Álgebra linear Exemplos

$2 x + b - 3 c = 12$ , $5 a - 4 b + 7 c = 27$ , $10 a + 3 b - c = 40$

Etapa 1

Escreva o sistema de equações em formato de matriz.

$[\begin{array}{c} 0 & 1 & - 3 & 2 & 12 \\ 5 & - 4 & 7 & 0 & 27 \\ 10 & 3 & - 1 & 0 & 40 \end{array}]$

Etapa 2

Encontre a forma escalonada reduzida por linhas.

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Etapa 2.1

Swap $R_{2}$ with $R_{1}$ to put a nonzero entry at $1, 1$ .

$[\begin{array}{c} 5 & - 4 & 7 & 0 & 27 \\ 0 & 1 & - 3 & 2 & 12 \\ 10 & 3 & - 1 & 0 & 40 \end{array}]$

Etapa 2.2

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

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Etapa 2.2.1

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

$[\begin{array}{c} \frac{5}{5} & \frac{- 4}{5} & \frac{7}{5} & \frac{0}{5} & \frac{27}{5} \\ 0 & 1 & - 3 & 2 & 12 \\ 10 & 3 & - 1 & 0 & 40 \end{array}]$

Etapa 2.2.2

Simplifique $R_{1}$ .

$[\begin{array}{c} 1 & - \frac{4}{5} & \frac{7}{5} & 0 & \frac{27}{5} \\ 0 & 1 & - 3 & 2 & 12 \\ 10 & 3 & - 1 & 0 & 40 \end{array}]$

Etapa 2.3

Perform the row operation $R_{3} = R_{3} - 10 R_{1}$ to make the entry at $3, 1$ a $0$ .

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Etapa 2.3.1

Perform the row operation $R_{3} = R_{3} - 10 R_{1}$ to make the entry at $3, 1$ a $0$ .

$[\begin{array}{c} 1 & - \frac{4}{5} & \frac{7}{5} & 0 & \frac{27}{5} \\ 0 & 1 & - 3 & 2 & 12 \\ 10 - 10 \cdot 1 & 3 - 10 (- \frac{4}{5}) & - 1 - 10 (\frac{7}{5}) & 0 - 10 \cdot 0 & 40 - 10 (\frac{27}{5}) \end{array}]$

Etapa 2.3.2

Simplifique $R_{3}$ .

$[\begin{array}{c} 1 & - \frac{4}{5} & \frac{7}{5} & 0 & \frac{27}{5} \\ 0 & 1 & - 3 & 2 & 12 \\ 0 & 11 & - 15 & 0 & - 14 \end{array}]$

Etapa 2.4

Perform the row operation $R_{3} = R_{3} - 11 R_{2}$ to make the entry at $3, 2$ a $0$ .

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Etapa 2.4.1

Perform the row operation $R_{3} = R_{3} - 11 R_{2}$ to make the entry at $3, 2$ a $0$ .

$[\begin{array}{c} 1 & - \frac{4}{5} & \frac{7}{5} & 0 & \frac{27}{5} \\ 0 & 1 & - 3 & 2 & 12 \\ 0 - 11 \cdot 0 & 11 - 11 \cdot 1 & - 15 - 11 \cdot - 3 & 0 - 11 \cdot 2 & - 14 - 11 \cdot 12 \end{array}]$

Etapa 2.4.2

Simplifique $R_{3}$ .

$[\begin{array}{c} 1 & - \frac{4}{5} & \frac{7}{5} & 0 & \frac{27}{5} \\ 0 & 1 & - 3 & 2 & 12 \\ 0 & 0 & 18 & - 22 & - 146 \end{array}]$

Etapa 2.5

Multiply each element of $R_{3}$ by $\frac{1}{18}$ to make the entry at $3, 3$ a $1$ .

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Etapa 2.5.1

Multiply each element of $R_{3}$ by $\frac{1}{18}$ to make the entry at $3, 3$ a $1$ .

$[\begin{array}{c} 1 & - \frac{4}{5} & \frac{7}{5} & 0 & \frac{27}{5} \\ 0 & 1 & - 3 & 2 & 12 \\ \frac{0}{18} & \frac{0}{18} & \frac{18}{18} & \frac{- 22}{18} & \frac{- 146}{18} \end{array}]$

Etapa 2.5.2

Simplifique $R_{3}$ .

$[\begin{array}{c} 1 & - \frac{4}{5} & \frac{7}{5} & 0 & \frac{27}{5} \\ 0 & 1 & - 3 & 2 & 12 \\ 0 & 0 & 1 & - \frac{11}{9} & - \frac{73}{9} \end{array}]$

Etapa 2.6

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

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Etapa 2.6.1

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

$[\begin{array}{c} 1 & - \frac{4}{5} & \frac{7}{5} & 0 & \frac{27}{5} \\ 0 + 3 \cdot 0 & 1 + 3 \cdot 0 & - 3 + 3 \cdot 1 & 2 + 3 (- \frac{11}{9}) & 12 + 3 (- \frac{73}{9}) \\ 0 & 0 & 1 & - \frac{11}{9} & - \frac{73}{9} \end{array}]$

Etapa 2.6.2

Simplifique $R_{2}$ .

$[\begin{array}{c} 1 & - \frac{4}{5} & \frac{7}{5} & 0 & \frac{27}{5} \\ 0 & 1 & 0 & - \frac{5}{3} & - \frac{37}{3} \\ 0 & 0 & 1 & - \frac{11}{9} & - \frac{73}{9} \end{array}]$

Etapa 2.7

Perform the row operation $R_{1} = R_{1} - \frac{7}{5} R_{3}$ to make the entry at $1, 3$ a $0$ .

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Etapa 2.7.1

Perform the row operation $R_{1} = R_{1} - \frac{7}{5} R_{3}$ to make the entry at $1, 3$ a $0$ .

$[\begin{array}{c} 1 - \frac{7}{5} \cdot 0 & - \frac{4}{5} - \frac{7}{5} \cdot 0 & \frac{7}{5} - \frac{7}{5} \cdot 1 & 0 - \frac{7}{5} (- \frac{11}{9}) & \frac{27}{5} - \frac{7}{5} (- \frac{73}{9}) \\ 0 & 1 & 0 & - \frac{5}{3} & - \frac{37}{3} \\ 0 & 0 & 1 & - \frac{11}{9} & - \frac{73}{9} \end{array}]$

Etapa 2.7.2

Simplifique $R_{1}$ .

$[\begin{array}{c} 1 & - \frac{4}{5} & 0 & \frac{77}{45} & \frac{754}{45} \\ 0 & 1 & 0 & - \frac{5}{3} & - \frac{37}{3} \\ 0 & 0 & 1 & - \frac{11}{9} & - \frac{73}{9} \end{array}]$

Etapa 2.8

Perform the row operation $R_{1} = R_{1} + \frac{4}{5} R_{2}$ to make the entry at $1, 2$ a $0$ .

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Etapa 2.8.1

Perform the row operation $R_{1} = R_{1} + \frac{4}{5} R_{2}$ to make the entry at $1, 2$ a $0$ .

$[\begin{array}{c} 1 + \frac{4}{5} \cdot 0 & - \frac{4}{5} + \frac{4}{5} \cdot 1 & 0 + \frac{4}{5} \cdot 0 & \frac{77}{45} + \frac{4}{5} (- \frac{5}{3}) & \frac{754}{45} + \frac{4}{5} (- \frac{37}{3}) \\ 0 & 1 & 0 & - \frac{5}{3} & - \frac{37}{3} \\ 0 & 0 & 1 & - \frac{11}{9} & - \frac{73}{9} \end{array}]$

Etapa 2.8.2

Simplifique $R_{1}$ .

$[\begin{array}{c} 1 & 0 & 0 & \frac{17}{45} & \frac{62}{9} \\ 0 & 1 & 0 & - \frac{5}{3} & - \frac{37}{3} \\ 0 & 0 & 1 & - \frac{11}{9} & - \frac{73}{9} \end{array}]$

Etapa 3

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

$a + \frac{17 x}{45} = \frac{62}{9}$

$b - \frac{5 x}{3} = - \frac{37}{3}$

$c - \frac{11 x}{9} = - \frac{73}{9}$

Etapa 4

Subtraia $\frac{17 x}{45}$ dos dois lados da equação.

$a = \frac{62}{9} - \frac{17 x}{45}$

$b - \frac{5 x}{3} = - \frac{37}{3}$

$c - \frac{11 x}{9} = - \frac{73}{9}$

Etapa 5

Some $\frac{5 x}{3}$ aos dois lados da equação.

$b = - \frac{37}{3} + \frac{5 x}{3}$

$a = \frac{62}{9} - \frac{17 x}{45}$

$c - \frac{11 x}{9} = - \frac{73}{9}$

Etapa 6

Some $\frac{11 x}{9}$ aos dois lados da equação.

$c = - \frac{73}{9} + \frac{11 x}{9}$

$a = \frac{62}{9} - \frac{17 x}{45}$

$b = - \frac{37}{3} + \frac{5 x}{3}$

Etapa 7

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

$(\frac{62}{9} - \frac{17 x}{45}, - \frac{37}{3} + \frac{5 x}{3}, - \frac{73}{9} + \frac{11 x}{9}, x)$

Etapa 8

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.

$X = [\begin{array}{c} a \\ b \\ c \\ x \end{array}] = [\begin{array}{c} \frac{62}{9} - \frac{17 x}{45} \\ - \frac{37}{3} + \frac{5 x}{3} \\ - \frac{73}{9} + \frac{11 x}{9} \\ x \end{array}]$