Matemática discreta Exemplos

$3 x + 5 y - z = 2$ , $4 x + 7 y + z = 4$

Etapa 1

Write the system as a matrix.

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

Etapa 2

Encontre a forma escalonada reduzida por linhas.

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Etapa 2.1

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

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Etapa 2.1.1

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

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

Etapa 2.1.2

Simplifique $R_{1}$ .

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

Etapa 2.2

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

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Etapa 2.2.1

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

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

Etapa 2.2.2

Simplifique $R_{2}$ .

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

Etapa 2.3

Multiply each element of $R_{2}$ by $3$ to make the entry at $2, 2$ a $1$ .

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Etapa 2.3.1

Multiply each element of $R_{2}$ by $3$ to make the entry at $2, 2$ a $1$ .

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

Etapa 2.3.2

Simplifique $R_{2}$ .

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

Etapa 2.4

Perform the row operation $R_{1} = R_{1} - \frac{5}{3} R_{2}$ to make the entry at $1, 2$ a $0$ .

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Etapa 2.4.1

Perform the row operation $R_{1} = R_{1} - \frac{5}{3} R_{2}$ to make the entry at $1, 2$ a $0$ .

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

Etapa 2.4.2

Simplifique $R_{1}$ .

$[\begin{array}{c} 1 & 0 & - 12 & - 6 \\ 0 & 1 & 7 & 4 \end{array}]$

Etapa 3

Use the result matrix to declare the final solution to the system of equations.

$x - 12 z = - 6$

$y + 7 z = 4$

Etapa 4

The solution is the set of ordered pairs that make the system true.

$(- 6 + 12 z, 4 - 7 z, z)$