Algebra
CURSO DE ENGENHARIA DE ALIMENTOS
ALGEBRA LINEAR E GEOMETRIA ANALITICA
ABISAGUE MARIA
GABRIELA TUZINO
HENRRIQUE ISSA MONTANHOLI
LUCAS VEIGA SABINO
MAIKON CRISTIANO BERNSTEIN
VINICIUS DA COSTA ARCA
ALGEBRA LINEAR E GEOMETRIA ANALITICA
CAMPO MOURÃO
2011
01- ) Dados os vetores u = (2 , 4), v = (-5 , 1) e w= (-12 , 6), determinar w = u + v. Solução:
W=
u+
v
(-12 , 6) =
2
(2 , 4) +
-4
(-5 , 1)
(-12 , 6) = (2
, -4
) + (-5
(-12 , 6) = (2
-5
, -4
,
+
-5
+
)
=6
- 10
-4
+
= -24
=6
-9
2
-5
2
= -12
- 10 = -12
2
= -18
- 5(2) = -12
2
tais que
= -12 . (2)
4
)
e
=2
= -2
= -1
02- ) Dados v = (1 , 2 , -4 , 0) e w = (-3 , 5 , 1 , 1), resolva as equações em x:
(a) 5x - 2v = 2 (w – 5x)
Solução:
(b) 5x – IivIIv = IiwII(w -5x)
(a) 5x – 2v = 2(w -5x)
5x – (2 , 4 , -8 , 0) = 2 [(-3 , 5 , 1 , 1) – 5x]
5x – (2 , 4 , -8 , 0) = (-6 , 10 , 2 , 2) – 10x
15x = (-6 , 10 , 2 , 2) + (2 , 4 , -8 , 0)
15x = (-4 , 14 , -6 , 2) x= (b) 5x – IivII . v = IiwII (w – 5x)
5x-
. (1 , 2 , -4 , 0)=
[(-3,5,1,1)-5x]
5x – (
,2
, -4
, 0) = (-18 , 30 , 6 , 6) – 30x
40x = (-18 , 30 , 6 , 6) + (
,2
, -4
, 0)
40x = (-18 +
, 30 + 2
,6-4
, 6) x=( ,
,
,
)
03- ) Dados os vetores U = (2 , -1 , 1), v = (1 , -1 , 0) e w = (-1 , 2 , 2), Calcular:
(a) w X v
(b) v X (u – v)
(d) (u X v) . (u X v)
Solução:
(c) (u + v)X(u – v)
(e) (u + v) . (u X w)
(a)
= (2
(b)
)-(-2
=(
)=(
) = (2 , 2 , -1)
) = (1 , -1 , 1)
(u – v) = (2 , -1 , 1)-(1 , -1 , 0) = (1 , 0 , 1)
(c)
= (-2
)-(
) = (-2
) = (-2 , -2, 2)
(u + v) = (2 , -1 , 1)+(1 , -1 , 0) = (3 , -2 , 1)
(u – v) = (2 , -1 , 1)-(1 , -1 , 0) = (1 , 0 , 1)
(d)
=(
)-(
)=(
) = (1 , 1 , -1)
(u X v) . (u X v) = (1 , 1 , -1) . (1 , 1 , -1) = 3
(e)
= (-2
)-(2
) = (0
) = (0 , -5 , 3)
(u