Matemática discreta Ejemplos

$x + 3 y + z = 1$ , $x - z = 0$ , $- 2 x + 3 y + 4 z = 1$

Paso 1

Write the system as a matrix.

$[\begin{array}{c} 1 & 3 & 1 & 1 \\ 1 & 0 & - 1 & 0 \\ - 2 & 3 & 4 & 1 \end{array}]$

Paso 2

Obtén la forma escalonada reducida por filas.

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Paso 2.1

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

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Paso 2.1.1

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

$[\begin{array}{c} 1 & 3 & 1 & 1 \\ 1 - 1 & 0 - 3 & - 1 - 1 & 0 - 1 \\ - 2 & 3 & 4 & 1 \end{array}]$

Paso 2.1.2

Simplifica $R_{2}$ .

$[\begin{array}{c} 1 & 3 & 1 & 1 \\ 0 & - 3 & - 2 & - 1 \\ - 2 & 3 & 4 & 1 \end{array}]$

Paso 2.2

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

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Paso 2.2.1

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

$[\begin{array}{c} 1 & 3 & 1 & 1 \\ 0 & - 3 & - 2 & - 1 \\ - 2 + 2 \cdot 1 & 3 + 2 \cdot 3 & 4 + 2 \cdot 1 & 1 + 2 \cdot 1 \end{array}]$

Paso 2.2.2

Simplifica $R_{3}$ .

$[\begin{array}{c} 1 & 3 & 1 & 1 \\ 0 & - 3 & - 2 & - 1 \\ 0 & 9 & 6 & 3 \end{array}]$

Paso 2.3

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

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Paso 2.3.1

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

$[\begin{array}{c} 1 & 3 & 1 & 1 \\ - \frac{1}{3} \cdot 0 & - \frac{1}{3} \cdot - 3 & - \frac{1}{3} \cdot - 2 & - \frac{1}{3} \cdot - 1 \\ 0 & 9 & 6 & 3 \end{array}]$

Paso 2.3.2

Simplifica $R_{2}$ .

$[\begin{array}{c} 1 & 3 & 1 & 1 \\ 0 & 1 & \frac{2}{3} & \frac{1}{3} \\ 0 & 9 & 6 & 3 \end{array}]$

Paso 2.4

Perform the row operation $R_{3} = R_{3} - 9 R_{2}$ to make the entry at $3, 2$ a $0$ .

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Paso 2.4.1

Perform the row operation $R_{3} = R_{3} - 9 R_{2}$ to make the entry at $3, 2$ a $0$ .

$[\begin{array}{c} 1 & 3 & 1 & 1 \\ 0 & 1 & \frac{2}{3} & \frac{1}{3} \\ 0 - 9 \cdot 0 & 9 - 9 \cdot 1 & 6 - 9 (\frac{2}{3}) & 3 - 9 (\frac{1}{3}) \end{array}]$

Paso 2.4.2

Simplifica $R_{3}$ .

$[\begin{array}{c} 1 & 3 & 1 & 1 \\ 0 & 1 & \frac{2}{3} & \frac{1}{3} \\ 0 & 0 & 0 & 0 \end{array}]$

Paso 2.5

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

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Paso 2.5.1

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

$[\begin{array}{c} 1 - 3 \cdot 0 & 3 - 3 \cdot 1 & 1 - 3 (\frac{2}{3}) & 1 - 3 (\frac{1}{3}) \\ 0 & 1 & \frac{2}{3} & \frac{1}{3} \\ 0 & 0 & 0 & 0 \end{array}]$

Paso 2.5.2

Simplifica $R_{1}$ .

$[\begin{array}{c} 1 & 0 & - 1 & 0 \\ 0 & 1 & \frac{2}{3} & \frac{1}{3} \\ 0 & 0 & 0 & 0 \end{array}]$

Paso 3

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

$x - z = 0$

$y + \frac{2}{3} z = \frac{1}{3}$

$0 = 0$

Paso 4

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

$(z, \frac{1}{3} - \frac{2 z}{3}, z)$