Álgebra lineal Ejemplos

$3 x - 12 y + 18 z = - 73$ , $2 x + 4 y - 3 z = 9$ , $x - 4 y + 6 z = - 24$

Paso 1

Escribe el sistema de ecuaciones en forma de matriz.

$[\begin{array}{c} 3 & - 12 & 18 & - 73 \\ 2 & 4 & - 3 & 9 \\ 1 & - 4 & 6 & - 24 \end{array}]$

Paso 2

Obtén la forma escalonada reducida por filas.

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Paso 2.1

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

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Paso 2.1.1

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

$[\begin{array}{c} \frac{3}{3} & \frac{- 12}{3} & \frac{18}{3} & \frac{- 73}{3} \\ 2 & 4 & - 3 & 9 \\ 1 & - 4 & 6 & - 24 \end{array}]$

Paso 2.1.2

Simplifica $R_{1}$ .

$[\begin{array}{c} 1 & - 4 & 6 & - \frac{73}{3} \\ 2 & 4 & - 3 & 9 \\ 1 & - 4 & 6 & - 24 \end{array}]$

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 $0$ .

$[\begin{array}{c} 1 & - 4 & 6 & - \frac{73}{3} \\ 2 - 2 \cdot 1 & 4 - 2 \cdot - 4 & - 3 - 2 \cdot 6 & 9 - 2 (- \frac{73}{3}) \\ 1 & - 4 & 6 & - 24 \end{array}]$

Paso 2.2.2

Simplifica $R_{2}$ .

$[\begin{array}{c} 1 & - 4 & 6 & - \frac{73}{3} \\ 0 & 12 & - 15 & \frac{173}{3} \\ 1 & - 4 & 6 & - 24 \end{array}]$

Paso 2.3

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

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Paso 2.3.1

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

$[\begin{array}{c} 1 & - 4 & 6 & - \frac{73}{3} \\ 0 & 12 & - 15 & \frac{173}{3} \\ 1 - 1 & - 4 + 4 & 6 - 6 & - 24 + \frac{73}{3} \end{array}]$

Paso 2.3.2

Simplifica $R_{3}$ .

$[\begin{array}{c} 1 & - 4 & 6 & - \frac{73}{3} \\ 0 & 12 & - 15 & \frac{173}{3} \\ 0 & 0 & 0 & \frac{1}{3} \end{array}]$

Paso 2.4

Multiply each element of $R_{2}$ by $\frac{1}{12}$ to make the entry at $2, 2$ a $1$ .

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Paso 2.4.1

Multiply each element of $R_{2}$ by $\frac{1}{12}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & - 4 & 6 & - \frac{73}{3} \\ \frac{0}{12} & \frac{12}{12} & \frac{- 15}{12} & \frac{\frac{173}{3}}{12} \\ 0 & 0 & 0 & \frac{1}{3} \end{array}]$

Paso 2.4.2

Simplifica $R_{2}$ .

$[\begin{array}{c} 1 & - 4 & 6 & - \frac{73}{3} \\ 0 & 1 & - \frac{5}{4} & \frac{173}{36} \\ 0 & 0 & 0 & \frac{1}{3} \end{array}]$

Paso 2.5

Multiply each element of $R_{3}$ by $3$ to make the entry at $3, 4$ a $1$ .

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Paso 2.5.1

Multiply each element of $R_{3}$ by $3$ to make the entry at $3, 4$ a $1$ .

$[\begin{array}{c} 1 & - 4 & 6 & - \frac{73}{3} \\ 0 & 1 & - \frac{5}{4} & \frac{173}{36} \\ 3 \cdot 0 & 3 \cdot 0 & 3 \cdot 0 & 3 (\frac{1}{3}) \end{array}]$

Paso 2.5.2

Simplifica $R_{3}$ .

$[\begin{array}{c} 1 & - 4 & 6 & - \frac{73}{3} \\ 0 & 1 & - \frac{5}{4} & \frac{173}{36} \\ 0 & 0 & 0 & 1 \end{array}]$

Paso 2.6

Perform the row operation $R_{2} = R_{2} - \frac{173}{36} R_{3}$ to make the entry at $2, 4$ a $0$ .

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Paso 2.6.1

Perform the row operation $R_{2} = R_{2} - \frac{173}{36} R_{3}$ to make the entry at $2, 4$ a $0$ .

$[\begin{array}{c} 1 & - 4 & 6 & - \frac{73}{3} \\ 0 - \frac{173}{36} \cdot 0 & 1 - \frac{173}{36} \cdot 0 & - \frac{5}{4} - \frac{173}{36} \cdot 0 & \frac{173}{36} - \frac{173}{36} \cdot 1 \\ 0 & 0 & 0 & 1 \end{array}]$

Paso 2.6.2

Simplifica $R_{2}$ .

$[\begin{array}{c} 1 & - 4 & 6 & - \frac{73}{3} \\ 0 & 1 & - \frac{5}{4} & 0 \\ 0 & 0 & 0 & 1 \end{array}]$

Paso 2.7

Perform the row operation $R_{1} = R_{1} + \frac{73}{3} R_{3}$ to make the entry at $1, 4$ a $0$ .

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Paso 2.7.1

Perform the row operation $R_{1} = R_{1} + \frac{73}{3} R_{3}$ to make the entry at $1, 4$ a $0$ .

$[\begin{array}{c} 1 + \frac{73}{3} \cdot 0 & - 4 + \frac{73}{3} \cdot 0 & 6 + \frac{73}{3} \cdot 0 & - \frac{73}{3} + \frac{73}{3} \cdot 1 \\ 0 & 1 & - \frac{5}{4} & 0 \\ 0 & 0 & 0 & 1 \end{array}]$

Paso 2.7.2

Simplifica $R_{1}$ .

$[\begin{array}{c} 1 & - 4 & 6 & 0 \\ 0 & 1 & - \frac{5}{4} & 0 \\ 0 & 0 & 0 & 1 \end{array}]$

Paso 2.8

Perform the row operation $R_{1} = R_{1} + 4 R_{2}$ to make the entry at $1, 2$ a $0$ .

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Paso 2.8.1

Perform the row operation $R_{1} = R_{1} + 4 R_{2}$ to make the entry at $1, 2$ a $0$ .

$[\begin{array}{c} 1 + 4 \cdot 0 & - 4 + 4 \cdot 1 & 6 + 4 (- \frac{5}{4}) & 0 + 4 \cdot 0 \\ 0 & 1 & - \frac{5}{4} & 0 \\ 0 & 0 & 0 & 1 \end{array}]$

Paso 2.8.2

Simplifica $R_{1}$ .

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

Paso 3

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

$x + z = 0$

$y - \frac{5 z}{4} = 0$

$0 = 1$

Paso 4

Resta $z$ de ambos lados de la ecuación.

$x = - z$

$y - \frac{5 z}{4} = 0$

$0 = 1$

Paso 5

Suma $\frac{5 z}{4}$ a ambos lados de la ecuación.

$y = \frac{5 z}{4}$

$x = - z$

$0 = 1$

Paso 6

Como $0 \neq 1$ , no hay soluciones.

No hay solución