1. **State the problem:** Find bases for the row space, column space, and null space of the matrix $$ A = \begin{bmatrix} 1 & 2 & -3 & 4 & 0 \\ -1 & 0 & 2 & -4 & 1 \\ -1 & 4 & 0 & -4 & 3 \end{bmatrix} $$ 2. **Recall definitions and formulas:** - The **row space** is the span of the rows of $A$. - The **column space** is the span of the columns of $A$. - The **null space** is the set of all vectors $x$ such that $Ax=0$. To find bases, we use row reduction to echelon form. 3. **Row reduce $A$:** Start with $$ \begin{bmatrix} 1 & 2 & -3 & 4 & 0 \\ -1 & 0 & 2 & -4 & 1 \\ -1 & 4 & 0 & -4 & 3 \end{bmatrix} $$ Add row 1 to row 2: $$ R_2 = R_2 + R_1 \Rightarrow \begin{bmatrix} 1 & 2 & -3 & 4 & 0 \\ 0 & 2 & -1 & 0 & 1 \\ -1 & 4 & 0 & -4 & 3 \end{bmatrix} $$ Add row 1 to row 3: $$ R_3 = R_3 + R_1 \Rightarrow \begin{bmatrix} 1 & 2 & -3 & 4 & 0 \\ 0 & 2 & -1 & 0 & 1 \\ 0 & 6 & -3 & 0 & 3 \end{bmatrix} $$ Replace row 3 by $R_3 - 3R_2$: $$ R_3 = R_3 - 3R_2 = \begin{bmatrix} 1 & 2 & -3 & 4 & 0 \\ 0 & 2 & -1 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 \end{bmatrix} $$ Divide row 2 by 2: $$ R_2 = \frac{1}{2} R_2 = \begin{bmatrix} 1 & 2 & -3 & 4 & 0 \\ 0 & 1 & -\frac{1}{2} & 0 & \frac{1}{2} \\ 0 & 0 & 0 & 0 & 0 \end{bmatrix} $$ Eliminate the 2 in row 1, column 2: $$ R_1 = R_1 - 2R_2 = \begin{bmatrix} 1 & 0 & -2 & 4 & -1 \\ 0 & 1 & -\frac{1}{2} & 0 & \frac{1}{2} \\ 0 & 0 & 0 & 0 & 0 \end{bmatrix} $$ 4. **Bases for row space:** The nonzero rows of the row echelon form form a basis: $$ \{ (1,0,-2,4,-1), (0,1,-\frac{1}{2},0,\frac{1}{2}) \} $$ 5. **Bases for column space:** Pivot columns in original matrix correspond to columns 1 and 2. Columns 1 and 2 of $A$ are: $$ \begin{bmatrix} 1 \\ -1 \\ -1 \end{bmatrix}, \quad \begin{bmatrix} 2 \\ 0 \\ 4 \end{bmatrix} $$ So a basis for column space is: $$ \left\{ \begin{bmatrix} 1 \\ -1 \\ -1 \end{bmatrix}, \begin{bmatrix} 2 \\ 0 \\ 4 \end{bmatrix} \right\} $$ 6. **Basis for null space:** Solve $Ax=0$ using reduced form: Let $x = (x_1,x_2,x_3,x_4,x_5)^T$. From row 2: $$ x_2 - \frac{1}{2} x_3 + \frac{1}{2} x_5 = 0 \Rightarrow x_2 = \frac{1}{2} x_3 - \frac{1}{2} x_5 $$ From row 1: $$ x_1 - 2 x_3 + 4 x_4 - x_5 = 0 \Rightarrow x_1 = 2 x_3 - 4 x_4 + x_5 $$ Variables $x_3, x_4, x_5$ are free. Write solution vector: $$ x = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \\ x_5 \end{bmatrix} = x_3 \begin{bmatrix} 2 \\ \frac{1}{2} \\ 1 \\ 0 \\ 0 \end{bmatrix} + x_4 \begin{bmatrix} -4 \\ 0 \\ 0 \\ 1 \\ 0 \end{bmatrix} + x_5 \begin{bmatrix} 1 \\ -\frac{1}{2} \\ 0 \\ 0 \\ 1 \end{bmatrix} $$ So a basis for null space is: $$ \left\{ \begin{bmatrix} 2 \\ \frac{1}{2} \\ 1 \\ 0 \\ 0 \end{bmatrix}, \begin{bmatrix} -4 \\ 0 \\ 0 \\ 1 \\ 0 \end{bmatrix}, \begin{bmatrix} 1 \\ -\frac{1}{2} \\ 0 \\ 0 \\ 1 \end{bmatrix} \right\} $$ **Final answers:** - Basis row space: $\{(1,0,-2,4,-1),(0,1,-\frac{1}{2},0,\frac{1}{2})\}$ - Basis column space: $\{(1,-1,-1),(2,0,4)\}$ - Basis null space: $\left\{ \begin{bmatrix} 2 \\ \frac{1}{2} \\ 1 \\ 0 \\ 0 \end{bmatrix}, \begin{bmatrix} -4 \\ 0 \\ 0 \\ 1 \\ 0 \end{bmatrix}, \begin{bmatrix} 1 \\ -\frac{1}{2} \\ 0 \\ 0 \\ 1 \end{bmatrix} \right\}$