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 $\mathbf{A} \mathbf{x} = \mathbf{b}$.