Find a vector r(t) = at^2 + bt + c such that the set B = {2t + 1, t^2 - 1, r(t)} is a basis for P2.
Find the values of a, b, c such that B is a basis. Find the general condition and then give a particular solution.
To determine the values of a, b, and c such that the set B is a basis for P2, we need to make sure that B is linearly independent and spans the entire space P2.
First, let's analyze the linear independence of B. The set B = {2t + 1, t^2 - 1, r(t)} will be linearly independent if and only if the coefficient of each term in r(t) is independent of the other two terms in B.
We can write the vector r(t) as r(t) = at^2 + bt + c. Now, let's consider the linear combination of the vectors in B:
k(2t + 1) + m(t^2 - 1) + n(at^2 + bt + c) = 0,
where k, m, and n are scalars.
Expanding this equation gives:
2kt + k + mt^2 - m + nat^2 + nbt + nc = 0.
Two polynomials are equal to each other if and only if the coefficients of their corresponding terms are equal. Therefore, we can equate the coefficients of each power of t to get a system of equations:
2k + na = 0 (coefficients of t^1)
1 - m + nb = 0 (coefficients of t^0)
0 + 0 + nc = 0 (constant term)
Simplifying these equations, we have:
na + 2k = 0 (equation 1)
nb - m + 1 = 0 (equation 2)
nc = 0 (equation 3)
Now, let's consider the conditions for B to span P2. In other words, any polynomial in P2 can be expressed as a linear combination of the vectors in B. A general polynomial in P2 can be written as p(t) = dt^2 + et + f.
To guarantee that B spans P2, we need to find values of a, b, and c that satisfy the following system of equations:
k(2t + 1) + m(t^2 - 1) + n(at^2 + bt + c) = dt^2 + et + f,
which expands to:
2kt + k + mt^2 - m + nat^2 + nbt + nc = dt^2 + et + f.
Equating the coefficients, we get:
2k + na = d,
k - m = e,
k + nb + 1 = f.
To find a particular solution, we can choose arbitrary values for two of the variables (a, b, and c) and then solve for the remaining variable. For simplicity, let's assume a = 1 and b = 0.
Using these assumed values, let's solve equations 1, 2, and 3:
n + 2k = 0 (equation 1)
m - 1 + nb = 0 (equation 2)
nc = 0 (equation 3)
Since b = 0, equation 2 becomes m - 1 = 0, which implies m = 1.
Solving equations 1 and 3 simultaneously yields:
n + 2k = 0
nc = 0.
Since nc = 0, we have two possibilities: n = 0 or c = 0.
Case 1: n = 0
Substituting n = 0 into equation 1 gives us 2k = 0, which implies k = 0.
Using these values, we have a particular solution for k, m, n, a, b, and c: k = 0, m = 1, n = 0, a = 1, b = 0, c is arbitrary.
Thus, a possible solution for r(t) is r(t) = t^2 - 1.
Case 2: c = 0
Substituting c = 0 into equation 3 gives us nc = 0, which implies n = 0.
Using these values, we have a particular solution for k, m, n, a, b, and c: k is arbitrary, m = -1, n = 0, a is arbitrary, b is arbitrary, c = 0.
Thus, a possible solution for r(t) is r(t) = at^2 + bt.
In conclusion, to guarantee that B = {2t + 1, t^2 - 1, r(t)} is a basis for P2, we can choose a particular vector r(t) from two possibilities: r(t) = t^2 - 1 or r(t) = at^2 + bt, where a and b can be any real numbers, and c is arbitrary (0 in the second case).