Consider the vectors u = (1, 2,-1) and v = (6, 4, 2) in r3. Show that w = (9, 2, 7) is a linear combination of u and v and that w= (4,-1, 8) is not a linear combination of u and v
Answers
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Answer:
1) w = (9, 2, 7) = -3u + 2v i.e. w is a linear combination of u and v.
2) w= (4,-1, 8) ≠ -5u + 3/2v i.e. w is not linear combination of u and v.
Step-by-step explanation:
Given u = (1, 2,-1) and v = (6, 4, 2)
Case-1: If w = (9, 2, 7) is a linear combination of u and v, then we can
write, w = (9, 2, 7) = αu + βv [ α and β are constants]
(9, 2, 7) = α(1, 2,-1) + β(6, 4, 2) [putting values of u and v]
(9, 2, 7) = (α, 2α,-α) + (6β, 4β, 2β)
(9, 2, 7) = (α + 6β, 2α + 4β,-α +2β ) ............(1)
Comparing the two sides , α + 6β = 9.......(2)
2α + 4β = 2 or, α + 2β = 1 .......(3)
-α +2β = 7..........(4)
Subtracting equation (3) from (2), (α + 6β) - (α + 2β) = 9 - 1
⇒ 6β - 2β = 8 ⇒ β = 2
From equation (2), α = 9 - 6β = 9 - 6(2) = -3 ⇒ α = -3
Putting the values of α and β in right hand side of equation (1),
(α + 6β, 2α + 4β,-α +2β ) = ( -3 + 6x2, 2x(-3) + 4x2, 3 + 2x2)
= (9, 2, 7) = w
∴ w is linear combination of u and v, w = -3u + 2v
Case-2: If w = (4,-1, 8)) is a linear combination of u and v, then we can
write, w = (4,-1, 8) = αu + βv [ α and β are constants]
(4,-1, 8) = α(1, 2,-1) + β(6, 4, 2) [putting values of u and v]
(4,-1, 8) = (α, 2α,-α) + (6β, 4β, 2β)
(4,-1, 8) = (α + 6β, 2α + 4β,-α +2β ) ............(1)
Comparing the two sides , α + 6β = 4.......(2)
2α + 4β = -1 or, α + 2β = 1 .......(3)
-α +2β = 8..........(4)
Adding equation (2) and (4), (α + 6β) + (-α + 2β) = 4 + 8
⇒ 6β + 2β = 12 ⇒ β = 3/2
From equation (2), α = 4 - 6β = 4 - 6(3/2) = -5 ⇒ α = -5
Putting the values of α and β in right hand side of equation (1),
(α + 6β, 2α + 4β,-α +2β ) = ( -5 + 6x3/2, 2x(-5) + 4x3/2, 5 + 2x3/2)
= (4, -4, 8) ≠ w
∴ w ≠ -5u + 3/2v
∴ w is not linear combination of u and v.