Describe all solutions of Ax = 0 in parametric vector form, where A is row equivalent to the given matrix. 1 041 3 -9 001 00-3 000 01 5 000 00 0 01
Describe all solutions of Ax = 0 in parametric vector form, where A is row equivalent to the given matrix. 1 041 3 -9 001 00-3 000 01 5 000 00 0 01
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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Question
![To describe all solutions of \( Ax = 0 \) in parametric vector form, we begin with the matrix row equivalent to \( A \):
\[
\begin{bmatrix}
1 & 0 & 4 & -1 & 3 & -9 \\
0 & 0 & 1 & 0 & 0 & -3 \\
0 & 0 & 0 & 0 & 1 & 5 \\
0 & 0 & 0 & 0 & 0 & 0 \\
\end{bmatrix}
\]
### Steps to Solve:
1. **Identify Pivot Columns:**
- Columns 1, 3, and 5 are pivot columns.
2. **Identify Free Variables:**
- The free variables correspond to the non-pivot columns: \( x_2 \), \( x_4 \), and \( x_6 \).
3. **Parametric Form:**
- Express every variable in terms of the free variables:
- From row 1: \( x_1 = -4x_3 + x_4 - 3x_5 + 9x_6 \)
- From row 2: \( x_3 = 3x_6 \)
- From row 3: \( x_5 = -5x_6 \)
4. **Parametric Vector:**
- Substituting back, express the solution in vector form:
\[
\begin{align*}
x &= \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \\ x_5 \\ x_6 \end{bmatrix} \\
&= \begin{bmatrix} -4x_3 + x_4 - 3x_5 + 9x_6 \\ x_2 \\ x_3 \\ x_4 \\ x_5 \\ x_6 \end{bmatrix} \\
&= x_2 \begin{bmatrix} 0 \\ 1 \\ 0 \\ 0 \\ 0 \\ 0 \end{bmatrix} + x_4 \begin{bmatrix} 1 \\ 0 \\ 0 \\ 1 \\ 0 \\ 0 \end{bmatrix} + x_6 \begin{bmatrix} 9 \\ 0 \\ 3 \\ 0 \\ -5 \\ 1](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F46029127-96a5-4dc2-997a-3c090d1aab2d%2F3f418559-d31d-4bd6-9dd6-3ce943a77f0a%2Fkapiqby_processed.png&w=3840&q=75)
Transcribed Image Text:To describe all solutions of \( Ax = 0 \) in parametric vector form, we begin with the matrix row equivalent to \( A \):
\[
\begin{bmatrix}
1 & 0 & 4 & -1 & 3 & -9 \\
0 & 0 & 1 & 0 & 0 & -3 \\
0 & 0 & 0 & 0 & 1 & 5 \\
0 & 0 & 0 & 0 & 0 & 0 \\
\end{bmatrix}
\]
### Steps to Solve:
1. **Identify Pivot Columns:**
- Columns 1, 3, and 5 are pivot columns.
2. **Identify Free Variables:**
- The free variables correspond to the non-pivot columns: \( x_2 \), \( x_4 \), and \( x_6 \).
3. **Parametric Form:**
- Express every variable in terms of the free variables:
- From row 1: \( x_1 = -4x_3 + x_4 - 3x_5 + 9x_6 \)
- From row 2: \( x_3 = 3x_6 \)
- From row 3: \( x_5 = -5x_6 \)
4. **Parametric Vector:**
- Substituting back, express the solution in vector form:
\[
\begin{align*}
x &= \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \\ x_5 \\ x_6 \end{bmatrix} \\
&= \begin{bmatrix} -4x_3 + x_4 - 3x_5 + 9x_6 \\ x_2 \\ x_3 \\ x_4 \\ x_5 \\ x_6 \end{bmatrix} \\
&= x_2 \begin{bmatrix} 0 \\ 1 \\ 0 \\ 0 \\ 0 \\ 0 \end{bmatrix} + x_4 \begin{bmatrix} 1 \\ 0 \\ 0 \\ 1 \\ 0 \\ 0 \end{bmatrix} + x_6 \begin{bmatrix} 9 \\ 0 \\ 3 \\ 0 \\ -5 \\ 1
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