1. Problem Statement: Find the reduced row echelon form (RREF) of each of the given matrices and record the row operations. 2. For matrix (a): $$A=\begin{bmatrix} 1 & 0 & -2 \\ -2 & 1 & 9 \\ 3 & 2 & 4 \end{bmatrix}$$ - Step 1: Use $R_1$ to eliminate below pivots: - $R_2 \to R_2 + 2R_1 = \begin{bmatrix} 0 & 1 & 5 \end{bmatrix}$ - $R_3 \to R_3 - 3R_1 = \begin{bmatrix} 0 & 2 & 10 \end{bmatrix}$ - Step 2: Make pivot in $R_2,2$ a leading 1 (already 1), use it to eliminate $R_3,2$: - $R_3 \to R_3 - 2R_2 = \begin{bmatrix} 0 & 0 & 0 \end{bmatrix}$ - Step 3: Use $R_2$ to eliminate above: - $R_1 \to R_1 - 0\cdot R_2$ (no change) RREF of (a): $$\begin{bmatrix} 1 & 0 & -2 \\ 0 & 1 & 5 \\ 0 & 0 & 0 \end{bmatrix}$$ 3. For matrix (b): $$A=\begin{bmatrix} 1 & 0 & 1 \\ -1 & 2 & -2 \\ 0 & 1 & 0 \\ -2 & 7 & -5 \end{bmatrix}$$ - Step 1: Use $R_1$ to eliminate first column below pivot: - $R_2 \to R_2 + R_1 = \begin{bmatrix} 0 & 2 & -1 \end{bmatrix}$ - $R_4 \to R_4 + 2R_1 = \begin{bmatrix} 0 & 7 & -3 \end{bmatrix}$ - Step 2: Make a pivot in $R_2,2$ by dividing $R_2$ by 2: - $R_2 \to \frac{1}{2} R_2 = \begin{bmatrix} 0 & 1 & -\frac{1}{2} \end{bmatrix}$ - Step 3: Eliminate other entries in second column: - $R_3 \to R_3 - R_2 = \begin{bmatrix} 0 & 0 & \frac{1}{2} \end{bmatrix}$ - $R_4 \to R_4 - 7R_2 = \begin{bmatrix} 0 & 0 & \frac{1}{2} \end{bmatrix}$ - $R_1 \to R_1 - 0 \cdot R_2$ (no change) - Step 4: Normalize third column pivot by dividing $R_3$ by $\frac{1}{2}$: - $R_3 \to 2R_3 = \begin{bmatrix} 0 & 0 & 1 \end{bmatrix}$ - Step 5: Use $R_3$ to eliminate third column entries: - $R_1 \to R_1 - R_3 = \begin{bmatrix} 1 & 0 & 0 \end{bmatrix}$ - $R_2 \to R_2 + \frac{1}{2} R_3 = \begin{bmatrix} 0 & 1 & 0 \end{bmatrix}$ - $R_4 \to R_4 - \frac{1}{2} R_3 = \begin{bmatrix} 0 & 0 & 0 \end{bmatrix}$ RREF of (b): $$\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix}$$ 4. For matrix 6 (a): $$A=\begin{bmatrix} -1 & 2 & -5 \\ 2 & -1 & 6 \end{bmatrix}$$ - Step 1: Make first pivot 1 by multiplying $R_1$ by $-1$: - $R_1 \to -1 \cdot R_1 = \begin{bmatrix} 1 & -2 & 5 \end{bmatrix}$ - Step 2: Eliminate below pivot: - $R_2 \to R_2 - 2R_1 = \begin{bmatrix} 0 & 3 & -4 \end{bmatrix}$ - Step 3: Make pivot in $R_2,2$ equal to 1 by dividing by 3: - $R_2 \to \frac{1}{3} R_2 = \begin{bmatrix} 0 & 1 & -\frac{4}{3} \end{bmatrix}$ - Step 4: Eliminate above using $R_2$: - $R_1 \to R_1 + 2R_2 = \begin{bmatrix} 1 & 0 & \frac{7}{3} \end{bmatrix}$ RREF of matrix 6 (a): $$\begin{bmatrix} 1 & 0 & \frac{7}{3} \\ 0 & 1 & -\frac{4}{3} \end{bmatrix}$$