1. **Problem Statement:** Find a spanning set for the nullspace of the matrix $$A = \begin{bmatrix}-3 & 6 & -1 & 1 & -7 \\ 1 & -2 & 2 & 3 & -1 \\ 2 & -4 & 5 & 8 & -4 \end{bmatrix}$$ 2. **Recall:** The nullspace of a matrix $A$ consists of all vectors $\mathbf{x}$ such that $A\mathbf{x} = \mathbf{0}$. To find a spanning set, we solve the homogeneous system: $$A\mathbf{x} = \mathbf{0}$$ where $\mathbf{x} = \begin{bmatrix}x_1 & x_2 & x_3 & x_4 & x_5\end{bmatrix}^T$. 3. **Set up the system:** $$\begin{cases} -3x_1 + 6x_2 - x_3 + x_4 - 7x_5 = 0 \\ x_1 - 2x_2 + 2x_3 + 3x_4 - x_5 = 0 \\ 2x_1 - 4x_2 + 5x_3 + 8x_4 - 4x_5 = 0 \end{cases}$$ 4. **Form the augmented matrix and perform row reduction:** $$\left[\begin{array}{ccccc|c} -3 & 6 & -1 & 1 & -7 & 0 \\ 1 & -2 & 2 & 3 & -1 & 0 \\ 2 & -4 & 5 & 8 & -4 & 0 \end{array}\right]$$ 5. **Row operations:** - Swap $R_1$ and $R_2$ to get a leading 1: $$\left[\begin{array}{ccccc|c} 1 & -2 & 2 & 3 & -1 & 0 \\ -3 & 6 & -1 & 1 & -7 & 0 \\ 2 & -4 & 5 & 8 & -4 & 0 \end{array}\right]$$ - Replace $R_2$ by $R_2 + 3R_1$: $$\left[\begin{array}{ccccc|c} 1 & -2 & 2 & 3 & -1 & 0 \\ 0 & 0 & 5 & 10 & -10 & 0 \\ 2 & -4 & 5 & 8 & -4 & 0 \end{array}\right]$$ - Replace $R_3$ by $R_3 - 2R_1$: $$\left[\begin{array}{ccccc|c} 1 & -2 & 2 & 3 & -1 & 0 \\ 0 & 0 & 5 & 10 & -10 & 0 \\ 0 & 0 & 1 & 2 & -2 & 0 \end{array}\right]$$ - Replace $R_2$ by $R_2 - 5R_3$: $$\left[\begin{array}{ccccc|c} 1 & -2 & 2 & 3 & -1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 2 & -2 & 0 \end{array}\right]$$ 6. **Back substitution:** From $R_3$: $$x_3 + 2x_4 - 2x_5 = 0 \implies x_3 = -2x_4 + 2x_5$$ From $R_1$: $$x_1 - 2x_2 + 2x_3 + 3x_4 - x_5 = 0$$ Substitute $x_3$: $$x_1 - 2x_2 + 2(-2x_4 + 2x_5) + 3x_4 - x_5 = 0$$ $$x_1 - 2x_2 - 4x_4 + 4x_5 + 3x_4 - x_5 = 0$$ $$x_1 - 2x_2 - x_4 + 3x_5 = 0$$ $$x_1 = 2x_2 + x_4 - 3x_5$$ 7. **Free variables:** Let $x_2 = s$, $x_4 = t$, $x_5 = u$ be free parameters. 8. **Express solution vector:** $$\mathbf{x} = \begin{bmatrix}x_1 \\ x_2 \\ x_3 \\ x_4 \\ x_5\end{bmatrix} = \begin{bmatrix}2s + t - 3u \\ s \\ -2t + 2u \\ t \\ u\end{bmatrix} = s \begin{bmatrix}2 \\ 1 \\ 0 \\ 0 \\ 0\end{bmatrix} + t \begin{bmatrix}1 \\ 0 \\ -2 \\ 1 \\ 0\end{bmatrix} + u \begin{bmatrix}-3 \\ 0 \\ 2 \\ 0 \\ 1\end{bmatrix}$$ 9. **Conclusion:** A spanning set for the nullspace of $A$ is: $$\left\{ \begin{bmatrix}2 \\ 1 \\ 0 \\ 0 \\ 0\end{bmatrix}, \begin{bmatrix}1 \\ 0 \\ -2 \\ 1 \\ 0\end{bmatrix}, \begin{bmatrix}-3 \\ 0 \\ 2 \\ 0 \\ 1\end{bmatrix} \right\}$$