To find a basis for the null space of a matrix, we typically start by setting up the equation $A \mathbf{x} = \mathbf{0}$, where $A$ is the given matrix and $\mathbf{x}$ is the vector of unknowns. However, in your case, it seems that the matrix has not been provided fully, but we can infer that $A$ is made up of the two vectors given:

0 & 2 & -2 & -1 \\ 0 & 4 & -4 & -2 \end{bmatrix}$$ ### Steps to find a basis for the null space: 1. Set up the system of equations represented by $$A \mathbf{x} = 0$$, where $$\mathbf{x} = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix}$$. $$\begin{bmatrix} 0 & 2 & -2 & -1 \\ 0 & 4 & -4 & -2 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}$$ This gives us two equations: $$2x_2 - 2x_3 - x_4 = 0$$ $$4x_2 - 4x_3 - 2x_4 = 0$$ 2. The second equation is simply twice the first, so we only need to solve the first equation: $$2x_2 - 2x_3 - x_4 = 0$$ Simplifying: $$x_2 - x_3 - \frac{1}{2}x_4 = 0$$ This can be rewritten as: $$x_2 = x_3 + \frac{1}{2}x_4$$ 3. Parametrize the solution: - Let $$x_3 = t$$ and $$x_4 = s$$, where $$t$$ and $$s$$ are free variables. - Then $$x_2 = t + \frac{1}{2}s$$, and $$x_1 = 0$$ (since the first column of $$A$$ is all zeros). Thus, the solution vector $$\mathbf{x}$$ can be written as: $$\mathbf{x} = \begin{bmatrix} 0 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix} = \begin{bmatrix} 0 \\ t + \frac{1}{2}s \\ t \\ s \end{bmatrix}$$ This can be split into two independent vectors by setting $$t = 1, s = 0$$ and $$t = 0, s = 1$$: $$\mathbf{x} = t \begin{bmatrix} 0 \\ 1 \\ 1 \\ 0 \end{bmatrix} + s \begin{bmatrix} 0 \\ \frac{1}{2} \\ 0 \\ 1 \end{bmatrix}$$ 4. Therefore, a basis for the null space is: $$\left\{ \begin{bmatrix} 0 \\ 1 \\ 1 \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ \frac{1}{2} \\ 0 \\ 1 \end{bmatrix} \right\}$$ This is the set of vectors that forms a basis for the null space of $$A$$. Would you like a detailed explanation on any of these steps? ### Related Questions: 1. What is the dimension of the null space (nullity) for this matrix? 2. How do you find the row space and column space of a matrix? 3. Can a matrix have more than one basis for its null space? 4. What is the significance of the rank-nullity theorem in this context? 5. How does the number of free variables relate to the nullity of a matrix? **Tip:** When dealing with free variables in null space, each free variable corresponds to a dimension in the null space, and the number of free variables gives the nullity.