Álgebra lineal Ejemplos

$S ([\begin{array}{c} a \\ b \\ c \end{array}]) = [\begin{array}{c} a - b - c \\ a - b - c \\ a - b + c \end{array}]$

Paso 1

El núcleo de una transformación es un vector que hace que la transformación sea igual al vector nulo (la imagen previa de la transformación).

$[\begin{array}{c} a - b - c \\ a - b - c \\ a - b + c \end{array}] = 0$

Paso 2

Crea un sistema de ecuaciones a partir de la ecuación vectorial.

$a - b - c = 0$

$a - b + c = 0$

Paso 3

Write the system as a matrix.

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

Paso 4

Obtén la forma escalonada reducida por filas.

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Paso 4.1

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

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Paso 4.1.1

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

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

Paso 4.1.2

Simplifica $R_{2}$ .

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

Paso 4.2

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

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Paso 4.2.1

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

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

Paso 4.2.2

Simplifica $R_{3}$ .

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

Paso 4.3

Swap $R_{3}$ with $R_{2}$ to put a nonzero entry at $2, 3$ .

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

Paso 4.4

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

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Paso 4.4.1

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

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

Paso 4.4.2

Simplifica $R_{2}$ .

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

Paso 4.5

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

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Paso 4.5.1

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

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

Paso 4.5.2

Simplifica $R_{1}$ .

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

Paso 5

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

$a - b = 0$

$c = 0$

$0 = 0$

Paso 6

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

$[\begin{array}{c} a \\ b \\ c \end{array}] = [\begin{array}{c} b \\ b \\ 0 \end{array}]$

Paso 7

Write the solution as a linear combination of vectors.

$[\begin{array}{c} a \\ b \\ c \end{array}] = b [\begin{array}{c} 1 \\ 1 \\ 0 \end{array}]$

Paso 8

Write as a solution set.

${b [\begin{array}{c} 1 \\ 1 \\ 0 \end{array}] | b \in R}$

Paso 9

The solution is the set of vectors created from the free variables of the system.

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

Paso 10

El núcleo de $S$ es el subespacio ${[\begin{array}{c} 1 \\ 1 \\ 0 \end{array}]}$ .

$K (S) = {[\begin{array}{c} 1 \\ 1 \\ 0 \end{array}]}$