Álgebra lineal Ejemplos

$s = {[\begin{array}{c} - 7 \\ 5 \\ 1 \end{array}], [\begin{array}{c} - 6 \\ 5 \\ 0 \end{array}]}$

Paso 1

To determine if the columns in the matrix are linearly dependent, determine if the equation $A x = 0$ has any non-trivial solutions.

Paso 2

Write as an augmented matrix for $A x = 0$ .

$[\begin{array}{c} - 7 & - 6 & 0 \\ 5 & 5 & 0 \\ 1 & 0 & 0 \end{array}]$

Paso 3

Obtén la forma escalonada reducida por filas.

Toca para ver más pasos...

Paso 3.1

Multiply each element of $R_{1}$ by $- \frac{1}{7}$ to make the entry at $1, 1$ a $1$ .

Toca para ver más pasos...

Paso 3.1.1

Multiply each element of $R_{1}$ by $- \frac{1}{7}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} - \frac{1}{7} \cdot - 7 & - \frac{1}{7} \cdot - 6 & - \frac{1}{7} \cdot 0 \\ 5 & 5 & 0 \\ 1 & 0 & 0 \end{array}]$

Paso 3.1.2

Simplifica $R_{1}$ .

$[\begin{array}{c} 1 & \frac{6}{7} & 0 \\ 5 & 5 & 0 \\ 1 & 0 & 0 \end{array}]$

Paso 3.2

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

Toca para ver más pasos...

Paso 3.2.1

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

$[\begin{array}{c} 1 & \frac{6}{7} & 0 \\ 5 - 5 \cdot 1 & 5 - 5 (\frac{6}{7}) & 0 - 5 \cdot 0 \\ 1 & 0 & 0 \end{array}]$

Paso 3.2.2

Simplifica $R_{2}$ .

$[\begin{array}{c} 1 & \frac{6}{7} & 0 \\ 0 & \frac{5}{7} & 0 \\ 1 & 0 & 0 \end{array}]$

Paso 3.3

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

Toca para ver más pasos...

Paso 3.3.1

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

$[\begin{array}{c} 1 & \frac{6}{7} & 0 \\ 0 & \frac{5}{7} & 0 \\ 1 - 1 & 0 - \frac{6}{7} & 0 - 0 \end{array}]$

Paso 3.3.2

Simplifica $R_{3}$ .

$[\begin{array}{c} 1 & \frac{6}{7} & 0 \\ 0 & \frac{5}{7} & 0 \\ 0 & - \frac{6}{7} & 0 \end{array}]$

Paso 3.4

Multiply each element of $R_{2}$ by $\frac{7}{5}$ to make the entry at $2, 2$ a $1$ .

Toca para ver más pasos...

Paso 3.4.1

Multiply each element of $R_{2}$ by $\frac{7}{5}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & \frac{6}{7} & 0 \\ \frac{7}{5} \cdot 0 & \frac{7}{5} \cdot \frac{5}{7} & \frac{7}{5} \cdot 0 \\ 0 & - \frac{6}{7} & 0 \end{array}]$

Paso 3.4.2

Simplifica $R_{2}$ .

$[\begin{array}{c} 1 & \frac{6}{7} & 0 \\ 0 & 1 & 0 \\ 0 & - \frac{6}{7} & 0 \end{array}]$

Paso 3.5

Perform the row operation $R_{3} = R_{3} + \frac{6}{7} R_{2}$ to make the entry at $3, 2$ a $0$ .

Toca para ver más pasos...

Paso 3.5.1

Perform the row operation $R_{3} = R_{3} + \frac{6}{7} R_{2}$ to make the entry at $3, 2$ a $0$ .

$[\begin{array}{c} 1 & \frac{6}{7} & 0 \\ 0 & 1 & 0 \\ 0 + \frac{6}{7} \cdot 0 & - \frac{6}{7} + \frac{6}{7} \cdot 1 & 0 + \frac{6}{7} \cdot 0 \end{array}]$

Paso 3.5.2

Simplifica $R_{3}$ .

$[\begin{array}{c} 1 & \frac{6}{7} & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{array}]$

Paso 3.6

Perform the row operation $R_{1} = R_{1} - \frac{6}{7} R_{2}$ to make the entry at $1, 2$ a $0$ .

Toca para ver más pasos...

Paso 3.6.1

Perform the row operation $R_{1} = R_{1} - \frac{6}{7} R_{2}$ to make the entry at $1, 2$ a $0$ .

$[\begin{array}{c} 1 - \frac{6}{7} \cdot 0 & \frac{6}{7} - \frac{6}{7} \cdot 1 & 0 - \frac{6}{7} \cdot 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{array}]$

Paso 3.6.2

Simplifica $R_{1}$ .

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

Paso 4

Remove rows that are all zeros.

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

Paso 5

Write the matrix as a system of linear equations.

$x = 0$

$y = 0$

Paso 6

Since the only solution to $A x = 0$ is the trivial solution, the vectors are linearly independent.

Linealmente independiente