Matemática discreta Ejemplos

$x + y - z = 5$ , $3 x - y + z = - 1$ , $x - 2 y + 4 z = 19$

Paso 1

Write the system as a matrix.

$[\begin{array}{c} 1 & 1 & - 1 & 5 \\ 3 & - 1 & 1 & - 1 \\ 1 & - 2 & 4 & 19 \end{array}]$

Paso 2

Obtén la forma escalonada reducida por filas.

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Paso 2.1

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

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Paso 2.1.1

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

$[\begin{array}{c} 1 & 1 & - 1 & 5 \\ 3 - 3 \cdot 1 & - 1 - 3 \cdot 1 & 1 - 3 \cdot - 1 & - 1 - 3 \cdot 5 \\ 1 & - 2 & 4 & 19 \end{array}]$

Paso 2.1.2

Simplifica $R_{2}$ .

$[\begin{array}{c} 1 & 1 & - 1 & 5 \\ 0 & - 4 & 4 & - 16 \\ 1 & - 2 & 4 & 19 \end{array}]$

Paso 2.2

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

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Paso 2.2.1

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

$[\begin{array}{c} 1 & 1 & - 1 & 5 \\ 0 & - 4 & 4 & - 16 \\ 1 - 1 & - 2 - 1 & 4 + 1 & 19 - 5 \end{array}]$

Paso 2.2.2

Simplifica $R_{3}$ .

$[\begin{array}{c} 1 & 1 & - 1 & 5 \\ 0 & - 4 & 4 & - 16 \\ 0 & - 3 & 5 & 14 \end{array}]$

Paso 2.3

Multiply each element of $R_{2}$ by $- \frac{1}{4}$ to make the entry at $2, 2$ a $1$ .

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Paso 2.3.1

Multiply each element of $R_{2}$ by $- \frac{1}{4}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & 1 & - 1 & 5 \\ - \frac{1}{4} \cdot 0 & - \frac{1}{4} \cdot - 4 & - \frac{1}{4} \cdot 4 & - \frac{1}{4} \cdot - 16 \\ 0 & - 3 & 5 & 14 \end{array}]$

Paso 2.3.2

Simplifica $R_{2}$ .

$[\begin{array}{c} 1 & 1 & - 1 & 5 \\ 0 & 1 & - 1 & 4 \\ 0 & - 3 & 5 & 14 \end{array}]$

Paso 2.4

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

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Paso 2.4.1

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

$[\begin{array}{c} 1 & 1 & - 1 & 5 \\ 0 & 1 & - 1 & 4 \\ 0 + 3 \cdot 0 & - 3 + 3 \cdot 1 & 5 + 3 \cdot - 1 & 14 + 3 \cdot 4 \end{array}]$

Paso 2.4.2

Simplifica $R_{3}$ .

$[\begin{array}{c} 1 & 1 & - 1 & 5 \\ 0 & 1 & - 1 & 4 \\ 0 & 0 & 2 & 26 \end{array}]$

Paso 2.5

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

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Paso 2.5.1

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

$[\begin{array}{c} 1 & 1 & - 1 & 5 \\ 0 & 1 & - 1 & 4 \\ \frac{0}{2} & \frac{0}{2} & \frac{2}{2} & \frac{26}{2} \end{array}]$

Paso 2.5.2

Simplifica $R_{3}$ .

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

Paso 2.6

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

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Paso 2.6.1

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

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

Paso 2.6.2

Simplifica $R_{2}$ .

$[\begin{array}{c} 1 & 1 & - 1 & 5 \\ 0 & 1 & 0 & 17 \\ 0 & 0 & 1 & 13 \end{array}]$

Paso 2.7

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

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Paso 2.7.1

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

$[\begin{array}{c} 1 + 0 & 1 + 0 & - 1 + 1 \cdot 1 & 5 + 1 \cdot 13 \\ 0 & 1 & 0 & 17 \\ 0 & 0 & 1 & 13 \end{array}]$

Paso 2.7.2

Simplifica $R_{1}$ .

$[\begin{array}{c} 1 & 1 & 0 & 18 \\ 0 & 1 & 0 & 17 \\ 0 & 0 & 1 & 13 \end{array}]$

Paso 2.8

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

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Paso 2.8.1

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

$[\begin{array}{c} 1 - 0 & 1 - 1 & 0 - 0 & 18 - 17 \\ 0 & 1 & 0 & 17 \\ 0 & 0 & 1 & 13 \end{array}]$

Paso 2.8.2

Simplifica $R_{1}$ .

$[\begin{array}{c} 1 & 0 & 0 & 1 \\ 0 & 1 & 0 & 17 \\ 0 & 0 & 1 & 13 \end{array}]$

Paso 3

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

$x = 1$

$y = 17$

$z = 13$

Paso 4

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

$(1, 17, 13)$