1. **Problem Statement:** Find the Reduced Row Echelon Form (RREF) of the matrix $$\begin{bmatrix}1 & 2 & 1 & 4 \\ 2 & 4 & -3 & 1 \\ 3 & 6 & -5 & 2\end{bmatrix}$$. 2. **Formula and Rules:** RREF is a form of a matrix where: - Each leading entry in a row is 1. - Each leading 1 is the only nonzero entry in its column. - Leading 1s move to the right as you go down rows. - Rows with all zeros are at the bottom. 3. **Step-by-step Solution:** - Start with the matrix: $$\begin{bmatrix}1 & 2 & 1 & 4 \\ 2 & 4 & -3 & 1 \\ 3 & 6 & -5 & 2\end{bmatrix}$$ - Use row 1 to eliminate entries below the leading 1: - Replace row 2 with row 2 minus 2 times row 1: $$R_2 = R_2 - 2R_1 = \begin{bmatrix}2 - 2(1) & 4 - 2(2) & -3 - 2(1) & 1 - 2(4)\end{bmatrix} = \begin{bmatrix}0 & 0 & -5 & -7\end{bmatrix}$$ - Replace row 3 with row 3 minus 3 times row 1: $$R_3 = R_3 - 3R_1 = \begin{bmatrix}3 - 3(1) & 6 - 3(2) & -5 - 3(1) & 2 - 3(4)\end{bmatrix} = \begin{bmatrix}0 & 0 & -8 & -10\end{bmatrix}$$ - The matrix is now: $$\begin{bmatrix}1 & 2 & 1 & 4 \\ 0 & 0 & -5 & -7 \\ 0 & 0 & -8 & -10\end{bmatrix}$$ - Make the leading coefficient in row 2 a 1 by dividing row 2 by -5: $$R_2 = \frac{1}{-5} R_2 = \begin{bmatrix}0 & 0 & 1 & \frac{7}{5}\end{bmatrix}$$ - Use row 2 to eliminate the entry in row 3, column 3: $$R_3 = R_3 - (-8)R_2 = R_3 + 8R_2 = \begin{bmatrix}0 & 0 & -8 + 8(1) & -10 + 8(\frac{7}{5})\end{bmatrix} = \begin{bmatrix}0 & 0 & 0 & \frac{6}{5}\end{bmatrix}$$ - The matrix is now: $$\begin{bmatrix}1 & 2 & 1 & 4 \\ 0 & 0 & 1 & \frac{7}{5} \\ 0 & 0 & 0 & \frac{6}{5}\end{bmatrix}$$ - Make the leading coefficient in row 3 a 1 by multiplying row 3 by $\frac{5}{6}$: $$R_3 = \frac{5}{6} R_3 = \begin{bmatrix}0 & 0 & 0 & 1\end{bmatrix}$$ - Use row 3 to eliminate the entry in row 1, column 4: $$R_1 = R_1 - 4R_3 = \begin{bmatrix}1 & 2 & 1 & 0\end{bmatrix}$$ - Use row 3 to eliminate the entry in row 2, column 4: $$R_2 = R_2 - \frac{7}{5} R_3 = \begin{bmatrix}0 & 0 & 1 & 0\end{bmatrix}$$ - Use row 2 to eliminate the entry in row 1, column 3: $$R_1 = R_1 - R_2 = \begin{bmatrix}1 & 2 & 0 & 0\end{bmatrix}$$ - The matrix is now: $$\begin{bmatrix}1 & 2 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1\end{bmatrix}$$ - Since the second column has no leading 1, swap rows or note it as a free variable. Here, the second column corresponds to a free variable. 4. **Final RREF matrix:** $$\begin{bmatrix}1 & 2 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1\end{bmatrix}$$ This is the Reduced Row Echelon Form of the original matrix.