Matemática discreta Ejemplos

$- 4 x + 5 y = 13$ , $- 12 x + 15 y = 39$

Paso 1

Write the system as a matrix.

$[\begin{array}{c} - 4 & 5 & 13 \\ - 12 & 15 & 39 \end{array}]$

Paso 2

Obtén la forma escalonada reducida por filas.

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Paso 2.1

Multiply each element of $R_{1}$ by $- \frac{1}{4}$ to make the entry at $1, 1$ a $1$ .

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Paso 2.1.1

Multiply each element of $R_{1}$ by $- \frac{1}{4}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} - \frac{1}{4} \cdot - 4 & - \frac{1}{4} \cdot 5 & - \frac{1}{4} \cdot 13 \\ - 12 & 15 & 39 \end{array}]$

Paso 2.1.2

Simplifica $R_{1}$ .

$[\begin{array}{c} 1 & - \frac{5}{4} & - \frac{13}{4} \\ - 12 & 15 & 39 \end{array}]$

Paso 2.2

Perform the row operation $R_{2} = R_{2} + 12 R_{1}$ to make the entry at $2, 1$ a $0$ .

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Paso 2.2.1

Perform the row operation $R_{2} = R_{2} + 12 R_{1}$ to make the entry at $2, 1$ a $0$ .

$[\begin{array}{c} 1 & - \frac{5}{4} & - \frac{13}{4} \\ - 12 + 12 \cdot 1 & 15 + 12 (- \frac{5}{4}) & 39 + 12 (- \frac{13}{4}) \end{array}]$

Paso 2.2.2

Simplifica $R_{2}$ .

$[\begin{array}{c} 1 & - \frac{5}{4} & - \frac{13}{4} \\ 0 & 0 & 0 \end{array}]$

Paso 3

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

$x - \frac{5}{4} y = - \frac{13}{4}$

$0 = 0$

Paso 4

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

$(- \frac{13}{4} + \frac{5 y}{4}, y)$