Álgebra lineal Ejemplos

S ([\begin{array}{c} a \\ b \\ c \end{array}]) = [\begin{array}{c} 2 a - 6 b + 6 c \\ a + 2 b + c \\ 2 a + b + 2 c \end{array}]

$S ([\begin{array}{c} a \\ b \\ c \end{array}]) = [\begin{array}{c} 2 a - 6 b + 6 c \\ a + 2 b + c \\ 2 a + b + 2 c \end{array}]$

Paso 1

La transformación define un mapa de

ℝ^{3}

$ℝ^{3}$ a

ℝ^{3}

$ℝ^{3}$ . Para probar que la transformación es lineal, esta debe conservar la multiplicación escalar, la suma y el vector cero.

ℝ^{3} \to ℝ^{3}

$ℝ^{3} \to ℝ^{3}$

Paso 2

Primero pruebe que la transformación conserva esta propiedad.

S (x + y) = S (x) + S (y)

$S (x + y) = S (x) + S (y)$

Paso 3

Establece dos matrices para comprobar que

S

$S$ conserva la propiedad de la suma.

S ([\begin{array}{c} x_{1} \\ x_{2} \\ x_{3} \end{array}] + [\begin{array}{c} y_{1} \\ y_{2} \\ y_{3} \end{array}])

$S ([\begin{array}{c} x_{1} \\ x_{2} \\ x_{3} \end{array}] + [\begin{array}{c} y_{1} \\ y_{2} \\ y_{3} \end{array}])$

Paso 4

Suma las dos matrices.

S [\begin{array}{c} x_{1} + y_{1} \\ x_{2} + y_{2} \\ x_{3} + y_{3} \end{array}]

$S [\begin{array}{c} x_{1} + y_{1} \\ x_{2} + y_{2} \\ x_{3} + y_{3} \end{array}]$

Paso 5

Aplica la transformación al vector.

S (x + y) = [\begin{array}{c} 2 (x_{1} + y_{1}) - 6 (x_{2} + y_{2}) + 6 (x_{3} + y_{3}) \\ x_{1} + y_{1} + 2 (x_{2} + y_{2}) + x_{3} + y_{3} \\ 2 (x_{1} + y_{1}) + x_{2} + y_{2} + 2 (x_{3} + y_{3}) \end{array}]

$S (x + y) = [\begin{array}{c} 2 (x_{1} + y_{1}) - 6 (x_{2} + y_{2}) + 6 (x_{3} + y_{3}) \\ x_{1} + y_{1} + 2 (x_{2} + y_{2}) + x_{3} + y_{3} \\ 2 (x_{1} + y_{1}) + x_{2} + y_{2} + 2 (x_{3} + y_{3}) \end{array}]$

Paso 6

Simplifica cada elemento de la matriz.

Toca para ver más pasos...

Paso 6.1

Reorganiza

2 (x_{1} + y_{1}) - 6 (x_{2} + y_{2}) + 6 (x_{3} + y_{3})

$2 (x_{1} + y_{1}) - 6 (x_{2} + y_{2}) + 6 (x_{3} + y_{3})$ .

S (x + y) = [\begin{array}{c} 2 x_{1} - 6 x_{2} + 6 x_{3} + 2 y_{1} - 6 y_{2} + 6 y_{3} \\ x_{1} + y_{1} + 2 (x_{2} + y_{2}) + x_{3} + y_{3} \\ 2 (x_{1} + y_{1}) + x_{2} + y_{2} + 2 (x_{3} + y_{3}) \end{array}]

$S (x + y) = [\begin{array}{c} 2 x_{1} - 6 x_{2} + 6 x_{3} + 2 y_{1} - 6 y_{2} + 6 y_{3} \\ x_{1} + y_{1} + 2 (x_{2} + y_{2}) + x_{3} + y_{3} \\ 2 (x_{1} + y_{1}) + x_{2} + y_{2} + 2 (x_{3} + y_{3}) \end{array}]$

Paso 6.2

Reorganiza

x_{1} + y_{1} + 2 (x_{2} + y_{2}) + x_{3} + y_{3}

$x_{1} + y_{1} + 2 (x_{2} + y_{2}) + x_{3} + y_{3}$ .

S (x + y) = [\begin{array}{c} 2 x_{1} - 6 x_{2} + 6 x_{3} + 2 y_{1} - 6 y_{2} + 6 y_{3} \\ x_{1} + 2 x_{2} + x_{3} + y_{1} + 2 y_{2} + y_{3} \\ 2 (x_{1} + y_{1}) + x_{2} + y_{2} + 2 (x_{3} + y_{3}) \end{array}]

$S (x + y) = [\begin{array}{c} 2 x_{1} - 6 x_{2} + 6 x_{3} + 2 y_{1} - 6 y_{2} + 6 y_{3} \\ x_{1} + 2 x_{2} + x_{3} + y_{1} + 2 y_{2} + y_{3} \\ 2 (x_{1} + y_{1}) + x_{2} + y_{2} + 2 (x_{3} + y_{3}) \end{array}]$

Paso 6.3

Reorganiza

2 (x_{1} + y_{1}) + x_{2} + y_{2} + 2 (x_{3} + y_{3})

$2 (x_{1} + y_{1}) + x_{2} + y_{2} + 2 (x_{3} + y_{3})$ .

S (x + y) = [\begin{array}{c} 2 x_{1} - 6 x_{2} + 6 x_{3} + 2 y_{1} - 6 y_{2} + 6 y_{3} \\ x_{1} + 2 x_{2} + x_{3} + y_{1} + 2 y_{2} + y_{3} \\ 2 x_{1} + x_{2} + 2 x_{3} + 2 y_{1} + y_{2} + 2 y_{3} \end{array}]

$S (x + y) = [\begin{array}{c} 2 x_{1} - 6 x_{2} + 6 x_{3} + 2 y_{1} - 6 y_{2} + 6 y_{3} \\ x_{1} + 2 x_{2} + x_{3} + y_{1} + 2 y_{2} + y_{3} \\ 2 x_{1} + x_{2} + 2 x_{3} + 2 y_{1} + y_{2} + 2 y_{3} \end{array}]$

S (x + y) = [\begin{array}{c} 2 x_{1} - 6 x_{2} + 6 x_{3} + 2 y_{1} - 6 y_{2} + 6 y_{3} \\ x_{1} + 2 x_{2} + x_{3} + y_{1} + 2 y_{2} + y_{3} \\ 2 x_{1} + x_{2} + 2 x_{3} + 2 y_{1} + y_{2} + 2 y_{3} \end{array}]

Paso 7

Divide el resultado en dos matrices mediante la agrupación de las variables.

S (x + y) = [\begin{array}{c} 2 x_{1} - 6 x_{2} + 6 x_{3} \\ x_{1} + 2 x_{2} + x_{3} \\ 2 x_{1} + x_{2} + 2 x_{3} \end{array}] + [\begin{array}{c} 2 y_{1} - 6 y_{2} + 6 y_{3} \\ y_{1} + 2 y_{2} + y_{3} \\ 2 y_{1} + y_{2} + 2 y_{3} \end{array}]

$S (x + y) = [\begin{array}{c} 2 x_{1} - 6 x_{2} + 6 x_{3} \\ x_{1} + 2 x_{2} + x_{3} \\ 2 x_{1} + x_{2} + 2 x_{3} \end{array}] + [\begin{array}{c} 2 y_{1} - 6 y_{2} + 6 y_{3} \\ y_{1} + 2 y_{2} + y_{3} \\ 2 y_{1} + y_{2} + 2 y_{3} \end{array}]$

Paso 8

La propiedad de la suma de la transformación se mantiene verdadera.

S (x + y) = S (x) + S (y)

$S (x + y) = S (x) + S (y)$

Paso 9

Para que una transformación sea lineal, debe mantener la multiplicación escalar.

S (p x) = T (p [\begin{array}{c} a \\ b \\ c \end{array}])

$S (p x) = T (p [\begin{array}{c} a \\ b \\ c \end{array}])$

Paso 10

Factoriza la

p

$p$ de cada elemento.

Toca para ver más pasos...

Paso 10.1

Multiplica

p

$p$ por cada elemento en la matriz.

S (p x) = S ([\begin{array}{c} p a \\ p b \\ p c \end{array}])

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

Paso 10.2

Aplica la transformación al vector.

S (p x) = [\begin{array}{c} 2 ((p a) - 6 (p b) + 6 (p c)) \\ (p a) + 2 (p b) + p c \\ 2 (p a + p b + 2 (p c)) \end{array}]

$S (p x) = [\begin{array}{c} 2 ((p a) - 6 (p b) + 6 (p c)) \\ (p a) + 2 (p b) + p c \\ 2 (p a + p b + 2 (p c)) \end{array}]$

Paso 10.3

Simplifica cada elemento de la matriz.

Toca para ver más pasos...

Paso 10.3.1

Reorganiza

2 ((p a) - 6 (p b) + 6 (p c))

$2 ((p a) - 6 (p b) + 6 (p c))$ .

S (p x) = [\begin{array}{c} 2 a p - 12 b p + 12 c p \\ (p a) + 2 (p b) + p c \\ 2 (p a + p b + 2 (p c)) \end{array}]

$S (p x) = [\begin{array}{c} 2 a p - 12 b p + 12 c p \\ (p a) + 2 (p b) + p c \\ 2 (p a + p b + 2 (p c)) \end{array}]$

Paso 10.3.2

Reorganiza

(p a) + 2 (p b) + p c

$(p a) + 2 (p b) + p c$ .

S (p x) = [\begin{array}{c} 2 a p - 12 b p + 12 c p \\ a p + 2 b p + c p \\ 2 (p a + p b + 2 (p c)) \end{array}]

$S (p x) = [\begin{array}{c} 2 a p - 12 b p + 12 c p \\ a p + 2 b p + c p \\ 2 (p a + p b + 2 (p c)) \end{array}]$

Paso 10.3.3

Reorganiza

2 (p a + p b + 2 (p c))

$2 (p a + p b + 2 (p c))$ .

S (p x) = [\begin{array}{c} 2 a p - 12 b p + 12 c p \\ a p + 2 b p + c p \\ 2 a p + 2 b p + 4 c p \end{array}]

$S (p x) = [\begin{array}{c} 2 a p - 12 b p + 12 c p \\ a p + 2 b p + c p \\ 2 a p + 2 b p + 4 c p \end{array}]$

S (p x) = [\begin{array}{c} 2 a p - 12 b p + 12 c p \\ a p + 2 b p + c p \\ 2 a p + 2 b p + 4 c p \end{array}]

$S (p x) = [\begin{array}{c} 2 a p - 12 b p + 12 c p \\ a p + 2 b p + c p \\ 2 a p + 2 b p + 4 c p \end{array}]$

Paso 10.4

Factoriza cada elemento de la matriz.

Toca para ver más pasos...

Paso 10.4.1

Factoriza el elemento

0, 0

$0, 0$ mediante la multiplicación de

2 a p - 12 b p + 12 c p

$2 a p - 12 b p + 12 c p$ .

S (p x) = [\begin{array}{c} p (2 a - 12 b + 12 c) \\ a p + 2 b p + c p \\ 2 a p + 2 b p + 4 c p \end{array}]

$S (p x) = [\begin{array}{c} p (2 a - 12 b + 12 c) \\ a p + 2 b p + c p \\ 2 a p + 2 b p + 4 c p \end{array}]$

Paso 10.4.2

Factoriza el elemento

1, 0

$1, 0$ mediante la multiplicación de

a p + 2 b p + c p

$a p + 2 b p + c p$ .

S (p x) = [\begin{array}{c} p (2 a - 12 b + 12 c) \\ p (a + 2 b + c) \\ 2 a p + 2 b p + 4 c p \end{array}]

$S (p x) = [\begin{array}{c} p (2 a - 12 b + 12 c) \\ p (a + 2 b + c) \\ 2 a p + 2 b p + 4 c p \end{array}]$

Paso 10.4.3

Factoriza el elemento

2, 0

$2, 0$ mediante la multiplicación de

2 a p + 2 b p + 4 c p

$2 a p + 2 b p + 4 c p$ .

S (p x) = [\begin{array}{c} p (2 a - 12 b + 12 c) \\ p (a + 2 b + c) \\ p (2 a + 2 b + 4 c) \end{array}]

$S (p x) = [\begin{array}{c} p (2 a - 12 b + 12 c) \\ p (a + 2 b + c) \\ p (2 a + 2 b + 4 c) \end{array}]$

S (p x) = [\begin{array}{c} p (2 a - 12 b + 12 c) \\ p (a + 2 b + c) \\ p (2 a + 2 b + 4 c) \end{array}]

$S (p x) = [\begin{array}{c} p (2 a - 12 b + 12 c) \\ p (a + 2 b + c) \\ p (2 a + 2 b + 4 c) \end{array}]$

S (p x) = [\begin{array}{c} p (2 a - 12 b + 12 c) \\ p (a + 2 b + c) \\ p (2 a + 2 b + 4 c) \end{array}]

$S (p x) = [\begin{array}{c} p (2 a - 12 b + 12 c) \\ p (a + 2 b + c) \\ p (2 a + 2 b + 4 c) \end{array}]$

Paso 11

La segunda propiedad de las transformaciones lineales se conserva en esta transformación.

S (p [\begin{array}{c} a \\ b \\ c \end{array}]) = p S (x)

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

Paso 12

Para que la transformación sea lineal, se debe conservar el vector cero.

S (0) = 0

$S (0) = 0$

Paso 13

Aplica la transformación al vector.

S (0) = [\begin{array}{c} 2 (0) - 6 \cdot 0 + 6 (0) \\ (0) + 2 (0) + 0 \\ 2 (0) + 0 + 2 (0) \end{array}]

$S (0) = [\begin{array}{c} 2 (0) - 6 \cdot 0 + 6 (0) \\ (0) + 2 (0) + 0 \\ 2 (0) + 0 + 2 (0) \end{array}]$

Paso 14

Simplifica cada elemento de la matriz.

Toca para ver más pasos...

Paso 14.1

Reorganiza

2 (0) - 6 \cdot 0 + 6 (0)

$2 (0) - 6 \cdot 0 + 6 (0)$ .

S (0) = [\begin{array}{c} 0 \\ (0) + 2 (0) + 0 \\ 2 (0) + 0 + 2 (0) \end{array}]

$S (0) = [\begin{array}{c} 0 \\ (0) + 2 (0) + 0 \\ 2 (0) + 0 + 2 (0) \end{array}]$

Paso 14.2

Reorganiza

(0) + 2 (0) + 0

$(0) + 2 (0) + 0$ .

S (0) = [\begin{array}{c} 0 \\ 0 \\ 2 (0) + 0 + 2 (0) \end{array}]

$S (0) = [\begin{array}{c} 0 \\ 0 \\ 2 (0) + 0 + 2 (0) \end{array}]$

Paso 14.3

Reorganiza

2 (0) + 0 + 2 (0)

$2 (0) + 0 + 2 (0)$ .

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

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

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

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

Paso 15

La transformación conserva el vector cero.

S (0) = 0

$S (0) = 0$

Paso 16

Como las tres propiedades de las transformaciones lineales no se cumplen, esta no es una transformación lineal.

Transformación lineal

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