Final answer:
The set p is linearly independent in the vector space of polynomials over R.
Step-by-step explanation:
The set p is considered linearly independent if the only solution to the equation a1p1(x) + a2p2(x) + a3p3(x) + a4p4(x) = 0
is when a
1
= a
2
= a
3
= a
4
= 0.
Let's try to find the coefficients a1, a2, a3, and a4 that make the equation true:
- 2a1 + xa2 + (10x²-5)a3 + (x³-3x)a4 = 0
- 2a1 + xa2 + 10a3x² - 5a3 + (a4x³ - 3a4x) = 0
- 2a1 + xa2 + 10a3x² - 5a3 + a4x³ - 3a4x = 0
We can see that we have a system of linear equations.
By equating the coefficients of like terms, we have the following equations:
- 2a1 = 0
- a2 = 0
- 10a3 = 0
- -5a3 = 0
- a4 = 0
- -3a4 = 0
These equations have a unique solution: a1 = a2 = a3 = a4 = 0. Therefore, the set p is linearly independent.