1. **Problem Statement:** Find the row space, column space, and null space of the matrix $$\mathbf{A} = \begin{pmatrix} 2 & 0 & -1 & 3 \\ -3 & 1 & -2 & 1 \\ 2 & 0 & -2 & 2 \end{pmatrix}$$ 2. **Row Space:** The row space is the span of the linearly independent rows of $\mathbf{A}$. We find the row echelon form (REF) of $\mathbf{A}$ by row reducing: Start with $\mathbf{A}$: $$\begin{pmatrix} 2 & 0 & -1 & 3 \\ -3 & 1 & -2 & 1 \\ 2 & 0 & -2 & 2 \end{pmatrix}$$ Divide row 1 by 2: $$\begin{pmatrix} 1 & 0 & -\frac{1}{2} & \frac{3}{2} \\ -3 & 1 & -2 & 1 \\ 2 & 0 & -2 & 2 \end{pmatrix}$$ Add 3 times row 1 to row 2: $$\begin{pmatrix} 1 & 0 & -\frac{1}{2} & \frac{3}{2} \\ 0 & 1 & -\frac{7}{2} & \frac{11}{2} \\ 2 & 0 & -2 & 2 \end{pmatrix}$$ Subtract 2 times row 1 from row 3: $$\begin{pmatrix} 1 & 0 & -\frac{1}{2} & \frac{3}{2} \\ 0 & 1 & -\frac{7}{2} & \frac{11}{2} \\ 0 & 0 & -1 & -1 \end{pmatrix}$$ Multiply row 3 by -1: $$\begin{pmatrix} 1 & 0 & -\frac{1}{2} & \frac{3}{2} \\ 0 & 1 & -\frac{7}{2} & \frac{11}{2} \\ 0 & 0 & 1 & 1 \end{pmatrix}$$ Add $\frac{1}{2}$ times row 3 to row 1: $$\begin{pmatrix} 1 & 0 & 0 & 2 \\ 0 & 1 & -\frac{7}{2} & \frac{11}{2} \\ 0 & 0 & 1 & 1 \end{pmatrix}$$ Add $\frac{7}{2}$ times row 3 to row 2: $$\begin{pmatrix} 1 & 0 & 0 & 2 \\ 0 & 1 & 0 & 9 \\ 0 & 0 & 1 & 1 \end{pmatrix}$$ The nonzero rows form a basis for the row space: $$\text{Row space} = \text{span} \left\{ \begin{pmatrix} 1 & 0 & 0 & 2 \end{pmatrix}, \begin{pmatrix} 0 & 1 & 0 & 9 \end{pmatrix}, \begin{pmatrix} 0 & 0 & 1 & 1 \end{pmatrix} \right\}$$ 3. **Column Space:** The column space is the span of the pivot columns of the original matrix $\mathbf{A}$. From the REF, pivots are in columns 1, 2, and 3. So the column space is spanned by columns 1, 2, and 3 of $\mathbf{A}$: $$\text{Col space} = \text{span} \left\{ \begin{pmatrix} 2 \\ -3 \\ 2 \end{pmatrix}, \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}, \begin{pmatrix} -1 \\ -2 \\ -2 \end{pmatrix} \right\}$$ 4. **Null Space:** Solve $\mathbf{A} \mathbf{x} = \mathbf{0}$. From REF, variables correspond to columns: - $x_1, x_2, x_3$ are pivots - $x_4$ is free From the REF system: \[ \begin{cases} x_1 + 2x_4 = 0 \\ x_2 + 9x_4 = 0 \\ x_3 + x_4 = 0 \end{cases} \] Express pivot variables in terms of free variable $x_4 = t$: $$x_1 = -2t, \quad x_2 = -9t, \quad x_3 = -t, \quad x_4 = t$$ Vector form: $$\mathbf{x} = t \begin{pmatrix} -2 \\ -9 \\ -1 \\ 1 \end{pmatrix}$$ So null space is: $$\text{Null space} = \text{span} \left\{ \begin{pmatrix} -2 \\ -9 \\ -1 \\ 1 \end{pmatrix} \right\}$$