Let p₁ (t) = 7+t², P₂ (t)=t-2t², P3(t) = 2+t-4t². Complete parts (a) and (b) below. a. Use coordinate vectors to show that these polynomials form a basis for P2. What are the coordinate vectors corresponding to P₁, P2, and p3? P₁ = 7 0 0 P₂= 1 P3= 1 -2 2 1 Place these coordinate vectors into the columns of a matrix A. What can be said about the matrix A? A. The matrix A is invertible because it is row equivalent to 13 and therefore the original columns of A form Theorem. OB. The matrix A is invertible because it is row equivalent to 13 and therefore the row reduced columns of A form a basis for R³ by the Invertible Matrix Theorem. OC. The matrix A is invertible because it is row equivalent to 13 and therefore the null space of A, denoted Nul A, forms a basis for R³ by the Invertible Theorem. OD. The matrix A forms a basis for R³ How does this show that the polynomials form a basis for P2? by the Invertible Matrix Theorem because all square matrices are row equivalent to 13. How does this show that the polynomials form a basis for P₂? basis for R³ by the Invertible Matrix A. The polynomials form a basis for P₂ because of the isomorphism between R³ and P2. OB. The polynomials form a basis for P₂ because the columns of A are linearly independent. c. The polynomials form a basis for P₂ because the matrix A is invertible. OD. The polynomials form a basis for P₂ because any basis in R is also a basis in P q(t)= (-3) + (3) t+ (-11) ₁² b. Consider the basis B = {P₁, P2, P3} for P₂. Find q in P₂, given that [q] = 1
Let p₁ (t) = 7+t², P₂ (t)=t-2t², P3(t) = 2+t-4t². Complete parts (a) and (b) below. a. Use coordinate vectors to show that these polynomials form a basis for P2. What are the coordinate vectors corresponding to P₁, P2, and p3? P₁ = 7 0 0 P₂= 1 P3= 1 -2 2 1 Place these coordinate vectors into the columns of a matrix A. What can be said about the matrix A? A. The matrix A is invertible because it is row equivalent to 13 and therefore the original columns of A form Theorem. OB. The matrix A is invertible because it is row equivalent to 13 and therefore the row reduced columns of A form a basis for R³ by the Invertible Matrix Theorem. OC. The matrix A is invertible because it is row equivalent to 13 and therefore the null space of A, denoted Nul A, forms a basis for R³ by the Invertible Theorem. OD. The matrix A forms a basis for R³ How does this show that the polynomials form a basis for P2? by the Invertible Matrix Theorem because all square matrices are row equivalent to 13. How does this show that the polynomials form a basis for P₂? basis for R³ by the Invertible Matrix A. The polynomials form a basis for P₂ because of the isomorphism between R³ and P2. OB. The polynomials form a basis for P₂ because the columns of A are linearly independent. c. The polynomials form a basis for P₂ because the matrix A is invertible. OD. The polynomials form a basis for P₂ because any basis in R is also a basis in P q(t)= (-3) + (3) t+ (-11) ₁² b. Consider the basis B = {P₁, P2, P3} for P₂. Find q in P₂, given that [q] = 1
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![Let p₁ (t)=1
=7+1²,
‚ P₂(t) = t−2t², p3(t) = 2 +t−4t². Complete parts (a) and (b) below.
a. Use coordinate vectors to show that these polynomials form a basis for P2.
What are the coordinate vectors corresponding to P₁, P2, and p3?
P₁ = 0 P₂ =
1
0
1 P3=
-2
2
Place these coordinate vectors into the columns of a matrix A. What can be said about the matrix A?
1
-4
A. The matrix A is invertible because it is row equivalent to 13 and therefore the original columns of A form a basis for R³ by the Invertible Matrix
Theorem.
OB. The matrix A is invertible because it is row equivalent to 13 and therefore the row reduced columns of A form a basis for R³ by the Invertible
Matrix Theorem.
q(t) =
OC. The matrix A is invertible because it is row equivalent to 13 and therefore the null space of A, denoted Nul A, forms a basis for R³ by the
Invertible Matrix Theorem.
OD. The matrix A forms a basis for R³ by the Invertible Matrix Theorem because all square matrices are row equivalent to 13.
How does this show that the polynomials form a basis for P2?
How does this show that the polynomials form a basis for P2?
The polynomials form a basis for P₂ because of the isomorphism between R³ and P2.
OB. The polynomials form a basis for P₂ because the columns of A are linearly independent.
C. The polynomials form a basis for P₂ because the matrix A is invertible.
OD. The polynomials form a basis for P₂ because any basis in R is also a basis in P
-1
b. Consider the basis B= (P₁, P2, P3} for P2. Find q in P2, given that [q] =
1
1
2
-3) + (3) t+ (-11) ₁²](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff413a13f-4ccc-4089-b9ed-1c21a38ff098%2F5464909e-f21d-4b08-9db1-e8f0c065bf9d%2Fyhfz8dd.jpeg&w=3840&q=75)
Transcribed Image Text:Let p₁ (t)=1
=7+1²,
‚ P₂(t) = t−2t², p3(t) = 2 +t−4t². Complete parts (a) and (b) below.
a. Use coordinate vectors to show that these polynomials form a basis for P2.
What are the coordinate vectors corresponding to P₁, P2, and p3?
P₁ = 0 P₂ =
1
0
1 P3=
-2
2
Place these coordinate vectors into the columns of a matrix A. What can be said about the matrix A?
1
-4
A. The matrix A is invertible because it is row equivalent to 13 and therefore the original columns of A form a basis for R³ by the Invertible Matrix
Theorem.
OB. The matrix A is invertible because it is row equivalent to 13 and therefore the row reduced columns of A form a basis for R³ by the Invertible
Matrix Theorem.
q(t) =
OC. The matrix A is invertible because it is row equivalent to 13 and therefore the null space of A, denoted Nul A, forms a basis for R³ by the
Invertible Matrix Theorem.
OD. The matrix A forms a basis for R³ by the Invertible Matrix Theorem because all square matrices are row equivalent to 13.
How does this show that the polynomials form a basis for P2?
How does this show that the polynomials form a basis for P2?
The polynomials form a basis for P₂ because of the isomorphism between R³ and P2.
OB. The polynomials form a basis for P₂ because the columns of A are linearly independent.
C. The polynomials form a basis for P₂ because the matrix A is invertible.
OD. The polynomials form a basis for P₂ because any basis in R is also a basis in P
-1
b. Consider the basis B= (P₁, P2, P3} for P2. Find q in P2, given that [q] =
1
1
2
-3) + (3) t+ (-11) ₁²
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