Algebra Lineal-EXAMEN T4- Espacios Vectoriales

Determinar el valor de x para que el vector (1, x, 5) ∈ R³ pertenezca al subespacio < (1, 2, 3),(1, 1, 1) >

• α = 2, β = −1 y x = 3

• α = 3, β = −1 y x = 2

• α = 2, β = 1 y x = -3

• α = 3, β = 2 y x = 1

Determinar los valores de a y b, si es que existen, para que

< (a, 1, −1, 2),(1, b, 0, 3) > = < (1, −1, 1, −2),(−2, 0, 0, −6) >

• a = −1 y b = 0

• a = 1 y b = 0

• a = −1 y b = 1

• a = 0 y b = 1

Utilizar el proceso de Gram-Schmidt para transformar la siguiente base en  en una base ortonormal
 $B=\left\{\left(5,\ 1,\ 1,\ -3\right),\ \left(9,\ 3,\ 3,\ -7\right),\ \left(-5,\ 5,\ -1,5\right)\right\}$  

•   $\left\{\left(\frac{5}{6},\ \frac{1}{6},\ \frac{1}{6},\ -\frac{1}{2}\right),\ \left(-\frac{1}{2},\ \frac{1}{2},\ \frac{1}{2},\ -\frac{1}{2}\right),\ \left(\frac{1}{6},\ \frac{5}{6},\ -\frac{1}{6},\ \frac{1}{2}\right)\right\}$  

•  $\left\{\left(\frac{5}{6},\ \frac{1}{6},\ \frac{1}{6},\ \frac{1}{2}\right),\ \left(-\frac{1}{2},\ \frac{1}{2},\ \frac{1}{2},\ -\frac{1}{2}\right),\ \left(\frac{1}{6},\ \frac{5}{6},\ -\frac{1}{6},\ \frac{1}{2}\right)\right\}$  

•  $\left\{\left(\frac{5}{6},\ \frac{1}{6},\ \frac{1}{6},\ -\frac{1}{2}\right),\ \left(\frac{1}{2},\ \frac{1}{2},\ \frac{1}{2},\ -\frac{1}{2}\right),\ \left(\frac{1}{6},\ \frac{5}{6},\ -\frac{1}{6},\ \frac{1}{2}\right)\right\}$  

•  $\left\{\left(\frac{5}{6},\ \frac{1}{6},\ \frac{1}{6},\ -\frac{1}{2}\right),\ \left(-\frac{1}{2},\ \frac{1}{2},\ \frac{1}{2},\ \frac{1}{2}\right),\ \left(\frac{1}{6},\ \frac{5}{6},\ -\frac{1}{6},\ \frac{1}{2}\right)\right\}$  

Hallar el vector $v⃗$ tal que  $w⃗=2u⃗+v⃗$   

 $u⃗=(4,-1)\ \ \ \ \ w⃗=(3,2)$  

•  $v⃗=(-5,4)$  

•  $v⃗=(5,4)$  

•  $v⃗=(5,-4)$  

•  $v⃗=(-5,-4)$  

Hallar las coordenadas del vector    $v⃗$ de tal manera que  $w⃗=3u⃗-1/5v⃗$   , siendo:  $u⃗=(1,2)\ \ \ \ \ w⃗=(-3,5)$  

•  $v⃗=(5,\ 30)$  

•  $v⃗=(-30,5)$  

•  $v⃗=(30,5)$  

•  $v⃗=(30,-5)$  

Utilizar el proceso de Gram-Schmidt para transformar la siguiente base en  en una base ortonormal:  $B=(0,-2,-3,-3,1),(3,-5,0,0,5),(2,1,4,1,3)$  

•  $\left\{\left(0,\ -\frac{2}{\sqrt{23}},\ -\frac{3}{\sqrt{23}},\ \frac{3}{\sqrt{23}}\right),\ \left(\frac{3\sqrt{\frac{23}{283}}}{2},\ -\frac{85}{2\sqrt{6509}},\ \frac{45}{2\sqrt{6509}},\ \frac{45}{2\sqrt{6509}},\ \frac{50}{\sqrt{6509}}\right),\ \left(\frac{53\sqrt{\frac{5}{154518}}}{2},\ \frac{9\sqrt{\frac{105}{7358}}}{2},\ \frac{19\sqrt{\frac{35}{22074}}}{2},\ -\frac{1033}{2\sqrt{772590}},\ 131\sqrt{\frac{3}{257530}}\right)\right\}$  

•  $\left\{\left(0,\ -\frac{2}{\sqrt{23}},\ \frac{3}{\sqrt{23}},\ -\frac{3}{\sqrt{23}}\right),\ \left(\frac{3\sqrt{\frac{23}{283}}}{2},\ -\frac{85}{2\sqrt{6509}},\ \frac{45}{2\sqrt{6509}},\ \frac{45}{2\sqrt{6509}},\ \frac{50}{\sqrt{6509}}\right),\ \left(\frac{53\sqrt{\frac{5}{154518}}}{2},\ \frac{9\sqrt{\frac{105}{7358}}}{2},\ \frac{19\sqrt{\frac{35}{22074}}}{2},\ -\frac{1033}{2\sqrt{772590}},\ 131\sqrt{\frac{3}{257530}}\right)\right\}$  

•  $\left\{\left(0,\ \frac{2}{\sqrt{23}},\ -\frac{3}{\sqrt{23}},\ -\frac{3}{\sqrt{23}}\right),\ \left(\frac{3\sqrt{\frac{23}{283}}}{2},\ -\frac{85}{2\sqrt{6509}},\ \frac{45}{2\sqrt{6509}},\ \frac{45}{2\sqrt{6509}},\ \frac{50}{\sqrt{6509}}\right),\ \left(\frac{53\sqrt{\frac{5}{154518}}}{2},\ \frac{9\sqrt{\frac{105}{7358}}}{2},\ \frac{19\sqrt{\frac{35}{22074}}}{2},\ -\frac{1033}{2\sqrt{772590}},\ 131\sqrt{\frac{3}{257530}}\right)\right\}$  

•  $B'=\left\{\left(0,\ -\frac{2}{\sqrt{23}},\ -\frac{3}{\sqrt{23}},\ -\frac{3}{\sqrt{23}}\right),\ \left(\frac{3\sqrt{\frac{23}{283}}}{2},\ -\frac{85}{2\sqrt{6509}},\ \frac{45}{2\sqrt{6509}},\ \frac{45}{2\sqrt{6509}},\ \frac{50}{\sqrt{6509}}\right),\ \left(\frac{53\sqrt{\frac{5}{154518}}}{2},\ \frac{9\sqrt{\frac{105}{7358}}}{2},\ \frac{19\sqrt{\frac{35}{22074}}}{2},\ -\frac{1033}{2\sqrt{772590}},\ 131\sqrt{\frac{3}{257530}}\right)\right\}$  

Comprobar que el vector  $w⃗=(4,7)$   es combinación lineal de los vectores:  $u⃗=(2,1)$   y  $v⃗=(0,5)$   y, ¿Qué combinación forman? 

•  $w⃗=2u⃗+v⃗$  

•  $w⃗=u⃗+2v⃗$  

•  $w⃗=2u⃗+3v⃗$  

•  $w⃗=u⃗+v⃗$  

Utilizar el proceso de Gram-Schmidt para transformar la siguiente base en  en una base ortonormal:

 $B=\left\{\left(\frac{\sqrt{2}}{3},\ 0,\ -\frac{\sqrt{2}}{6}\right),\ \left(0,\ \frac{2\sqrt{5}}{5},\ -\frac{\sqrt{5}}{5}\right),\ \left(\frac{\sqrt{5}}{5},\ 0,\ \frac{1}{2}\right)\right\}$  

•  $\left(0.894,\ 0,\ -0.447\right),\ \left(-0.183,\ 0.913,\ -0.365\right),\ \left(0.408,\ 0.408,\ .816\right)$  

•  $\left(0.894,\ 0,\ 0.447\right),\ \left(-0.183,\ 0.913,\ -0.365\right),\ \left(0.408,\ 0.408,\ .816\right)$  

•  $\left(0.894,\ 0,\ -0.447\right),\ \left(-0.183,\ 0.913,\ 0.365\right),\ \left(0.408,\ 0.408,\ .816\right)$  

•  $\left(0.894,\ 0,\ -0.447\right),\ \left(-0.183,\ 0.913,\ -0.365\right),\ \left(-0.408,\ -0.408,\ -0.816\right)$  

Determine si el siguiente conjunto es una BASE en    $R^3$  

 $S=(1,2,3),(0,1,2),(-2,0,1)$  

• Si es una base

• No es una Base

Utilizar el proceso de Gram-Schmidt para transformar la siguiente base en    $R^3$   en una base ortonormal:

 $B=\left\{(1,1,0),\ (0,1,1),\ (1,0,1))\right\}$  

•  $\left(\frac{1}{\sqrt{2}},\ \frac{1}{\sqrt{2}},\ 0\right),\ \left(-\frac{1}{\sqrt{6}},\ \frac{1}{\sqrt{6}},\ \sqrt{\frac{2}{3}}\right),\ \left(\frac{1}{\sqrt{3}},\ -\frac{1}{\sqrt{3}},\ \frac{1}{\sqrt{3}}\right)$  

•  $\left(\frac{1}{\sqrt{2}},\ \frac{1}{\sqrt{2}},\ 0\right),\ \left(\frac{1}{\sqrt{6}},\ \frac{1}{\sqrt{6}},\ \sqrt{\frac{2}{3}}\right),\ \left(\frac{1}{\sqrt{3}},\ -\frac{1}{\sqrt{3}},\ \frac{1}{\sqrt{3}}\right)$  

•  $\left(\frac{1}{\sqrt{2}},\ \frac{1}{\sqrt{2}},\ 0\right),\ \left(-\frac{1}{\sqrt{6}},\ \frac{1}{\sqrt{6}},\ \sqrt{\frac{2}{3}}\right),\ \left(\frac{1}{\sqrt{3}},\ \frac{1}{\sqrt{3}},\ \frac{1}{\sqrt{3}}\right)$  

•  $\left(\frac{1}{\sqrt{2}},\ \frac{1}{\sqrt{2}},\ 0\right),\ \left(\frac{1}{\sqrt{6}},\ -\frac{1}{\sqrt{6}},\ \sqrt{\frac{2}{3}}\right),\ \left(\frac{1}{\sqrt{3}},\ -\frac{1}{\sqrt{3}},\ \frac{1}{\sqrt{3}}\right)$