Matemática discreta Ejemplos

$- x - 3 y = 12$ , $2 x + 6 y = - 24$

Paso 1

Write the system as a matrix.

$[\begin{array}{c} - 1 & - 3 & 12 \\ 2 & 6 & - 24 \end{array}]$

Paso 2

Obtén la forma escalonada reducida por filas.

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Paso 2.1

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

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Paso 2.1.1

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

$[\begin{array}{c} - - 1 & - - 3 & - 1 \cdot 12 \\ 2 & 6 & - 24 \end{array}]$

Paso 2.1.2

Simplifica $R_{1}$ .

$[\begin{array}{c} 1 & 3 & - 12 \\ 2 & 6 & - 24 \end{array}]$

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 $0$ .

$[\begin{array}{c} 1 & 3 & - 12 \\ 2 - 2 \cdot 1 & 6 - 2 \cdot 3 & - 24 - 2 \cdot - 12 \end{array}]$

Paso 2.2.2

Simplifica $R_{2}$ .

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

Paso 3

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

$x + 3 y = - 12$

$0 = 0$

Paso 4

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

$(- 12 - 3 y, y)$