Answer:
The equations are linearly independent in vector space R²
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