**Title: Describing All Solutions to $Ax = 0$ in Parametric Vector Form**To find the solutions for the equation $Ax = 0$ in parametric vector form, where the matrix $A$ is given as row equivalent to:\[\begin{bmatrix}1 & 3 & 0 & -2 \\4 & 12 & 0 & -8\end{bmatrix}\]We need to express $x$ in terms of the free variables.The augmented matrix to solve the system is:\[\left[\begin{array}{cccc|c}1 & 3 & 0 & -2 & 0 \\4 & 12 & 0 & -8 & 0 \end{array}\right]\]Perform row operations to get the matrix in row echelon form:1. $R2 \leftarrow R2 - 4R1$ to simplify the second row.2. The result is:\[\left[\begin{array}{cccc|c}1 & 3 & 0 & -2 & 0 \\0 & 0 & 0 & 0 & 0 \end{array}\right]\]This indicates that the system of equations has infinitely many solutions, with free variables corresponding to $x_2$, $x_3$, and $x_4$.From the simplified matrix, the equation $x_1 + 3x_2 - 2x_4 = 0$ is derived. Solving for $x_1$:\[ x_1 = -3x_2 + 2x_4 \]The solutions can be expressed in parametric vector form:\[ \mathbf{x} = \begin{bmatrix}x_1 \\x_2 \\x_3 \\x_4\end{bmatrix} = x_2 \begin{bmatrix}-3 \\1 \\0 \\0\end{bmatrix} + x_3 \begin{bmatrix}0 \\0 \\1 \\0\end{bmatrix} + x_4 \begin{bmatrix}2 \\0 \\0 \\1\end{bmatrix}\]In

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Describe all solutions of Ax = 0 in parametric vector form, where A is row equivalent to the given matrix. 1 30 -2 4 12 0 -8 X= x, + X3 + X4 (Type an integer or fraction for each matrix element.)

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

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Question

**Title: Describing All Solutions to $Ax = 0$ in Parametric Vector Form**

To find the solutions for the equation $Ax = 0$ in parametric vector form, where the matrix $A$ is given as row equivalent to:

\[
\begin{bmatrix}
1 & 3 & 0 & -2 \\
4 & 12 & 0 & -8
\end{bmatrix}
\]

We need to express $x$ in terms of the free variables.

The augmented matrix to solve the system is:

\[
\left[\begin{array}{cccc|c}
1 & 3 & 0 & -2 & 0 \\
4 & 12 & 0 & -8 & 0
\end{array}\right]
\]

Perform row operations to get the matrix in row echelon form:

1. $R2 \leftarrow R2 - 4R1$ to simplify the second row.
2. The result is:

\[
\left[\begin{array}{cccc|c}
1 & 3 & 0 & -2 & 0 \\
0 & 0 & 0 & 0 & 0
\end{array}\right]
\]

This indicates that the system of equations has infinitely many solutions, with free variables corresponding to $x_2$, $x_3$, and $x_4$.

From the simplified matrix, the equation $x_1 + 3x_2 - 2x_4 = 0$ is derived.

Solving for $x_1$:

\[
x_1 = -3x_2 + 2x_4
\]

The solutions can be expressed in parametric vector form:

\[
\mathbf{x} =
\begin{bmatrix}
x_1 \\
x_2 \\
x_3 \\
x_4
\end{bmatrix}
= x_2
\begin{bmatrix}
-3 \\
1 \\
0 \\
0
\end{bmatrix}
+ x_3
\begin{bmatrix}
0 \\
0 \\
1 \\
0
\end{bmatrix}
+ x_4
\begin{bmatrix}
2 \\
0 \\
0 \\
1
\end{bmatrix}
\]

In

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