Step-by-step explanation:

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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