Microsoft Microsoft Equation 3.0
1 ) Classificar os seguintes subconjuntos do R 3 em LI ou LD:
{ ( 2, −1,3)} a) Solução: a ( 2,
−1,3) = ( 0,0,0) ⇒ ( 2a,−a,3a ) = (0,0,0)
2 a = 0 − a = 0 3a = 0
2 −1 3
0
0 1 0 L1 ↔ 2 L
1
1 −1 3
0
0 L2 + L1 0 L3 ↔ L3 − 3 L1
A= P ( A)
= 1, P ( A′) = 1
P ( A)
= P ( A! ) ⇒ SP
n
∃a ∈ R 3
= 1, P ( A) = 1 ⇒ SPD
a
= 0 ⇒ Li
#(1,−1,1), (−1,1,1)"
b) Solução: a (1,−1,1) + b( − 1,1,1)
= (0,0,0)
( a,−a, a ) + ( −b, b, b) = (0,0,0) ( a − b,− a + b, a + b ) = (0,0,0)
1 − 1 −1 1 1 1
a − b = 0 a + b = 0 3a = 0
0
0 0
A=
1 − 1 −1 1 1 1
1 − 1 0 0 0 2 1 0 0
0
0
↔ L2 + L1 3 ↔ L3 − L1
L 0 L
2
0
0 L2 ↔ L3 0
0
0
1
0 a = 0
0
0 b = 0
1 − 1 0 2 0 0
0
1 0 L2 ↔ L2 2 0
1 − 1 0 1 0 0
0
0 L1 0
↔ L1 + L2
1 0 0
0
0
0 a = 0
P ( A)
= 2, P ( A′) = 2
P ( A)
= P ( A′) ⇒ SP
n
= 2, P ( A) = 2 ⇒ SPD
a
= b = 0 ⇒ Li
∃a, b ∈ R 3
Classificar os seguintes subconjuntos do R 3 em LI ou LD #( 2,−1,0), ( −1,3,1), (3,$,0)"
c) Solução: a ( 2,
−1,0) + b( − 1,3,1) + c(3,$,0) = (0,0,0)
( 2a,−a,0) + (−b,3b,0) + (3c,$c,0) = (0,0,0) ( 2a,−a,0) + (−b,3b,0) + (3c,$c,0) = (0,0,0)
2 − 1 − 1 3
2a − b + 3c = 0 − a + 3b + $c = 0
3 0
$ 0
A=
2 − 1 − 1 3 1 0
12
1 0
0
1 − 1 2 − 1 3
3 0
1 L2 ↔ L2 0 2
$
0
32
$ 2 13 2
1$
2 L3 ↔ L3 0 $
1 − 1 0 1
3% 2 13 % $
32 $
0
0 L3
↔ L3 + L1
0
1 L1 ↔ L1 − L2 0 2
0
1 13 $ 0 P ( A)
= 2, P ( A′) = 2
a+1c =0 $ 13 b + c = 0 ∃a, b, c ∈ R $
P ( A)
n
= P ( A′) ⇒ SP
= 3, P ( A) < 3 ⇒ SPI , Logo LD
#(2,1,3), (0,0,0), (1,$,2)"
d) Solução: a ( 2,1,3) + b(0,0,0) + c(1,$,2) = (0,0,0)
( 2a, a,3a ) + (0,0,0) + (c,$c,2c) = (0,0,0)
2a + c = 0 a + $c = 0 3a + 2c = 0
2 1 3
1
0
$
0
2 0
A=
2 1 3
1 $ 2
0
0 0 L1 ↔ L2
1 $ 0 1 0 − 13 n
1 2 3
0
0 L3 ↔ L3 − $ L2 0 L3 ↔ L3 − xL2
$ 1 2
0
0 0 L2 ↔ −2 L2
1 0 0
0
0
1
0
0
1 $ 0 − & 3 − 13
0 1 L2 ↔ L2 − 0 &
0 P ( A)
= 2, P ( A′) = 2
P ( A)
,
= 3, P ( A) < 3 ⇒ SPI
∃a ≠ 0 logo
, LD
Classificar os seguintes subconjuntos do R 3 em LI ou LD c)
0
= P ( A′) ⇒ SP
