Álgebra lineal Ejemplos

$x ([\begin{array}{c} a \\ b \\ d \end{array}]) = [\begin{array}{c} 5 a + 3 b - 2 d \\ 0 a + 4 b - d \end{array}]$ , $[\begin{array}{c} 14 \\ 2 \end{array}]$

Paso 1

Write as an augmented matrix for $x ([\begin{array}{c} a \\ b \\ d \end{array}]) x = [\begin{array}{c} 14 \\ 2 \end{array}]$ .

$[\begin{array}{c} 5 a + 3 b - 2 d & 14 \\ 0 a + 4 b - d & 2 \end{array}]$

Paso 2

Write as a linear system of equations.

$5 a + 3 b - 2 d = 14$

$0 a + 4 b - d = 2$

Paso 3

Resuelve el sistema de ecuaciones.

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Paso 3.1

Move variables to the left and constant terms to the right.

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Paso 3.1.1

Reordena los factores en $0 a + 4 b - d = 2$ .

$5 a + 3 b - 2 d = 14$

$a \cdot 0 + 4 b - d = 2$

Paso 3.1.2

Mueve $a \cdot 0$ .

$5 a + 3 b - 2 d = 14$

$4 b - d + a \cdot 0 = 2$

$5 a + 3 b - 2 d = 14$

$4 b - d + a \cdot 0 = 2$

Paso 3.2

Write the system as a matrix.

$[\begin{array}{c} 5 & 3 & - 2 & 14 \\ 4 & - 1 & 0 & 2 \end{array}]$

Paso 3.3

Obtén la forma escalonada reducida por filas.

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Paso 3.3.1

Multiply each element of $R_{1}$ by $\frac{1}{5}$ to make the entry at $1, 1$ a $1$ .

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Paso 3.3.1.1

Multiply each element of $R_{1}$ by $\frac{1}{5}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} \frac{5}{5} & \frac{3}{5} & \frac{- 2}{5} & \frac{14}{5} \\ 4 & - 1 & 0 & 2 \end{array}]$

Paso 3.3.1.2

Simplifica $R_{1}$ .

$[\begin{array}{c} 1 & \frac{3}{5} & - \frac{2}{5} & \frac{14}{5} \\ 4 & - 1 & 0 & 2 \end{array}]$

Paso 3.3.2

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

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Paso 3.3.2.1

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

$[\begin{array}{c} 1 & \frac{3}{5} & - \frac{2}{5} & \frac{14}{5} \\ 4 - 4 \cdot 1 & - 1 - 4 (\frac{3}{5}) & 0 - 4 (- \frac{2}{5}) & 2 - 4 (\frac{14}{5}) \end{array}]$

Paso 3.3.2.2

Simplifica $R_{2}$ .

$[\begin{array}{c} 1 & \frac{3}{5} & - \frac{2}{5} & \frac{14}{5} \\ 0 & - \frac{17}{5} & \frac{8}{5} & - \frac{46}{5} \end{array}]$

Paso 3.3.3

Multiply each element of $R_{2}$ by $- \frac{5}{17}$ to make the entry at $2, 2$ a $1$ .

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Paso 3.3.3.1

Multiply each element of $R_{2}$ by $- \frac{5}{17}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & \frac{3}{5} & - \frac{2}{5} & \frac{14}{5} \\ - \frac{5}{17} \cdot 0 & - \frac{5}{17} (- \frac{17}{5}) & - \frac{5}{17} \cdot \frac{8}{5} & - \frac{5}{17} (- \frac{46}{5}) \end{array}]$

Paso 3.3.3.2

Simplifica $R_{2}$ .

$[\begin{array}{c} 1 & \frac{3}{5} & - \frac{2}{5} & \frac{14}{5} \\ 0 & 1 & - \frac{8}{17} & \frac{46}{17} \end{array}]$

Paso 3.3.4

Perform the row operation $R_{1} = R_{1} - \frac{3}{5} R_{2}$ to make the entry at $1, 2$ a $0$ .

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Paso 3.3.4.1

Perform the row operation $R_{1} = R_{1} - \frac{3}{5} R_{2}$ to make the entry at $1, 2$ a $0$ .

$[\begin{array}{c} 1 - \frac{3}{5} \cdot 0 & \frac{3}{5} - \frac{3}{5} \cdot 1 & - \frac{2}{5} - \frac{3}{5} (- \frac{8}{17}) & \frac{14}{5} - \frac{3}{5} \cdot \frac{46}{17} \\ 0 & 1 & - \frac{8}{17} & \frac{46}{17} \end{array}]$

Paso 3.3.4.2

Simplifica $R_{1}$ .

$[\begin{array}{c} 1 & 0 & - \frac{2}{17} & \frac{20}{17} \\ 0 & 1 & - \frac{8}{17} & \frac{46}{17} \end{array}]$

Paso 3.4

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

$a - \frac{2}{17} d = \frac{20}{17}$

$b - \frac{8}{17} d = \frac{46}{17}$

Paso 3.5

Write a solution vector by solving in terms of the free variables in each row.

$[\begin{array}{c} a \\ b \\ d \end{array}] = [\begin{array}{c} \frac{20}{17} + \frac{2 d}{17} \\ \frac{46}{17} + \frac{8 d}{17} \\ d \end{array}]$

Paso 3.6

Write the solution as a linear combination of vectors.

$[\begin{array}{c} a \\ b \\ d \end{array}] = [\begin{array}{c} \frac{20}{17} \\ \frac{46}{17} \\ 0 \end{array}] + d [\begin{array}{c} \frac{2}{17} \\ \frac{8}{17} \\ 1 \end{array}]$

Paso 3.7

Write as a solution set.

${[\begin{array}{c} \frac{20}{17} \\ \frac{46}{17} \\ 0 \end{array}] + d [\begin{array}{c} \frac{2}{17} \\ \frac{8}{17} \\ 1 \end{array}] | d \in R}$