Matemática discreta Ejemplos

$- 6 x + y = 5$ , $2 x - 9 y = 5$

Paso 1

Write the system as a matrix.

$[\begin{array}{c} - 6 & 1 & 5 \\ 2 & - 9 & 5 \end{array}]$

Paso 2

Obtén la forma escalonada reducida por filas.

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Paso 2.1

Multiply each element of $R_{1}$ by $- \frac{1}{6}$ to make the entry at $1, 1$ a $1$ .

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Paso 2.1.1

Multiply each element of $R_{1}$ by $- \frac{1}{6}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} - \frac{1}{6} \cdot - 6 & - \frac{1}{6} \cdot 1 & - \frac{1}{6} \cdot 5 \\ 2 & - 9 & 5 \end{array}]$

Paso 2.1.2

Simplifica $R_{1}$ .

$[\begin{array}{c} 1 & - \frac{1}{6} & - \frac{5}{6} \\ 2 & - 9 & 5 \end{array}]$

Paso 2.2

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

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Paso 2.2.1

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

$[\begin{array}{c} 1 & - \frac{1}{6} & - \frac{5}{6} \\ 2 - 2 \cdot 1 & - 9 - 2 (- \frac{1}{6}) & 5 - 2 (- \frac{5}{6}) \end{array}]$

Paso 2.2.2

Simplifica $R_{2}$ .

$[\begin{array}{c} 1 & - \frac{1}{6} & - \frac{5}{6} \\ 0 & - \frac{26}{3} & \frac{20}{3} \end{array}]$

Paso 2.3

Multiply each element of $R_{2}$ by $- \frac{3}{26}$ to make the entry at $2, 2$ a $1$ .

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Paso 2.3.1

Multiply each element of $R_{2}$ by $- \frac{3}{26}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & - \frac{1}{6} & - \frac{5}{6} \\ - \frac{3}{26} \cdot 0 & - \frac{3}{26} (- \frac{26}{3}) & - \frac{3}{26} \cdot \frac{20}{3} \end{array}]$

Paso 2.3.2

Simplifica $R_{2}$ .

$[\begin{array}{c} 1 & - \frac{1}{6} & - \frac{5}{6} \\ 0 & 1 & - \frac{10}{13} \end{array}]$

Paso 2.4

Perform the row operation $R_{1} = R_{1} + \frac{1}{6} R_{2}$ to make the entry at $1, 2$ a $0$ .

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Paso 2.4.1

Perform the row operation $R_{1} = R_{1} + \frac{1}{6} R_{2}$ to make the entry at $1, 2$ a $0$ .

$[\begin{array}{c} 1 + \frac{1}{6} \cdot 0 & - \frac{1}{6} + \frac{1}{6} \cdot 1 & - \frac{5}{6} + \frac{1}{6} (- \frac{10}{13}) \\ 0 & 1 & - \frac{10}{13} \end{array}]$

Paso 2.4.2

Simplifica $R_{1}$ .

$[\begin{array}{c} 1 & 0 & - \frac{25}{26} \\ 0 & 1 & - \frac{10}{13} \end{array}]$

Paso 3

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

$x = - \frac{25}{26}$

$y = - \frac{10}{13}$

Paso 4

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

$(- \frac{25}{26}, - \frac{10}{13})$