प्रश्न 25. दिखाइये कि सदिश (1, 0, 0), (1, 1,0), (1,1,1), R' (R) का एक आधार है।
Answers
SOLUTION
TO PROVE
(1, 0, 0), (1, 1,0), (1,1,1) is basis of R³
PROOF
Let (a, b, c) ∈ R³
Suppose there exists x , y , z such that
(a, b, c) = x(1, 0, 0) + y(1, 1,0) + z(1,1,1)
Which gives
x + y + z = a - - - - (1)
y + z = b - - - - - (2)
z = c - - - - - (3)
From Equation 2 we get
y = b - c
From Equation 3 we get
x = a - b
Thus
(a, b, c) = (a-b) (1, 0, 0) + (b - c) (1, 1,0) + c(1,1,1)
So (1, 0, 0), (1, 1,0), (1,1,1) generates R³
Suppose there exists x , y , z such that
x(1, 0, 0) + y(1, 1,0) + z(1,1,1) = (0,0,0)
x + y + z = 0 - - - - (4)
y + z = 0 - - - - - (5)
z = 0 - - - - - (6)
From Equation 5 we get
y = 0
From Equation 4 we get
x = 0
Thus (1, 0, 0), (1, 1,0), (1,1,1) is linear independent
Hence (1, 0, 0), (1, 1,0), (1,1,1) is basis of R³
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