Exercise 6 Is the set of vectors V = {i² + 2t – 1,3t + 5, -t² + t + 6} linearly dependent?
Exercise 6 Is the set of vectors V = {i² + 2t – 1,3t + 5, -t² + t + 6} linearly dependent?
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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Help with exercise 6 please?
![**Exercise 6**
Is the set of vectors
\[ V = \{t^2 + 2t - 1, 3t + 5, -t^2 + t + 6\} \]
linearly dependent?
**Example 6**
What about this set of vectors?
\[ O = \{t^2 + 2t + 1, t^2 - t + 3, 3t + 1\} \]
Is it linearly dependent? Let's check. First, set it as a linear combination of the vectors set equal to the zero vector, \( \mathbf{0} \).
\[ a(t^2 + 2t + 1) + b(t^2 - t + 3) + c(3t + 1) = 0 \]
\[ (a + b)t^2 + (2a - b + 3c)t + (a + 3b + c) = 0 \]
Again, make sure you see how this works algebraically. This yields the following system of equations, augmented matrix, and resulting RREF matrix.
\[
\begin{align*}
a + b &= 0 \\
2a - b + 3c &= 0 \\
a + 3b + c &= 0
\end{align*}
\]
\[
\begin{bmatrix}
1 & 1 & 0 & : & 0 \\
2 & -1 & 3 & : & 0 \\
1 & 3 & 1 & : & 0
\end{bmatrix}
\quad \overset{RREF}{\longrightarrow} \quad
\begin{bmatrix}
1 & 0 & 0 & : & 0 \\
0 & 1 & 0 & : & 0 \\
0 & 0 & 1 & : & 0
\end{bmatrix}
\]
Remember from previous tutorials that this means that this system only has one solution. That solution is \((a, b, c) = (0, 0, 0)\). This is the trivial solution! Because this is the only solution, we do **NOT** say that the set of vectors \( O \) is linearly dependent. In fact, we say that the set of vectors \( O \) is linearly independent. Going back to](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F2c762e73-cba5-4fbf-869f-39b9ab2042ca%2Fcfbff989-64f3-4545-b97b-74ef820e72d9%2Fv9e0mdt_processed.png&w=3840&q=75)
Transcribed Image Text:**Exercise 6**
Is the set of vectors
\[ V = \{t^2 + 2t - 1, 3t + 5, -t^2 + t + 6\} \]
linearly dependent?
**Example 6**
What about this set of vectors?
\[ O = \{t^2 + 2t + 1, t^2 - t + 3, 3t + 1\} \]
Is it linearly dependent? Let's check. First, set it as a linear combination of the vectors set equal to the zero vector, \( \mathbf{0} \).
\[ a(t^2 + 2t + 1) + b(t^2 - t + 3) + c(3t + 1) = 0 \]
\[ (a + b)t^2 + (2a - b + 3c)t + (a + 3b + c) = 0 \]
Again, make sure you see how this works algebraically. This yields the following system of equations, augmented matrix, and resulting RREF matrix.
\[
\begin{align*}
a + b &= 0 \\
2a - b + 3c &= 0 \\
a + 3b + c &= 0
\end{align*}
\]
\[
\begin{bmatrix}
1 & 1 & 0 & : & 0 \\
2 & -1 & 3 & : & 0 \\
1 & 3 & 1 & : & 0
\end{bmatrix}
\quad \overset{RREF}{\longrightarrow} \quad
\begin{bmatrix}
1 & 0 & 0 & : & 0 \\
0 & 1 & 0 & : & 0 \\
0 & 0 & 1 & : & 0
\end{bmatrix}
\]
Remember from previous tutorials that this means that this system only has one solution. That solution is \((a, b, c) = (0, 0, 0)\). This is the trivial solution! Because this is the only solution, we do **NOT** say that the set of vectors \( O \) is linearly dependent. In fact, we say that the set of vectors \( O \) is linearly independent. Going back to
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