Álgebra lineal Ejemplos

$- x + y + z = 0$ , $12 - 250 y + 680 z = 0$ , $10 - 100 x - 680 = 0$

Paso 1

Resta $12$ de ambos lados de la ecuación.

$- x + y + z = 0, - 250 y + 680 z = - 12, 10 - 100 x - 680 = 0$

Paso 2

Resta $680$ de $10$ .

$- x + y + z = 0, - 250 y + 680 z = - 12, - 100 x - 670 = 0$

Paso 3

Suma $670$ a ambos lados de la ecuación.

$- x + y + z = 0, - 250 y + 680 z = - 12, - 100 x = 670$

Paso 4

Escribe el sistema de ecuaciones en forma de matriz.

$[\begin{array}{c} - 1 & 1 & 1 & 0 \\ 0 & - 250 & 680 & - 12 \\ - 100 & 0 & 0 & 670 \end{array}]$

Paso 5

Obtén la forma escalonada reducida por filas.

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Paso 5.1

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

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Paso 5.1.1

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

$[\begin{array}{c} - - 1 & - 1 \cdot 1 & - 1 \cdot 1 & - 0 \\ 0 & - 250 & 680 & - 12 \\ - 100 & 0 & 0 & 670 \end{array}]$

Paso 5.1.2

Simplifica $R_{1}$ .

$[\begin{array}{c} 1 & - 1 & - 1 & 0 \\ 0 & - 250 & 680 & - 12 \\ - 100 & 0 & 0 & 670 \end{array}]$

Paso 5.2

Perform the row operation $R_{3} = R_{3} + 100 R_{1}$ to make the entry at $3, 1$ a $0$ .

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Paso 5.2.1

Perform the row operation $R_{3} = R_{3} + 100 R_{1}$ to make the entry at $3, 1$ a $0$ .

$[\begin{array}{c} 1 & - 1 & - 1 & 0 \\ 0 & - 250 & 680 & - 12 \\ - 100 + 100 \cdot 1 & 0 + 100 \cdot - 1 & 0 + 100 \cdot - 1 & 670 + 100 \cdot 0 \end{array}]$

Paso 5.2.2

Simplifica $R_{3}$ .

$[\begin{array}{c} 1 & - 1 & - 1 & 0 \\ 0 & - 250 & 680 & - 12 \\ 0 & - 100 & - 100 & 670 \end{array}]$

Paso 5.3

Multiply each element of $R_{2}$ by $- \frac{1}{250}$ to make the entry at $2, 2$ a $1$ .

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Paso 5.3.1

Multiply each element of $R_{2}$ by $- \frac{1}{250}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & - 1 & - 1 & 0 \\ - \frac{1}{250} \cdot 0 & - \frac{1}{250} \cdot - 250 & - \frac{1}{250} \cdot 680 & - \frac{1}{250} \cdot - 12 \\ 0 & - 100 & - 100 & 670 \end{array}]$

Paso 5.3.2

Simplifica $R_{2}$ .

$[\begin{array}{c} 1 & - 1 & - 1 & 0 \\ 0 & 1 & - \frac{68}{25} & \frac{6}{125} \\ 0 & - 100 & - 100 & 670 \end{array}]$

Paso 5.4

Perform the row operation $R_{3} = R_{3} + 100 R_{2}$ to make the entry at $3, 2$ a $0$ .

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Paso 5.4.1

Perform the row operation $R_{3} = R_{3} + 100 R_{2}$ to make the entry at $3, 2$ a $0$ .

$[\begin{array}{c} 1 & - 1 & - 1 & 0 \\ 0 & 1 & - \frac{68}{25} & \frac{6}{125} \\ 0 + 100 \cdot 0 & - 100 + 100 \cdot 1 & - 100 + 100 (- \frac{68}{25}) & 670 + 100 (\frac{6}{125}) \end{array}]$

Paso 5.4.2

Simplifica $R_{3}$ .

$[\begin{array}{c} 1 & - 1 & - 1 & 0 \\ 0 & 1 & - \frac{68}{25} & \frac{6}{125} \\ 0 & 0 & - 372 & \frac{3374}{5} \end{array}]$

Paso 5.5

Multiply each element of $R_{3}$ by $- \frac{1}{372}$ to make the entry at $3, 3$ a $1$ .

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Paso 5.5.1

Multiply each element of $R_{3}$ by $- \frac{1}{372}$ to make the entry at $3, 3$ a $1$ .

$[\begin{array}{c} 1 & - 1 & - 1 & 0 \\ 0 & 1 & - \frac{68}{25} & \frac{6}{125} \\ - \frac{1}{372} \cdot 0 & - \frac{1}{372} \cdot 0 & - \frac{1}{372} \cdot - 372 & - \frac{1}{372} \cdot \frac{3374}{5} \end{array}]$

Paso 5.5.2

Simplifica $R_{3}$ .

$[\begin{array}{c} 1 & - 1 & - 1 & 0 \\ 0 & 1 & - \frac{68}{25} & \frac{6}{125} \\ 0 & 0 & 1 & - \frac{1687}{930} \end{array}]$

Paso 5.6

Perform the row operation $R_{2} = R_{2} + \frac{68}{25} R_{3}$ to make the entry at $2, 3$ a $0$ .

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Paso 5.6.1

Perform the row operation $R_{2} = R_{2} + \frac{68}{25} R_{3}$ to make the entry at $2, 3$ a $0$ .

$[\begin{array}{c} 1 & - 1 & - 1 & 0 \\ 0 + \frac{68}{25} \cdot 0 & 1 + \frac{68}{25} \cdot 0 & - \frac{68}{25} + \frac{68}{25} \cdot 1 & \frac{6}{125} + \frac{68}{25} (- \frac{1687}{930}) \\ 0 & 0 & 1 & - \frac{1687}{930} \end{array}]$

Paso 5.6.2

Simplifica $R_{2}$ .

$[\begin{array}{c} 1 & - 1 & - 1 & 0 \\ 0 & 1 & 0 & - \frac{2272}{465} \\ 0 & 0 & 1 & - \frac{1687}{930} \end{array}]$

Paso 5.7

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

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Paso 5.7.1

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

$[\begin{array}{c} 1 + 0 & - 1 + 0 & - 1 + 1 \cdot 1 & 0 - \frac{1687}{930} \\ 0 & 1 & 0 & - \frac{2272}{465} \\ 0 & 0 & 1 & - \frac{1687}{930} \end{array}]$

Paso 5.7.2

Simplifica $R_{1}$ .

$[\begin{array}{c} 1 & - 1 & 0 & - \frac{1687}{930} \\ 0 & 1 & 0 & - \frac{2272}{465} \\ 0 & 0 & 1 & - \frac{1687}{930} \end{array}]$

Paso 5.8

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

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Paso 5.8.1

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

$[\begin{array}{c} 1 + 0 & - 1 + 1 \cdot 1 & 0 + 0 & - \frac{1687}{930} - \frac{2272}{465} \\ 0 & 1 & 0 & - \frac{2272}{465} \\ 0 & 0 & 1 & - \frac{1687}{930} \end{array}]$

Paso 5.8.2

Simplifica $R_{1}$ .

$[\begin{array}{c} 1 & 0 & 0 & - \frac{67}{10} \\ 0 & 1 & 0 & - \frac{2272}{465} \\ 0 & 0 & 1 & - \frac{1687}{930} \end{array}]$

Paso 6

Usa la matriz de resultados para declarar las soluciones finales en el sistema de ecuaciones.

$x = - \frac{67}{10}$

$y = - \frac{2272}{465}$

$z = - \frac{1687}{930}$

Paso 7

La solución es el conjunto de pares ordenados que hacen que el sistema sea verdadero.

$(- \frac{67}{10}, - \frac{2272}{465}, - \frac{1687}{930})$

Paso 8

Descompone un vector de solución mediante la reorganización de cada ecuación representada en la forma reducida de fila de la matriz aumentada, a través de la resolución para la variable dependiente en cada fila, se obtiene la igualdad del vector.

$X = [\begin{array}{c} x \\ y \\ z \end{array}] = [\begin{array}{c} - \frac{67}{10} \\ - \frac{2272}{465} \\ - \frac{1687}{930} \end{array}]$