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We are given a matrix \( A \) and asked to find a basis for the null space \( V \), which consists of the solutions to the equation \( AX = 0 \), where \( A \in \mathbb{R}^{4 \times 4} \) and \( m = 1 \). Let me extract the matrix and solve the system step by step.

### Matrix \( A \) is:

\[
A = \begin{bmatrix}
1 & 2 & -2 & -1 \\
2 & 1 & 1 & 3 \\
1 & 5 & -7 & -6 \\
-3 & 0 & -4 & 1
\end{bmatrix}
\]

We need to find the null space of this matrix, which is the set of all vectors \( X \in \mathbb{R}^4 \) such that \( AX = 0 \).

### Step 1: Solve \( AX = 0 \)
The system of equations from \( AX = 0 \) is equivalent to solving this matrix equation. We will perform Gaussian elimination (or row reduction) to bring the matrix \( A \) to its reduced row echelon form (RREF) and then find the solution.

Let's proceed with the calculation:

1. Subtract 2 times row 1 from row 2, subtract 1 times row 1 from row 3, and add 3 times row 1 to row 4.
   
   Resulting matrix:
   \[
   \begin{bmatrix}
   1 & 2 & -2 & -1 \\
   0 & -3 & 5 & 5 \\
   0 & 3 & -5 & -5 \\
   0 & 6 & -10 & -2
   \end{bmatrix}
   \]

2. Now, subtract row 2 from row 3 and subtract 2 times row 2 from row 4.

   Resulting matrix:
   \[
   \begin{bmatrix}
   1 & 2 & -2 & -1 \\
   0 & -3 & 5 & 5 \\
   0 & 0 & 0 & 0 \\
   0 & 0 & 0 & -12
   \end{bmatrix}
   \]

3. Divide row 2 by -3 and row 4 by -12.

   Final matrix in RREF:
   \[
   \begin{bmatrix}
   1 & 2 & -2 & -1 \\
   0 & 1 & -\frac{5}{3} & -\frac{5}{3} \\
   0 & 0 & 0 & 0 \\
   0 & 0 & 0 & 1
   \end{bmatrix}
   \]

### Step 2: Interpretation
From the RREF, we can derive the following system of equations:
\[
\begin{aligned}
x_1 + 2x_2 - 2x_3 - x_4 &= 0 \\
x_2 - \frac{5}{3}x_3 - \frac{5}{3}x_4 &= 0 \\
x_4 &= 0
\end{aligned}
\]

Since \( x_4 = 0 \), substitute it into the other equations to simplify:
\[
\begin{aligned}
x_1 + 2x_2 - 2x_3 &= 0 \\
x_2 - \frac{5}{3}x_3 &= 0
\end{aligned}
\]

From \( x_2 = \frac{5}{3}x_3 \), substitute this into the first equation:
\[
x_1 + 2\left(\frac{5}{3}x_3\right) - 2x_3 = 0 \implies x_1 + \frac{10}{3}x_3 - 2x_3 = 0 \implies x_1 = -\frac{4}{3}x_3
\]

Thus, the general solution for the system is:
\[
x_1 = -\frac{4}{3}x_3, \quad x_2 = \frac{5}{3}x_3, \quad x_4 = 0
\]

Therefore, the solution vector is:
\[
X = x_3 \begin{bmatrix} -\frac{4}{3} \\ \frac{5}{3} \\ 1 \\ 0 \end{bmatrix}
\]

### Step 3: Find a basis for \( V \)
The null space is spanned by the vector:
\[
\left\{ \begin{bmatrix} -4 \\ 5 \\ 3 \\ 0 \end{bmatrix} \right\}
\]

Thus, a basis for \( V \) is \( \left\{ (-4, 5, 3, 0) \right\} \), corresponding to option **B** in the multiple-choice answers.

### Conclusion
The correct answer is **B: \( \left\{ (-4, 5, 3, 0) \right\} \)**.

Would you like more details on any step, or do you have any further questions?

Here are five related questions to deepen your understanding:
1. How is the null space of a matrix defined?
2. Why do we use Gaussian elimination to solve systems of linear equations?
3. What are the steps involved in row reduction to get a matrix into RREF?
4. How can we verify if a vector is in the null space of a matrix?
5. What does it mean if the null space of a matrix is trivial?

**Tip:** Always double-check row operations during Gaussian elimination to avoid calculation errors.

Consider the case where m = 1, and in the vector space R4, the subspace V is the set of solutions to the system AX = 0. Find a basis for V.

Find the basis of the null space for a 4x4 matrix A where m = 1. This problem involves solving the homogeneous system AX = 0 using Gaussian elimination and row reduction. Suitable for students studying linear algebra.

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