Question

Use Coordinate Vectors To Determine Whether The Given Polynomials Are Linearly Dependent In P2. Let B Be The Standard Basis Of The Space P2 Of Polynomials, That Is, Let B = {1, t, t^2)

a) 1+2t, 3 +6t^2, 1 +3t +4t^2
b) 1+ 2t + t^2, 3 – 9t^2, 1 + 4t + 5t^2

Answer 1

Explanation:

a) To determine if the polynomials 1+2t, 3+6t^2, 1+3t+4t^2 are linearly dependent in P2, we need to check if there exist constants c1, c2, and c3 such that c1(1+2t) + c2(3+6t^2) + c3(1+3t+4t^2) = 0, where 0 is the zero polynomial in P2.

Rewriting this equation in terms of the standard basis B = {1, t, t^2}, we have:

(c1 + c3) + (2c1 + 3c3)t + (4c2 + 3c3)t^2 = 0

This gives us the system of equations:

c1 + c3 = 0

2c1 + 3c3 = 0

4c2 + 3c3 = 0

Solving this system of equations, we get c1 = -3c3/2, c2 = -3c3/4. Therefore, any choice of c3 that is not equal to zero would give us a nontrivial solution, which implies that the polynomials are linearly dependent in P2.

b) To determine if the polynomials 1+2t+t^2, 3-9t^2, 1+4t+5t^2 are linearly dependent in P2, we need to check if there exist constants c1, c2, and c3 such that c1(1+2t+t^2) + c2(3-9t^2) + c3(1+4t+5t^2) = 0, where 0 is the zero polynomial in P2.

Rewriting this equation in terms of the standard basis B = {1, t, t^2}, we have:

(c1 + c3) + (2c1 + 4c3)t + (c1 + 5c3)t^2 - 9c2t^2 = 0

This gives us the system of equations:

c1 + c3 = 0

2c1 + 4c3 = 0

c1 + 5c3 - 9c2 = 0

Solving this system of equations, we get c1 = -2c3, c2 = (1/9)(c1 + 5c3). Therefore, any choice of c3 that is not equal to zero would give us a nontrivial solution, which implies that the polynomials are linearly dependent in P2.