1. **State the problem:** We need to transform the matrix $$A=\begin{pmatrix} -\frac{5}{2} & -\frac{7}{4} & -5 & \frac{35}{3} \\ -1 & \frac{3}{4} & -2 & 3 \\ \frac{1}{4} & -2 & \frac{1}{2} & \frac{4}{3} \end{pmatrix}$$ into its reduced row echelon form (RREF). 2. **Recall the RREF definition:** - 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. **Apply Gaussian elimination steps:** - Use row operations (swap, multiply, add multiples) to create leading 1s and zeros below and above them. 4. **Step-by-step transformation:** - Multiply Row 1 by $-\frac{2}{5}$ to get leading 1: $$R1 \to -\frac{2}{5} R1 = \left(1, \frac{7}{10}, 2, -\frac{14}{15}\right)$$ - Eliminate first column entries in Rows 2 and 3: $$R2 \to R2 + R1 = (0, \frac{17}{20}, 0, \frac{31}{15})$$ $$R3 \to R3 - \frac{1}{4} R1 = (0, -\frac{51}{40}, 0, \frac{59}{30})$$ - Multiply Row 2 by $\frac{20}{17}$ to get leading 1 in second column: $$R2 \to \left(0, 1, 0, \frac{62}{51}\right)$$ - Eliminate second column entry in Row 3: $$R3 \to R3 + \frac{51}{40} R2 = (0, 0, 0, \frac{59}{30} + \frac{51}{40} \times \frac{62}{51}) = (0, 0, 0, \frac{59}{30} + \frac{62}{40}) = (0, 0, 0, \frac{59}{30} + \frac{31}{20})$$ Calculate common denominator 60: $$\frac{59}{30} = \frac{118}{60}, \quad \frac{31}{20} = \frac{93}{60}$$ Sum: $$\frac{118}{60} + \frac{93}{60} = \frac{211}{60}$$ - Since Row 3 has zeros in first three columns but nonzero last entry, the system is inconsistent if this were an augmented matrix for equations. But here, we continue to get RREF. - Eliminate second column entry in Row 1: $$R1 \to R1 - \frac{7}{10} R2 = \left(1, 0, 2, -\frac{14}{15} - \frac{7}{10} \times \frac{62}{51}\right)$$ Calculate: $$\frac{7}{10} \times \frac{62}{51} = \frac{434}{510} = \frac{217}{255}$$ Convert $-\frac{14}{15}$ to denominator 255: $$-\frac{14}{15} = -\frac{238}{255}$$ Sum: $$-\frac{238}{255} - \frac{217}{255} = -\frac{455}{255} = -\frac{91}{51}$$ So $$R1 = (1, 0, 2, -\frac{91}{51})$$ - Since third column has no pivot, and Row 3 is zero in first three columns, the RREF is: $$\begin{pmatrix} 1 & 0 & 2 & -\frac{91}{51} \\ 0 & 1 & 0 & \frac{62}{51} \\ 0 & 0 & 0 & \frac{211}{60} \end{pmatrix}$$ 5. **Interpretation:** The matrix is in RREF except the last row indicates inconsistency if this is an augmented matrix for a system of equations. If just matrix transformation, this is the RREF. **Final RREF matrix:** $$\begin{pmatrix} 1 & 0 & 2 & -\frac{91}{51} \\ 0 & 1 & 0 & \frac{62}{51} \\ 0 & 0 & 0 & \frac{211}{60} \end{pmatrix}$$