To find a basis for the null space of a matrix, we need to solve the equation:

$A \mathbf{x} = 0$

where $A$ is the given matrix, and $\mathbf{x}$ is the vector of unknowns $x_1, x_2, x_3, x_4$. Let's start by setting up the matrix $A$:

$A = \begin{bmatrix} 1 & 0 & -3 & -4 \\ 0 & 1 & 3 & -4 \\ 0 & 0 & 0 & 0 \end{bmatrix}$

Step 1: Write the equation $A \mathbf{x} = 0$

Let $\mathbf{x} = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix}$.

The equation becomes:

$\begin{bmatrix} 1 & 0 & -3 & -4 \\ 0 & 1 & 3 & -4 \\ 0 & 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$

This gives the following system of linear equations:

1. $x_1 - 3x_3 - 4x_4 = 0$
2. $x_2 + 3x_3 - 4x_4 = 0$
3. $0 = 0$ (which is trivially satisfied)

Step 2: Solve the system

From the first equation:

$x_1 = 3x_3 + 4x_4$

From the second equation:

$x_2 = -3x_3 + 4x_4$

Step 3: Parametrize the solution

Let $x_3 = t$ and $x_4 = s$, where $t$ and $s$ are free variables. Then we can express the solution in terms of $t$ and $s$:

$x_1 = 3t + 4s$ $x_2 = -3t + 4s$ $x_3 = t$ $x_4 = s$

Thus, the general solution for $\mathbf{x}$ is:

$\mathbf{x} = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix} = \begin{bmatrix} 3t + 4s \\ -3t + 4s \\ t \\ s \end{bmatrix}$

This can be written as:

$\mathbf{x} = t \begin{bmatrix} 3 \\ -3 \\ 1 \\ 0 \end{bmatrix} + s \begin{bmatrix} 4 \\ 4 \\ 0 \\ 1 \end{bmatrix}$

Step 4: Conclusion

The null space of the matrix $A$ is spanned by the vectors:

$\begin{bmatrix} 3 \\ -3 \\ 1 \\ 0 \end{bmatrix} \quad \text{and} \quad \begin{bmatrix} 4 \\ 4 \\ 0 \\ 1 \end{bmatrix}$

Thus, a basis for the null space of $A$ is:

$\left\{ \begin{bmatrix} 3 \\ -3 \\ 1 \\ 0 \end{bmatrix}, \begin{bmatrix} 4 \\ 4 \\ 0 \\ 1 \end{bmatrix} \right\}$

Let me know if you'd like further details or clarification!

Further questions to explore:

1. How do you check if a set of vectors forms a basis for a subspace?
2. What is the rank-nullity theorem, and how does it relate to this problem?
3. How would you solve for the null space of a matrix using row reduction?
4. Can you find the null space of a matrix that has no solution or is inconsistent?
5. What happens if the matrix has more rows than columns in terms of the null space?

Tip:

To find the null space of a matrix, always start by writing down the system of equations and then use row reduction or substitution to solve for the free variables.