Matemática discreta Exemplos

$x = \frac{30}{7}$ , $y = \frac{16}{7}$ , $u = \frac{30}{7}$ , $y = \frac{16}{7}$ , $p = \frac{220}{7}$

Etapa 1

Write the system as a matrix.

$[\begin{array}{c} 0 & 0 & 1 & 0 & \frac{30}{7} \\ 0 & 0 & 0 & 1 & \frac{16}{7} \\ 0 & 1 & 0 & 0 & \frac{30}{7} \\ 0 & 0 & 0 & 1 & \frac{16}{7} \\ 1 & 0 & 0 & 0 & \frac{220}{7} \end{array}]$

Etapa 2

Encontre a forma escalonada reduzida por linhas.

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Etapa 2.1

Swap $R_{5}$ with $R_{1}$ to put a nonzero entry at $1, 1$ .

$[\begin{array}{c} 1 & 0 & 0 & 0 & \frac{220}{7} \\ 0 & 0 & 0 & 1 & \frac{16}{7} \\ 0 & 1 & 0 & 0 & \frac{30}{7} \\ 0 & 0 & 0 & 1 & \frac{16}{7} \\ 0 & 0 & 1 & 0 & \frac{30}{7} \end{array}]$

Etapa 2.2

Swap $R_{3}$ with $R_{2}$ to put a nonzero entry at $2, 2$ .

$[\begin{array}{c} 1 & 0 & 0 & 0 & \frac{220}{7} \\ 0 & 1 & 0 & 0 & \frac{30}{7} \\ 0 & 0 & 0 & 1 & \frac{16}{7} \\ 0 & 0 & 0 & 1 & \frac{16}{7} \\ 0 & 0 & 1 & 0 & \frac{30}{7} \end{array}]$

Etapa 2.3

Swap $R_{5}$ with $R_{3}$ to put a nonzero entry at $3, 3$ .

$[\begin{array}{c} 1 & 0 & 0 & 0 & \frac{220}{7} \\ 0 & 1 & 0 & 0 & \frac{30}{7} \\ 0 & 0 & 1 & 0 & \frac{30}{7} \\ 0 & 0 & 0 & 1 & \frac{16}{7} \\ 0 & 0 & 0 & 1 & \frac{16}{7} \end{array}]$

Etapa 2.4

Perform the row operation $R_{5} = R_{5} - R_{4}$ to make the entry at $5, 4$ a $0$ .

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Etapa 2.4.1

Perform the row operation $R_{5} = R_{5} - R_{4}$ to make the entry at $5, 4$ a $0$ .

$[\begin{array}{c} 1 & 0 & 0 & 0 & \frac{220}{7} \\ 0 & 1 & 0 & 0 & \frac{30}{7} \\ 0 & 0 & 1 & 0 & \frac{30}{7} \\ 0 & 0 & 0 & 1 & \frac{16}{7} \\ 0 - 0 & 0 - 0 & 0 - 0 & 1 - 1 & \frac{16}{7} - \frac{16}{7} \end{array}]$

Etapa 2.4.2

Simplifique $R_{5}$ .

$[\begin{array}{c} 1 & 0 & 0 & 0 & \frac{220}{7} \\ 0 & 1 & 0 & 0 & \frac{30}{7} \\ 0 & 0 & 1 & 0 & \frac{30}{7} \\ 0 & 0 & 0 & 1 & \frac{16}{7} \\ 0 & 0 & 0 & 0 & 0 \end{array}]$

Etapa 3

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

$p = \frac{220}{7}$

$u = \frac{30}{7}$

$x = \frac{30}{7}$

$y = \frac{16}{7}$

$0 = 0$

Etapa 4

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

$(\frac{220}{7}, \frac{30}{7}, \frac{30}{7}, \frac{16}{7})$