Prove or disprove that the vector V1=(1,2,3) v2=(1,1,1) and V3=(1,1,0) are linearly indepwndent in vector space R^2?

The equations are linearly independent in vector space R²

Step-by-step explanation:

The given vectors are;

V₁ = (1, 2, 3), V₂ = (1, 1, 1), V₃ = (1, 1, 0)

For the vectors to be linearly dependent, we have;

a₁·V₁ + a₂·V₂ + a₃·V₃ = 0

Where;

a₁, a₂, and a₃ are not all zero

We get;

The vectors are linear dependent, if the following system of equation has a unique set of solution

x + y + z = 0...(1)

2·x + y + z = 0...(2)

3·x + y = 0...(3)

From equation (1), we have;

x = -(y + z)

∴ y + z = -x

Plugging in the value of y + z in equation (2) gives;

2·x + y + z = 0

2·x + y + z = 2·x - x = x = 0

x = 0

From equation (3), we get;

3·x + y = 0, where x = 0, therefore;

3 × 0 + y = 0

∴ y = 0

From equation (1), we have;

x + y + z = 0

Where, x = 0, and y = 0, we get;

0 + 0 + z = 0

z = 0

The vectors, V₁, V₂, and V₃ are linear independent in R³

In two dimensional space, we have

x + y = 0...(1)

2·x + y = 0...(2)

x + y = 0...(3)

Subtracting equation (1) from equation (2) gives;

2·x + y - (x + y) = 2·x - x + y - y = x = 0

∴ x = 0

From equation (1), we get

x + y = 0

∴ 0 + y = 0

∴ y = 0

Therefore, the equations are also linearly independent in vector space R²

answer: i guess 5 units?

step-by-step explanation:

answer:   b

step-by-step explanation:

x2 = 16

x can not be a negative number because 16 is positive

the only answer would be positive 8