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The problem asks for three vectors \( \mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3 \), distinct from \( \mathbf{v} = \begin{bmatrix} -2 \\ 1 \\ -3 \end{bmatrix} \), that also solve the matrix equation \( \mathbf{A} \mathbf{x} = \begin{bmatrix} 44 \\ -44 \\ 88 \end{bmatrix} \). Given the null space (kernel) of the matrix \( \mathbf{A} \), \( \text{Nul}(\mathbf{A}) \), is spanned by two vectors:

\[
\text{Nul}(\mathbf{A}) = \text{Span} \left( \begin{bmatrix} -1 \\ -3 \\ 3 \end{bmatrix}, \begin{bmatrix} -2 \\ 2 \\ 2 \end{bmatrix} \right)
\]

This suggests that any solution to \( \mathbf{A} \mathbf{x} = \begin{bmatrix} 44 \\ -44 \\ 88 \end{bmatrix} \) is of the form:

\[
\mathbf{x} = \mathbf{v} + c_1 \begin{bmatrix} -1 \\ -3 \\ 3 \end{bmatrix} + c_2 \begin{bmatrix} -2 \\ 2 \\ 2 \end{bmatrix}
\]

where \( \mathbf{v} = \begin{bmatrix} -2 \\ 1 \\ -3 \end{bmatrix} \), and \( c_1 \) and \( c_2 \) are arbitrary constants.

To find three distinct vectors \( \mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3 \), we need to choose different values of \( c_1 \) and \( c_2 \). Let's compute:

1. For \( \mathbf{v}_1 \), set \( c_1 = 1 \) and \( c_2 = 0 \):

\[
\mathbf{v}_1 = \mathbf{v} + \begin{bmatrix} -1 \\ -3 \\ 3 \end{bmatrix} = \begin{bmatrix} -2 \\ 1 \\ -3 \end{bmatrix} + \begin{bmatrix} -1 \\ -3 \\ 3 \end{bmatrix} = \begin{bmatrix} -3 \\ -2 \\ 0 \end{bmatrix}
\]

2. For \( \mathbf{v}_2 \), set \( c_1 = 0 \) and \( c_2 = 1 \):

\[
\mathbf{v}_2 = \mathbf{v} + \begin{bmatrix} -2 \\ 2 \\ 2 \end{bmatrix} = \begin{bmatrix} -2 \\ 1 \\ -3 \end{bmatrix} + \begin{bmatrix} -2 \\ 2 \\ 2 \end{bmatrix} = \begin{bmatrix} -4 \\ 3 \\ -1 \end{bmatrix}
\]

3. For \( \mathbf{v}_3 \), set \( c_1 = 1 \) and \( c_2 = 1 \):

\[
\mathbf{v}_3 = \mathbf{v} + \begin{bmatrix} -1 \\ -3 \\ 3 \end{bmatrix} + \begin{bmatrix} -2 \\ 2 \\ 2 \end{bmatrix} = \begin{bmatrix} -2 \\ 1 \\ -3 \end{bmatrix} + \begin{bmatrix} -3 \\ -1 \\ 5 \end{bmatrix} = \begin{bmatrix} -5 \\ 0 \\ 2 \end{bmatrix}
\]

Thus, the solutions are:

\[
\mathbf{v}_1 = \begin{bmatrix} -3 \\ -2 \\ 0 \end{bmatrix}, \quad \mathbf{v}_2 = \begin{bmatrix} -4 \\ 3 \\ -1 \end{bmatrix}, \quad \mathbf{v}_3 = \begin{bmatrix} -5 \\ 0 \\ 2 \end{bmatrix}
\]

Do you want further clarification on any part of the process?

### Relative Questions:
1. How is the null space of a matrix used in solving systems of linear equations?
2. What role do the arbitrary constants \( c_1 \) and \( c_2 \) play in finding solutions?
3. Can you explain how to determine if a system of equations has infinitely many solutions?
4. What happens when the null space consists of more than two vectors?
5. How does changing the values of \( c_1 \) and \( c_2 \) affect the solution set?

### Tip:
To verify if a vector is a solution, you can substitute it back into the matrix equation to see if it satisfies \( \mathbf{A} \mathbf{x} = \mathbf{b} \).

Let A be a 3x3 matrix such that Nul(A) = Span([[-1], [-3], [3]], [[-2], [2], [2]]). Assume that the vector v = [-2, 1, -3] is a solution of the matrix equation Ax = [44, -44, 88]. Find three vectors v1, v2, v3, different from v, which are also solutions of this equation.

Solve for three vectors distinct from v = [-2, 1, -3] using the null space of a 3x3 matrix A and the equation Ax = [44, -44, 88]. This problem explores the concept of null space, vector addition, and matrix solutions.

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