Question

Let p, (t) 6+t, P2(t) =t-3t, p3 (t) = 1 +t-2t. Complete parts (a) and (b) below. Use coordinate vectors to show that these polynomials form a basis for P2. What are the coordinate vectors corresponding to p, p2, and pa? P- Place these coordinate vectors into the columns fa matrix A. What can be said about the matrix A? O A. The matrix A forms a basis for R3 by the Invertible Matrix Theorem because all square matrices are row equivalent to I3. O B. The matrix A is invertible because it is row equivalent to I and therefore the row reduced columns of A forma basis for R3 by the Invertible Matrix Theorem. O C. The matrix A is invertible because it is row equivalent to I3 and therefore the null space of A, denoted Nul A, forms a basis for R3 by the Invertible Matrix Theorem. O D. The matrix A is invertible because it is row equivalent to l3 and therefore the original columns of A form a basis for R3 by the Invertible Matrix Theorem. How does this show that the polynomials form a basis for P2? O A. The polynom ials form a basis for P2 because of the isomorphism between R3 and P2. O B. The polynomials form a basis for P2 because the columns of A are linearly independent. O C. The polynomials form a basis for P2 because the matrix A is invertible. O D. The polynomials form a basis for P2 because any basis in Rn is also a basis in P Consider the basis B= P1, P2. P3 for P2. Find q in P2. given that [qlB 1 2 q(t)= (+ t+ (D2

Answer 1

a. We know that the set {1,t,t²} is the standard basis of P₂ and dim(P₂) = 3.

Let M=

6	0	1	1	0	0
0	1	1	0	1	0
1	-3	-2	0	0	1

It may be observed that the entries in the columns of A are the   scalar multiples of 1 and the coefficients of t,t² in p₁(t), p₂(t) , p₃(t) and 1, t,t².

The RREF of M is

1	0	0	1/5	-3/5	-1/5
0	1	0	1/5	-13/5	-6/5
0	0	1	-1/5	18/5	6/5

This implies that the vectors/polynomials p₁(t), p₂(t) , p₃(t) are linearly independent and the vectors/polynomials 1, t,t² are linear combinations of p₁(t), p₂(t) , p₃(t).

Hence the vectors/polynomials p₁(t), p₂(t) , p₃(t) form a basis for P₂.

The coordinate vectors corresponding to p₁, p₂ , p₃ are (6,0,1)^T, (0,1,-3)^T and (1,1,-2)^T respectively.

Option B is the correct answer.

Option B is the correct answer.

b. If [q]_B = (-3,1,2)^T, then q = -3(6+t²)+1(t-3t²)+2(1+t-2t²) = -16+3t-10t².