa 3.3.1

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

$[\begin{array}{c} 1 & - \frac{2}{15} & 0 & - \frac{1}{15} \\ 0 & - \frac{1}{5} & 4 & - \frac{8}{5} \\ 1 - 1 & 0 + \frac{2}{15} & 1 - 0 & 3 + \frac{1}{15} \end{array}]$

Etapa 3.3.2

Simplifique $R_{3}$ .

$[\begin{array}{c} 1 & - \frac{2}{15} & 0 & - \frac{1}{15} \\ 0 & - \frac{1}{5} & 4 & - \frac{8}{5} \\ 0 & \frac{2}{15} & 1 & \frac{46}{15} \end{array}]$

Etapa 3.4

Multiply each element of $R_{2}$ by $- 5$ to make the entry at $2, 2$ a $1$ .

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Etapa 3.4.1

Multiply each element of $R_{2}$ by $- 5$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & - \frac{2}{15} & 0 & - \frac{1}{15} \\ - 5 \cdot 0 & - 5 (- \frac{1}{5}) & - 5 \cdot 4 & - 5 (- \frac{8}{5}) \\ 0 & \frac{2}{15} & 1 & \frac{46}{15} \end{array}]$

Etapa 3.4.2

Simplifique $R_{2}$ .

$[\begin{array}{c} 1 & - \frac{2}{15} & 0 & - \frac{1}{15} \\ 0 & 1 & - 20 & 8 \\ 0 & \frac{2}{15} & 1 & \frac{46}{15} \end{array}]$

Etapa 3.5

Perform the row operation $R_{3} = R_{3} - \frac{2}{15} R_{2}$ to make the entry at $3, 2$ a $0$ .

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Etapa 3.5.1

Perform the row operation $R_{3} = R_{3} - \frac{2}{15} R_{2}$ to make the entry at $3, 2$ a $0$ .

$[\begin{array}{c} 1 & - \frac{2}{15} & 0 & - \frac{1}{15} \\ 0 & 1 & - 20 & 8 \\ 0 - \frac{2}{15} \cdot 0 & \frac{2}{15} - \frac{2}{15} \cdot 1 & 1 - \frac{2}{15} \cdot - 20 & \frac{46}{15} - \frac{2}{15} \cdot 8 \end{array}]$

Etapa 3.5.2

Simplifique $R_{3}$ .

$[\begin{array}{c} 1 & - \frac{2}{15} & 0 & - \frac{1}{15} \\ 0 & 1 & - 20 & 8 \\ 0 & 0 & \frac{11}{3} & 2 \end{array}]$

Etapa 3.6

Multiply each element of $R_{3}$ by $\frac{3}{11}$ to make the entry at $3, 3$ a $1$ .

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Etapa 3.6.1

Multiply each element of $R_{3}$ by $\frac{3}{11}$ to make the entry at $3, 3$ a $1$ .

$[\begin{array}{c} 1 & - \frac{2}{15} & 0 & - \frac{1}{15} \\ 0 & 1 & - 20 & 8 \\ \frac{3}{11} \cdot 0 & \frac{3}{11} \cdot 0 & \frac{3}{11} \cdot \frac{11}{3} & \frac{3}{11} \cdot 2 \end{array}]$

Etapa 3.6.2

Simplifique $R_{3}$ .

$[\begin{array}{c} 1 & - \frac{2}{15} & 0 & - \frac{1}{15} \\ 0 & 1 & - 20 & 8 \\ 0 & 0 & 1 & \frac{6}{11} \end{array}]$

Etapa 3.7

Perform the row operation $R_{2} = R_{2} + 20 R_{3}$ to make the entry at $2, 3$ a $0$ .

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Etapa 3.7.1

Perform the row operation $R_{2} = R_{2} + 20 R_{3}$ to make the entry at $2, 3$ a $0$ .

$[\begin{array}{c} 1 & - \frac{2}{15} & 0 & - \frac{1}{15} \\ 0 + 20 \cdot 0 & 1 + 20 \cdot 0 & - 20 + 20 \cdot 1 & 8 + 20 (\frac{6}{11}) \\ 0 & 0 & 1 & \frac{6}{11} \end{array}]$

Etapa 3.7.2

Simplifique $R_{2}$ .

$[\begin{array}{c} 1 & - \frac{2}{15} & 0 & - \frac{1}{15} \\ 0 & 1 & 0 & \frac{208}{11} \\ 0 & 0 & 1 & \frac{6}{11} \end{array}]$

Etapa 3.8

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

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Etapa 3.8.1

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

$[\begin{array}{c} 1 + \frac{2}{15} \cdot 0 & - \frac{2}{15} + \frac{2}{15} \cdot 1 & 0 + \frac{2}{15} \cdot 0 & - \frac{1}{15} + \frac{2}{15} \cdot \frac{208}{11} \\ 0 & 1 & 0 & \frac{208}{11} \\ 0 & 0 & 1 & \frac{6}{11} \end{array}]$

Etapa 3.8.2

Simplifique $R_{1}$ .

$[\begin{array}{c} 1 & 0 & 0 & \frac{27}{11} \\ 0 & 1 & 0 & \frac{208}{11} \\ 0 & 0 & 1 & \frac{6}{11} \end{array}]$

Etapa 4

Use a matriz de resultados para declarar as soluções finais ao sistema de equações.

$x = \frac{27}{11}$

$y = \frac{208}{11}$

$z = \frac{6}{11}$

Etapa 5

A solução é o conjunto de pares ordenados que tornam o sistema verdadeiro.

$(\frac{27}{11}, \frac{208}{11}, \frac{6}{11})$

Etapa 6

Para decompor um vetor da solução, reorganize cada equação representada na forma de linha reduzida da matriz aumentada resolvendo a variável dependente em cada linha que produz igualdade vetorial.

$X = [\begin{array}{c} x \\ y \\ z \end{array}] = [\begin{array}{c} \frac{27}{11} \\ \frac{208}{11} \\ \frac{6}{11} \end{array}]$