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To find a basis for the solution space of the given homogeneous system, we need to solve the system of linear equations:

\[
\begin{bmatrix}
-2 & 0 & -10 & 4 \\
1 & -2 & 3 & -1 \\
1 & -4 & 1 & 0
\end{bmatrix}
\begin{bmatrix}
x_1 \\
x_2 \\
x_3 \\
x_4
\end{bmatrix}
= 
\begin{bmatrix}
0 \\
0 \\
0
\end{bmatrix}.
\]

### Step 1: Row Reduction
To solve this system, we’ll row reduce the matrix to its reduced row echelon form (RREF) to identify the free variables.

1. Start with the matrix:
   \[
   \begin{bmatrix}
   -2 & 0 & -10 & 4 \\
   1 & -2 & 3 & -1 \\
   1 & -4 & 1 & 0
   \end{bmatrix}.
   \]

2. Perform row operations to simplify this matrix.

### Step 2: Identify Free Variables
Once in RREF, we can determine which variables are leading (basic) and which are free. The free variables will correspond to the parameters in our general solution, forming the basis vectors.

### Step 3: Form Basis Vectors
The basis vectors for the solution space will be the vectors corresponding to each free variable in terms of the leading variables.

Let's complete the row reduction and solve this to give the final basis vectors. I'll proceed with the row reduction now.After row reducing, we obtain the matrix in reduced row echelon form (RREF):

\[
\begin{bmatrix}
1 & 0 & 5 & -2 \\
0 & 1 & 1 & -\frac{1}{2} \\
0 & 0 & 0 & 0
\end{bmatrix}.
\]

From this, we can identify the pivot columns (leading variables) as \(x_1\) and \(x_2\), while \(x_3\) and \(x_4\) are free variables.

### Expressing Variables in Terms of Free Variables
1. From the first row: \( x_1 = -5x_3 + 2x_4 \).
2. From the second row: \( x_2 = -x_3 + \frac{1}{2}x_4 \).

Letting \(x_3 = t\) and \(x_4 = s\), where \(t\) and \(s\) are parameters, we can express the solution vector as:

\[
\begin{bmatrix}
x_1 \\
x_2 \\
x_3 \\
x_4
\end{bmatrix}
=
t \begin{bmatrix}
-5 \\
-1 \\
1 \\
0
\end{bmatrix}
+
s \begin{bmatrix}
2 \\
\frac{1}{2} \\
0 \\
1
\end{bmatrix}.
\]

### Basis for the Solution Space
The basis for the solution space is thus:

\[
\left\{
\begin{bmatrix}
-5 \\
-1 \\
1 \\
0
\end{bmatrix},
\begin{bmatrix}
2 \\
\frac{1}{2} \\
0 \\
1
\end{bmatrix}
\right\}.
\]

Would you like further details on each step or have any questions?

### Additional Questions to Consider
1. How does row reduction help in determining the basis of a solution space?
2. Why are some variables chosen as free, while others are pivot variables?
3. What is the significance of a basis in a vector space?
4. How can this solution be interpreted geometrically?
5. How would the basis change if we altered the homogeneous system?

### Tip
In linear algebra, the number of free variables in a homogeneous system is equal to the dimension of the solution space, which can help predict the number of basis vectors.

Find a basis for the solution space of the given homogeneous system of equations:

\[
\begin{bmatrix}
-2 & 0 & -10 & 4 \\
1 & -2 & 3 & -1 \\
1 & -4 & 1 & 0
\end{bmatrix}
\begin{bmatrix}
x_1 \\
x_2 \\
x_3 \\
x_4
\end{bmatrix}
= 
\begin{bmatrix}
0 \\
0 \\
0
\end{bmatrix}.
\]

This problem demonstrates finding a basis for the solution space of a homogeneous system of equations using matrix row reduction. It covers essential concepts in linear algebra, including basis vectors, row reduction, and homogeneous solutions. Suitable for undergraduate students studying linear algebra.

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