https://file.aiquickdraw.com/mathbot/file/1726454713447z3yrwetj.jpg

The problem in the image asks to find a set of basic vectors \( \{ \mathbf{u}, \mathbf{v} \} \) in \( \mathbb{R}^4 \) that spans the solution set of the system of equations:

\[
\begin{cases}
w - x + 3z = 0 \\
4w + 2x + y + 2z = 0
\end{cases}
\]

### Step 1: Set up the system in matrix form
We can express this system as an augmented matrix:

\[
\begin{pmatrix}
1 & -1 & 0 & 3 & | & 0 \\
4 & 2 & 1 & 2 & | & 0
\end{pmatrix}
\]

This corresponds to the system:

\[
\begin{aligned}
1w - 1x + 0y + 3z &= 0 \\
4w + 2x + 1y + 2z &= 0
\end{aligned}
\]

### Step 2: Row reduce the system
Now, we row-reduce the augmented matrix.

1. The matrix is:
   \[
   \begin{pmatrix}
   1 & -1 & 0 & 3 \\
   4 & 2 & 1 & 2
   \end{pmatrix}
   \]
   
2. Subtract 4 times the first row from the second row:
   \[
   \begin{pmatrix}
   1 & -1 & 0 & 3 \\
   0 & 6 & 1 & -10
   \end{pmatrix}
   \]

3. Divide the second row by 6:
   \[
   \begin{pmatrix}
   1 & -1 & 0 & 3 \\
   0 & 1 & \frac{1}{6} & -\frac{5}{3}
   \end{pmatrix}
   \]

4. Add the first row to the second row multiplied by 1 (to eliminate the second element in the first row):
   \[
   \begin{pmatrix}
   1 & 0 & \frac{1}{6} & -\frac{2}{3} \\
   0 & 1 & \frac{1}{6} & -\frac{5}{3}
   \end{pmatrix}
   \]

### Step 3: Find the solution
Now that the matrix is in row-reduced echelon form, we can solve for \(w, x, y, z\). We can express the free variables in terms of the others:

- Let \(z = t\) and \(y = s\).
- From the first row: \( w = -\frac{1}{6}s + \frac{2}{3}t \).
- From the second row: \( x = -\frac{1}{6}s + \frac{5}{3}t \).

So the general solution is:

\[
\begin{pmatrix} w \\ x \\ y \\ z \end{pmatrix} = s \begin{pmatrix} -\frac{1}{6} \\ -\frac{1}{6} \\ 1 \\ 0 \end{pmatrix} + t \begin{pmatrix} \frac{2}{3} \\ \frac{5}{3} \\ 0 \\ 1 \end{pmatrix}
\]

Thus, the basic vectors are:

\[
\mathbf{u} = \begin{pmatrix} -\frac{1}{6} \\ -\frac{1}{6} \\ 1 \\ 0 \end{pmatrix}, \quad \mathbf{v} = \begin{pmatrix} \frac{2}{3} \\ \frac{5}{3} \\ 0 \\ 1 \end{pmatrix}
\]

### Conclusion
The solution set is spanned by the vectors \( \mathbf{u} \) and \( \mathbf{v} \) as:

\[
\mathbf{u} = \begin{pmatrix} -\frac{1}{6} \\ -\frac{1}{6} \\ 1 \\ 0 \end{pmatrix}, \quad \mathbf{v} = \begin{pmatrix} \frac{2}{3} \\ \frac{5}{3} \\ 0 \\ 1 \end{pmatrix}
\]

---

Would you like more details on any of the steps? Here are some related questions you might find useful:

1. How can we interpret the geometric meaning of the vectors \( \mathbf{u} \) and \( \mathbf{v} \)?
2. What would change in the solution if we had more than two equations?
3. How does row reduction help in solving systems of equations?
4. What is the rank of the coefficient matrix in this system?
5. How can we check if the vectors \( \mathbf{u} \) and \( \mathbf{v} \) are linearly independent?

**Tip**: Always check if your system has free variables to determine the dimension of the solution space.

Find a set of basic vectors {u, v} in R^4 so that every vector in the solution set of the equations can be written as a linear combination of the basic vectors. The system of equations is: w - x + 3z = 0 and 4w + 2x + y + 2z = 0.

Solve the system of equations w - x + 3z = 0 and 4w + 2x + y + 2z = 0 by finding a set of basic vectors that span the solution space. This detailed solution walks through the row-reduction process and highlights key concepts in linear algebra, suitable for undergraduate students.

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