Parametrization Nullspace F0A072

1. **State the problem:** We are given a parametrization of the solution set to $Ax = b$: $$\begin{bmatrix} x \\ y \\ z \\ 1 \end{bmatrix} = \begin{bmatrix} -6 \\ 4 \\ 1 \end{bmatrix} + x_2 \begin{bmatrix} -7 \\ 9 \\ 8 \end{bmatrix} + x_3 \begin{bmatrix} 5 \\ 0 \\ 9 \end{bmatrix}$$ where $x_2$ and $x_3$ are free variables. We want to find a parametrization for the solution set to $Ax = 0$ using $x_2$ and $x_3$. 2. **Recall the theory:** The general solution to $Ax = b$ can be written as: $$x = x_p + x_h$$ where $x_p$ is a particular solution to $Ax = b$, and $x_h$ is the general solution to the homogeneous system $Ax = 0$. 3. **Identify the particular solution and homogeneous solutions:** From the given parametrization: - The vector $$\begin{bmatrix} -6 \\ 4 \\ 1 \end{bmatrix}$$ is a particular solution $x_p$. - The vectors $$\begin{bmatrix} -7 \\ 9 \\ 8 \end{bmatrix}$$ and $$\begin{bmatrix} 5 \\ 0 \\ 9 \end{bmatrix}$$ span the null space (solutions to $Ax=0$). 4. **Write the parametrization for $Ax=0$:** The solution set to $Ax=0$ is all linear combinations of the null space vectors: $$\begin{bmatrix} x \\ y \\ z \\ 1 \end{bmatrix} = x_2 \begin{bmatrix} -7 \\ 9 \\ 8 \end{bmatrix} + x_3 \begin{bmatrix} 5 \\ 0 \\ 9 \end{bmatrix}$$ Since the last coordinate is $1$ in the original parametrization but the homogeneous solutions correspond to the vector part without the constant term, the last coordinate for the homogeneous solution is $0$ (not shown explicitly in the problem but implied). Therefore, the parametrization for $Ax=0$ is: $$\begin{bmatrix} x \\ y \\ z \end{bmatrix} = x_2 \begin{bmatrix} -7 \\ 9 \\ 8 \end{bmatrix} + x_3 \begin{bmatrix} 5 \\ 0 \\ 9 \end{bmatrix}$$ 5. **Final answer:** $$\boxed{\begin{bmatrix} x \\ y \\ z \end{bmatrix} = x_2 \begin{bmatrix} -7 \\ 9 \\ 8 \end{bmatrix} + x_3 \begin{bmatrix} 5 \\ 0 \\ 9 \end{bmatrix}}$$