1. Show that the given vector functions are linearly independent on (-∞, ∞). X₁ (t) = = 0 X₂ (t) H 2t 3t² X3 (t) = 0 3+3
1. Show that the given vector functions are linearly independent on (-∞, ∞). X₁ (t) = = 0 X₂ (t) H 2t 3t² X3 (t) = 0 3+3
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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![**Problem: Linear Independence of Vector Functions**
1. Show that the given vector functions are linearly independent on \((-\infty, \infty)\).
The vector functions are:
\[
\mathbf{x}_1(t) = \begin{bmatrix} t \\ 0 \\ -t^3 \end{bmatrix}, \quad \mathbf{x}_2(t) = \begin{bmatrix} 2t \\ 3t^2 \\ 0 \end{bmatrix}, \quad \mathbf{x}_3(t) = \begin{bmatrix} -t \\ 0 \\ 3t^3 \end{bmatrix}.
\]
To prove that the vector functions \(\mathbf{x}_1(t)\), \(\mathbf{x}_2(t)\), and \(\mathbf{x}_3(t)\) are linearly independent, we need to ensure that the only solution to:
\[
c_1\begin{bmatrix} t \\ 0 \\ -t^3 \end{bmatrix} + c_2\begin{bmatrix} 2t \\ 3t^2 \\ 0 \end{bmatrix} + c_3\begin{bmatrix} -t \\ 0 \\ 3t^3 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}
\]
is \(c_1 = c_2 = c_3 = 0\).
**Explanation:**
To check linear independence, we form a linear combination of the given vectors equating to the zero vector. The constants \(c_1\), \(c_2\), and \(c_3\) must satisfy:
1. \(c_1 t + c_2 (2t) + c_3 (-t) = 0\)
2. \(c_2 (3t^2) = 0\)
3. \(c_1 (-t^3) + c_3 (3t^3) = 0\)
By examining these equations, determine if the only solution is the trivial one. These vectors are linearly independent if this is the case.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F67429e19-8d15-47bc-bbfe-dfa923849540%2F3ac72893-a810-44a5-ac60-5bc1e41339e9%2Fcqkmyf9_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem: Linear Independence of Vector Functions**
1. Show that the given vector functions are linearly independent on \((-\infty, \infty)\).
The vector functions are:
\[
\mathbf{x}_1(t) = \begin{bmatrix} t \\ 0 \\ -t^3 \end{bmatrix}, \quad \mathbf{x}_2(t) = \begin{bmatrix} 2t \\ 3t^2 \\ 0 \end{bmatrix}, \quad \mathbf{x}_3(t) = \begin{bmatrix} -t \\ 0 \\ 3t^3 \end{bmatrix}.
\]
To prove that the vector functions \(\mathbf{x}_1(t)\), \(\mathbf{x}_2(t)\), and \(\mathbf{x}_3(t)\) are linearly independent, we need to ensure that the only solution to:
\[
c_1\begin{bmatrix} t \\ 0 \\ -t^3 \end{bmatrix} + c_2\begin{bmatrix} 2t \\ 3t^2 \\ 0 \end{bmatrix} + c_3\begin{bmatrix} -t \\ 0 \\ 3t^3 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}
\]
is \(c_1 = c_2 = c_3 = 0\).
**Explanation:**
To check linear independence, we form a linear combination of the given vectors equating to the zero vector. The constants \(c_1\), \(c_2\), and \(c_3\) must satisfy:
1. \(c_1 t + c_2 (2t) + c_3 (-t) = 0\)
2. \(c_2 (3t^2) = 0\)
3. \(c_1 (-t^3) + c_3 (3t^3) = 0\)
By examining these equations, determine if the only solution is the trivial one. These vectors are linearly independent if this is the case.
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