Matemática discreta Exemplos

$x + 3 y = - 12$ , $- x + 7 y = - 412$

Etapa 1

Write the system as a matrix.

$[\begin{array}{c} 1 & 3 & - 12 \\ - 1 & 7 & - 412 \end{array}]$

Etapa 2

Encontre a forma escalonada reduzida por linhas.

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Etapa 2.1

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

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Etapa 2.1.1

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

$[\begin{array}{c} 1 & 3 & - 12 \\ - 1 + 1 \cdot 1 & 7 + 1 \cdot 3 & - 412 - 12 \end{array}]$

Etapa 2.1.2

Simplifique $R_{2}$ .

$[\begin{array}{c} 1 & 3 & - 12 \\ 0 & 10 & - 424 \end{array}]$

Etapa 2.2

Multiply each element of $R_{2}$ by $\frac{1}{10}$ to make the entry at $2, 2$ a $1$ .

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Etapa 2.2.1

Multiply each element of $R_{2}$ by $\frac{1}{10}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & 3 & - 12 \\ \frac{0}{10} & \frac{10}{10} & \frac{- 424}{10} \end{array}]$

Etapa 2.2.2

Simplifique $R_{2}$ .

$[\begin{array}{c} 1 & 3 & - 12 \\ 0 & 1 & - \frac{212}{5} \end{array}]$

Etapa 2.3

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

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Etapa 2.3.1

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

$[\begin{array}{c} 1 - 3 \cdot 0 & 3 - 3 \cdot 1 & - 12 - 3 (- \frac{212}{5}) \\ 0 & 1 & - \frac{212}{5} \end{array}]$

Etapa 2.3.2

Simplifique $R_{1}$ .

$[\begin{array}{c} 1 & 0 & \frac{576}{5} \\ 0 & 1 & - \frac{212}{5} \end{array}]$

Etapa 3

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

$x = \frac{576}{5}$

$y = - \frac{212}{5}$

Etapa 4

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

$(\frac{576}{5}, - \frac{212}{5})$