Álgebra linear Exemplos

$41 a - 4 b + 5 c + d = 0$ , $53 a - 2 b + 7 c + d = 0$ , $25 a + 4 b - 3 c + d = 0$

Etapa 1

Escreva o sistema de equações em formato de matriz.

$[\begin{array}{c} 41 & - 4 & 5 & 1 & 0 \\ 53 & - 2 & 7 & 1 & 0 \\ 25 & 4 & - 3 & 1 & 0 \end{array}]$

Etapa 2

Encontre a forma escalonada reduzida por linhas.

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Etapa 2.1

Multiply each element of $R_{1}$ by $\frac{1}{41}$ to make the entry at $1, 1$ a $1$ .

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Etapa 2.1.1

Multiply each element of $R_{1}$ by $\frac{1}{41}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} \frac{41}{41} & \frac{- 4}{41} & \frac{5}{41} & \frac{1}{41} & \frac{0}{41} \\ 53 & - 2 & 7 & 1 & 0 \\ 25 & 4 & - 3 & 1 & 0 \end{array}]$

Etapa 2.1.2

Simplifique $R_{1}$ .

$[\begin{array}{c} 1 & - \frac{4}{41} & \frac{5}{41} & \frac{1}{41} & 0 \\ 53 & - 2 & 7 & 1 & 0 \\ 25 & 4 & - 3 & 1 & 0 \end{array}]$

Etapa 2.2

Perform the row operation $R_{2} = R_{2} - 53 R_{1}$ to make the entry at $2, 1$ a $0$ .

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Etapa 2.2.1

Perform the row operation $R_{2} = R_{2} - 53 R_{1}$ to make the entry at $2, 1$ a $0$ .

$[\begin{array}{c} 1 & - \frac{4}{41} & \frac{5}{41} & \frac{1}{41} & 0 \\ 53 - 53 \cdot 1 & - 2 - 53 (- \frac{4}{41}) & 7 - 53 (\frac{5}{41}) & 1 - 53 (\frac{1}{41}) & 0 - 53 \cdot 0 \\ 25 & 4 & - 3 & 1 & 0 \end{array}]$

Etapa 2.2.2

Simplifique $R_{2}$ .

$[\begin{array}{c} 1 & - \frac{4}{41} & \frac{5}{41} & \frac{1}{41} & 0 \\ 0 & \frac{130}{41} & \frac{22}{41} & - \frac{12}{41} & 0 \\ 25 & 4 & - 3 & 1 & 0 \end{array}]$

Etapa 2.3

Perform the row operation $R_{3} = R_{3} - 25 R_{1}$ to make the entry at $3, 1$ a $0$ .

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Etapa 2.3.1

Perform the row operation $R_{3} = R_{3} - 25 R_{1}$ to make the entry at $3, 1$ a $0$ .

$[\begin{array}{c} 1 & - \frac{4}{41} & \frac{5}{41} & \frac{1}{41} & 0 \\ 0 & \frac{130}{41} & \frac{22}{41} & - \frac{12}{41} & 0 \\ 25 - 25 \cdot 1 & 4 - 25 (- \frac{4}{41}) & - 3 - 25 (\frac{5}{41}) & 1 - 25 (\frac{1}{41}) & 0 - 25 \cdot 0 \end{array}]$

Etapa 2.3.2

Simplifique $R_{3}$ .

$[\begin{array}{c} 1 & - \frac{4}{41} & \frac{5}{41} & \frac{1}{41} & 0 \\ 0 & \frac{130}{41} & \frac{22}{41} & - \frac{12}{41} & 0 \\ 0 & \frac{264}{41} & - \frac{248}{41} & \frac{16}{41} & 0 \end{array}]$

Etapa 2.4

Multiply each element of $R_{2}$ by $\frac{41}{130}$ to make the entry at $2, 2$ a $1$ .

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Etapa 2.4.1

Multiply each element of $R_{2}$ by $\frac{41}{130}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & - \frac{4}{41} & \frac{5}{41} & \frac{1}{41} & 0 \\ \frac{41}{130} \cdot 0 & \frac{41}{130} \cdot \frac{130}{41} & \frac{41}{130} \cdot \frac{22}{41} & \frac{41}{130} (- \frac{12}{41}) & \frac{41}{130} \cdot 0 \\ 0 & \frac{264}{41} & - \frac{248}{41} & \frac{16}{41} & 0 \end{array}]$

Etapa 2.4.2

Simplifique $R_{2}$ .

$[\begin{array}{c} 1 & - \frac{4}{41} & \frac{5}{41} & \frac{1}{41} & 0 \\ 0 & 1 & \frac{11}{65} & - \frac{6}{65} & 0 \\ 0 & \frac{264}{41} & - \frac{248}{41} & \frac{16}{41} & 0 \end{array}]$

Etapa 2.5

Perform the row operation $R_{3} = R_{3} - \frac{264}{41} R_{2}$ to make the entry at $3, 2$ a $0$ .

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Etapa 2.5.1

Perform the row operation $R_{3} = R_{3} - \frac{264}{41} R_{2}$ to make the entry at $3, 2$ a $0$ .

$[\begin{array}{c} 1 & - \frac{4}{41} & \frac{5}{41} & \frac{1}{41} & 0 \\ 0 & 1 & \frac{11}{65} & - \frac{6}{65} & 0 \\ 0 - \frac{264}{41} \cdot 0 & \frac{264}{41} - \frac{264}{41} \cdot 1 & - \frac{248}{41} - \frac{264}{41} \cdot \frac{11}{65} & \frac{16}{41} - \frac{264}{41} (- \frac{6}{65}) & 0 - \frac{264}{41} \cdot 0 \end{array}]$

Etapa 2.5.2

Simplifique $R_{3}$ .

$[\begin{array}{c} 1 & - \frac{4}{41} & \frac{5}{41} & \frac{1}{41} & 0 \\ 0 & 1 & \frac{11}{65} & - \frac{6}{65} & 0 \\ 0 & 0 & - \frac{464}{65} & \frac{64}{65} & 0 \end{array}]$

Etapa 2.6

Multiply each element of $R_{3}$ by $- \frac{65}{464}$ to make the entry at $3, 3$ a $1$ .

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Etapa 2.6.1

Multiply each element of $R_{3}$ by $- \frac{65}{464}$ to make the entry at $3, 3$ a $1$ .

$[\begin{array}{c} 1 & - \frac{4}{41} & \frac{5}{41} & \frac{1}{41} & 0 \\ 0 & 1 & \frac{11}{65} & - \frac{6}{65} & 0 \\ - \frac{65}{464} \cdot 0 & - \frac{65}{464} \cdot 0 & - \frac{65}{464} (- \frac{464}{65}) & - \frac{65}{464} \cdot \frac{64}{65} & - \frac{65}{464} \cdot 0 \end{array}]$

Etapa 2.6.2

Simplifique $R_{3}$ .

$[\begin{array}{c} 1 & - \frac{4}{41} & \frac{5}{41} & \frac{1}{41} & 0 \\ 0 & 1 & \frac{11}{65} & - \frac{6}{65} & 0 \\ 0 & 0 & 1 & - \frac{4}{29} & 0 \end{array}]$

Etapa 2.7

Perform the row operation $R_{2} = R_{2} - \frac{11}{65} R_{3}$ to make the entry at $2, 3$ a $0$ .

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Etapa 2.7.1

Perform the row operation $R_{2} = R_{2} - \frac{11}{65} R_{3}$ to make the entry at $2, 3$ a $0$ .

$[\begin{array}{c} 1 & - \frac{4}{41} & \frac{5}{41} & \frac{1}{41} & 0 \\ 0 - \frac{11}{65} \cdot 0 & 1 - \frac{11}{65} \cdot 0 & \frac{11}{65} - \frac{11}{65} \cdot 1 & - \frac{6}{65} - \frac{11}{65} (- \frac{4}{29}) & 0 - \frac{11}{65} \cdot 0 \\ 0 & 0 & 1 & - \frac{4}{29} & 0 \end{array}]$

Etapa 2.7.2

Simplifique $R_{2}$ .

$[\begin{array}{c} 1 & - \frac{4}{41} & \frac{5}{41} & \frac{1}{41} & 0 \\ 0 & 1 & 0 & - \frac{2}{29} & 0 \\ 0 & 0 & 1 & - \frac{4}{29} & 0 \end{array}]$

Etapa 2.8

Perform the row operation $R_{1} = R_{1} - \frac{5}{41} R_{3}$ to make the entry at $1, 3$ a $0$ .

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Etapa 2.8.1

Perform the row operation $R_{1} = R_{1} - \frac{5}{41} R_{3}$ to make the entry at $1, 3$ a $0$ .

$[\begin{array}{c} 1 - \frac{5}{41} \cdot 0 & - \frac{4}{41} - \frac{5}{41} \cdot 0 & \frac{5}{41} - \frac{5}{41} \cdot 1 & \frac{1}{41} - \frac{5}{41} (- \frac{4}{29}) & 0 - \frac{5}{41} \cdot 0 \\ 0 & 1 & 0 & - \frac{2}{29} & 0 \\ 0 & 0 & 1 & - \frac{4}{29} & 0 \end{array}]$

Etapa 2.8.2

Simplifique $R_{1}$ .

$[\begin{array}{c} 1 & - \frac{4}{41} & 0 & \frac{49}{1189} & 0 \\ 0 & 1 & 0 & - \frac{2}{29} & 0 \\ 0 & 0 & 1 & - \frac{4}{29} & 0 \end{array}]$

Etapa 2.9

Perform the row operation $R_{1} = R_{1} + \frac{4}{41} R_{2}$ to make the entry at $1, 2$ a $0$ .

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Etapa 2.9.1

Perform the row operation $R_{1} = R_{1} + \frac{4}{41} R_{2}$ to make the entry at $1, 2$ a $0$ .

$[\begin{array}{c} 1 + \frac{4}{41} \cdot 0 & - \frac{4}{41} + \frac{4}{41} \cdot 1 & 0 + \frac{4}{41} \cdot 0 & \frac{49}{1189} + \frac{4}{41} (- \frac{2}{29}) & 0 + \frac{4}{41} \cdot 0 \\ 0 & 1 & 0 & - \frac{2}{29} & 0 \\ 0 & 0 & 1 & - \frac{4}{29} & 0 \end{array}]$

Etapa 2.9.2

Simplifique $R_{1}$ .

$[\begin{array}{c} 1 & 0 & 0 & \frac{1}{29} & 0 \\ 0 & 1 & 0 & - \frac{2}{29} & 0 \\ 0 & 0 & 1 & - \frac{4}{29} & 0 \end{array}]$

Etapa 3

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

$a + \frac{d}{29} = 0$

$b - \frac{2 d}{29} = 0$

$c - \frac{4 d}{29} = 0$

Etapa 4

Subtraia $\frac{d}{29}$ dos dois lados da equação.

$a = - \frac{d}{29}$

$b - \frac{2 d}{29} = 0$

$c - \frac{4 d}{29} = 0$

Etapa 5

Some $\frac{2 d}{29}$ aos dois lados da equação.

$b = \frac{2 d}{29}$

$a = - \frac{d}{29}$

$c - \frac{4 d}{29} = 0$

Etapa 6

Some $\frac{4 d}{29}$ aos dois lados da equação.

$c = \frac{4 d}{29}$

$a = - \frac{d}{29}$

$b = \frac{2 d}{29}$

Etapa 7

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

$(- \frac{d}{29}, \frac{2 d}{29}, \frac{4 d}{29}, d)$

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.

$X = [\begin{array}{c} a \\ b \\ c \\ d \end{array}] = [\begin{array}{c} - \frac{d}{29} \\ \frac{2 d}{29} \\ \frac{4 d}{29} \\ d \end{array}]$