Question

1. दाइये कि सदिश (1.0.0), (1.1.0), (I.I.I). R(R) के लिए आधार निर्मित करते है bsc3year ​

Answer 1

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

━━━━━━━━━━━━━━━━

Learn more from Brainly :-

$1.$ A subset B of a vector space V over F is called a basis of V, if:

(A) B is linearly independent set only

(B) B spans...

https://brainly.in/question/30125898

2. The basis {(1,0,0),(0,1,0),(0,0,1)} of the vector space R³(R) is known as

https://brainly.in/question/24574737