5. Is the following set of polynomials linearly independent in P3? Explain your work. 1+ 2t³, 2 +t – 3t², –t + 2t² – t³
5. Is the following set of polynomials linearly independent in P3? Explain your work. 1+ 2t³, 2 +t – 3t², –t + 2t² – t³
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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please solve both im having troublem on them
![**Problem 5: Linearly Independent Polynomials**
Is the following set of polynomials linearly independent in \( P_3 \)? Explain your work.
\[
1 + 2t^3, \quad 2 + t - 3t^2, \quad -t + 2t^2 - t^3
\]
**Problem 6: Subspace and Basis in \( M_{2 \times 2} \)**
Let \( H \) be a subspace of \( M_{2 \times 2} \) whose vectors are of the form
\[
\begin{bmatrix} a & b \\ c & 0 \end{bmatrix}
\]
Then, \( \mathcal{B} = \left\{ \begin{bmatrix} 1 & 0 \\ -1 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}, \begin{bmatrix} 2 & -1 \\ 1 & 0 \end{bmatrix} \right\} \) is a basis for \( H \).
Find the coordinate vector of \( \mathbf{v} = \begin{bmatrix} 7 & -1 \\ 0 & 0 \end{bmatrix} \) according to the basis, \( \mathcal{B} \).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa51cb2c6-68d7-460b-b6c2-8a3d6ada438a%2F91daa027-bb62-4a2c-897b-01dd46a1b82c%2F1sjq8hp_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem 5: Linearly Independent Polynomials**
Is the following set of polynomials linearly independent in \( P_3 \)? Explain your work.
\[
1 + 2t^3, \quad 2 + t - 3t^2, \quad -t + 2t^2 - t^3
\]
**Problem 6: Subspace and Basis in \( M_{2 \times 2} \)**
Let \( H \) be a subspace of \( M_{2 \times 2} \) whose vectors are of the form
\[
\begin{bmatrix} a & b \\ c & 0 \end{bmatrix}
\]
Then, \( \mathcal{B} = \left\{ \begin{bmatrix} 1 & 0 \\ -1 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}, \begin{bmatrix} 2 & -1 \\ 1 & 0 \end{bmatrix} \right\} \) is a basis for \( H \).
Find the coordinate vector of \( \mathbf{v} = \begin{bmatrix} 7 & -1 \\ 0 & 0 \end{bmatrix} \) according to the basis, \( \mathcal{B} \).
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