p1, p2, p3 are elements of P_2
p1(x)= 1+x-2*x^2
p2(x)= 2+3x-x^2
p3(x)= 3+5x-x^2
Is S = {p1, p2, p3}
linearly independent?
To determine if the set S = {p1, p2, p3} is linearly independent, we need to check if the only solution to the equation a1p1 + a2p2 + a3p3 = 0 is a1 = a2 = a3 = 0.
Let's set up the equation and solve for a1, a2, and a3:
a1 * p1(x) + a2 * p2(x) + a3 * p3(x) = 0
a1(1+x-2x^2) + a2(2+3x-x^2) + a3(3+5x-x^2) = 0
Let's collect like terms:
(a1 + 2a2 + 3a3) + (a1 + 3a2 + 5a3)x + (-2a1 - a2 - a3)x^2 = 0
Now, we can equate the coefficients of each term to zero:
a1 + 2a2 + 3a3 = 0 ....(1)
a1 + 3a2 + 5a3 = 0 ....(2)
-2a1 - a2 - a3 = 0 ....(3)
We now have three equations with three unknowns. To find the solution, we can solve the system of linear equations formed by these three equations.
Let's solve the equations (1), (2), and (3) using a matrix representation:
| 1 2 3 | | a1 | | 0 |
| 1 3 5 | * | a2 | = | 0 |
|-2 -1 -1 | | a3 | | 0 |
By row-reducing the augmented matrix, we get:
| 1 2 3 | | a1 | | 0 |
| 0 1 2 | * | a2 | = | 0 |
| 0 0 0 | | a3 | | 0 |
From the row-reduced matrix, we can see that the third row is all zeros, indicating the system has infinitely many solutions. This means that there are multiple solutions to the equation a1p1 + a2p2 + a3p3 = 0, other than a1 = a2 = a3 = 0.
Therefore, the set S = {p1, p2, p3} is linearly dependent since there exists a nontrivial solution to the equation.