1. **Problem Statement:** Solve matrices B and C using the row reduction method (Gaussian elimination) to find their reduced row echelon forms (RREF). 2. **Row Reduction Method:** Row reduction transforms a matrix into a simpler form using elementary row operations: swapping rows, multiplying a row by a nonzero scalar, and adding multiples of one row to another. 3. **Matrix B:** $$B=\begin{bmatrix}14 & 0 & 6 & 0 \\ 0 & 9 & 8 & 12 \\ 13 & 7 & 0 & 11 \\ 10 & 0 & 8 & 10\end{bmatrix}$$ Step 1: Make the first pivot 1 by dividing row 1 by 14: $$R_1 \to \frac{1}{14}R_1 = \begin{bmatrix}1 & 0 & \frac{6}{14} & 0\end{bmatrix} = \begin{bmatrix}1 & 0 & \frac{3}{7} & 0\end{bmatrix}$$ Step 2: Eliminate first column entries below pivot: $$R_3 \to R_3 - 13R_1 = \begin{bmatrix}13 - 13(1), 7 - 13(0), 0 - 13(\frac{3}{7}), 11 - 13(0)\end{bmatrix} = \begin{bmatrix}0, 7, -\frac{39}{7}, 11\end{bmatrix}$$ $$R_4 \to R_4 - 10R_1 = \begin{bmatrix}10 - 10(1), 0 - 10(0), 8 - 10(\frac{3}{7}), 10 - 10(0)\end{bmatrix} = \begin{bmatrix}0, 0, 8 - \frac{30}{7}, 10\end{bmatrix} = \begin{bmatrix}0, 0, \frac{26}{7}, 10\end{bmatrix}$$ Matrix now: $$\begin{bmatrix}1 & 0 & \frac{3}{7} & 0 \\ 0 & 9 & 8 & 12 \\ 0 & 7 & -\frac{39}{7} & 11 \\ 0 & 0 & \frac{26}{7} & 10\end{bmatrix}$$ Step 3: Make pivot in row 2, column 2 equal to 1 by dividing row 2 by 9: $$R_2 \to \frac{1}{9}R_2 = \begin{bmatrix}0, 1, \frac{8}{9}, \frac{12}{9}\end{bmatrix} = \begin{bmatrix}0, 1, \frac{8}{9}, \frac{4}{3}\end{bmatrix}$$ Step 4: Eliminate second column entries above and below pivot: $$R_3 \to R_3 - 7R_2 = \begin{bmatrix}0, 7 - 7(1), -\frac{39}{7} - 7(\frac{8}{9}), 11 - 7(\frac{4}{3})\end{bmatrix} = \begin{bmatrix}0, 0, -\frac{39}{7} - \frac{56}{9}, 11 - \frac{28}{3}\end{bmatrix}$$ Calculate: $$-\frac{39}{7} - \frac{56}{9} = -\frac{351}{63} - \frac{392}{63} = -\frac{743}{63}$$ $$11 - \frac{28}{3} = \frac{33}{3} - \frac{28}{3} = \frac{5}{3}$$ So, $$R_3 = \begin{bmatrix}0, 0, -\frac{743}{63}, \frac{5}{3}\end{bmatrix}$$ Step 5: Eliminate second column entry in row 4 (already zero, no change). Step 6: Make pivot in row 4, column 3 equal to 1 by dividing row 4 by $\frac{26}{7}$: $$R_4 \to \frac{7}{26} R_4 = \begin{bmatrix}0, 0, 1, \frac{70}{26}\end{bmatrix} = \begin{bmatrix}0, 0, 1, \frac{35}{13}\end{bmatrix}$$ Step 7: Eliminate third column entries above pivot in rows 1 and 3: $$R_1 \to R_1 - \frac{3}{7} R_4 = \begin{bmatrix}1, 0, 0, 0 - \frac{3}{7} \times \frac{35}{13}\end{bmatrix} = \begin{bmatrix}1, 0, 0, -\frac{15}{26}\end{bmatrix}$$ $$R_3 \to R_3 + \frac{743}{63} R_4 = \begin{bmatrix}0, 0, 0, \frac{5}{3} + \frac{743}{63} \times \frac{35}{13}\end{bmatrix}$$ Calculate: $$\frac{743}{63} \times \frac{35}{13} = \frac{743 \times 35}{63 \times 13} = \frac{26005}{819}$$ $$\frac{5}{3} = \frac{1365}{819}$$ Sum: $$\frac{1365}{819} + \frac{26005}{819} = \frac{27370}{819}$$ So, $$R_3 = \begin{bmatrix}0, 0, 0, \frac{27370}{819}\end{bmatrix}$$ Step 8: Make pivot in row 3, column 4 equal to 1 by dividing row 3 by $\frac{27370}{819}$: $$R_3 \to \frac{819}{27370} R_3 = \begin{bmatrix}0, 0, 0, 1\end{bmatrix}$$ Step 9: Eliminate fourth column entries above pivot in rows 2 and 4: $$R_2 \to R_2 - \frac{4}{3} R_3 = \begin{bmatrix}0, 1, \frac{8}{9}, 0\end{bmatrix}$$ $$R_4 \to R_4 - \frac{35}{13} R_3 = \begin{bmatrix}0, 0, 1, 0\end{bmatrix}$$ **Final RREF of B:** $$\begin{bmatrix}1 & 0 & 0 & -\frac{15}{26} \\ 0 & 1 & \frac{8}{9} & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0\end{bmatrix}$$ --- 4. **Matrix C:** $$C=\begin{bmatrix}0 & 8 & 7 & 15 \\ 13 & 9 & 0 & 14 \\ 14 & 0 & 6 & 12 \\ 0 & 3 & 8 & 10\end{bmatrix}$$ Step 1: Swap row 1 and row 2 to get a nonzero pivot in first row: $$R_1 \leftrightarrow R_2$$ New matrix: $$\begin{bmatrix}13 & 9 & 0 & 14 \\ 0 & 8 & 7 & 15 \\ 14 & 0 & 6 & 12 \\ 0 & 3 & 8 & 10\end{bmatrix}$$ Step 2: Make pivot in row 1, column 1 equal to 1 by dividing row 1 by 13: $$R_1 \to \frac{1}{13} R_1 = \begin{bmatrix}1, \frac{9}{13}, 0, \frac{14}{13}\end{bmatrix}$$ Step 3: Eliminate first column entries below pivot: $$R_3 \to R_3 - 14 R_1 = \begin{bmatrix}14 - 14(1), 0 - 14(\frac{9}{13}), 6 - 14(0), 12 - 14(\frac{14}{13})\end{bmatrix} = \begin{bmatrix}0, -\frac{126}{13}, 6, 12 - \frac{196}{13}\end{bmatrix}$$ Calculate: $$12 - \frac{196}{13} = \frac{156}{13} - \frac{196}{13} = -\frac{40}{13}$$ So, $$R_3 = \begin{bmatrix}0, -\frac{126}{13}, 6, -\frac{40}{13}\end{bmatrix}$$ Rows 2 and 4 have zero in first column, no change. Step 4: Make pivot in row 2, column 2 equal to 1 by dividing row 2 by 8: $$R_2 \to \frac{1}{8} R_2 = \begin{bmatrix}0, 1, \frac{7}{8}, \frac{15}{8}\end{bmatrix}$$ Step 5: Eliminate second column entries in rows 1, 3, and 4: $$R_1 \to R_1 - \frac{9}{13} R_2 = \begin{bmatrix}1, 0, 0 - \frac{9}{13} \times \frac{7}{8}, \frac{14}{13} - \frac{9}{13} \times \frac{15}{8}\end{bmatrix}$$ Calculate: $$0 - \frac{9}{13} \times \frac{7}{8} = -\frac{63}{104}$$ $$\frac{14}{13} - \frac{9}{13} \times \frac{15}{8} = \frac{14}{13} - \frac{135}{104} = \frac{112}{104} - \frac{135}{104} = -\frac{23}{104}$$ So, $$R_1 = \begin{bmatrix}1, 0, -\frac{63}{104}, -\frac{23}{104}\end{bmatrix}$$ $$R_3 \to R_3 + \frac{126}{13} R_2 = \begin{bmatrix}0, 0, 6 + \frac{126}{13} \times \frac{7}{8}, -\frac{40}{13} + \frac{126}{13} \times \frac{15}{8}\end{bmatrix}$$ Calculate: $$\frac{126}{13} \times \frac{7}{8} = \frac{882}{104} = \frac{441}{52}$$ $$6 = \frac{312}{52}$$ Sum: $$\frac{312}{52} + \frac{441}{52} = \frac{753}{52}$$ $$-\frac{40}{13} = -\frac{160}{52}$$ $$\frac{126}{13} \times \frac{15}{8} = \frac{1890}{104} = \frac{945}{52}$$ Sum: $$-\frac{160}{52} + \frac{945}{52} = \frac{785}{52}$$ So, $$R_3 = \begin{bmatrix}0, 0, \frac{753}{52}, \frac{785}{52}\end{bmatrix}$$ $$R_4 \to R_4 - 3 R_2 = \begin{bmatrix}0, 0, 8 - 3 \times \frac{7}{8}, 10 - 3 \times \frac{15}{8}\end{bmatrix} = \begin{bmatrix}0, 0, 8 - \frac{21}{8}, 10 - \frac{45}{8}\end{bmatrix}$$ Calculate: $$8 = \frac{64}{8}$$ $$\frac{64}{8} - \frac{21}{8} = \frac{43}{8}$$ $$10 = \frac{80}{8}$$ $$\frac{80}{8} - \frac{45}{8} = \frac{35}{8}$$ So, $$R_4 = \begin{bmatrix}0, 0, \frac{43}{8}, \frac{35}{8}\end{bmatrix}$$ Step 6: Make pivot in row 3, column 3 equal to 1 by dividing row 3 by $\frac{753}{52}$: $$R_3 \to \frac{52}{753} R_3 = \begin{bmatrix}0, 0, 1, \frac{785}{52} \times \frac{52}{753}\end{bmatrix} = \begin{bmatrix}0, 0, 1, \frac{785}{753}\end{bmatrix}$$ Step 7: Eliminate third column entries in rows 1 and 4: $$R_1 \to R_1 + \frac{63}{104} R_3 = \begin{bmatrix}1, 0, 0, -\frac{23}{104} + \frac{63}{104} \times \frac{785}{753}\end{bmatrix}$$ Calculate: $$\frac{63}{104} \times \frac{785}{753} = \frac{49455}{78312} \approx 0.631$$ $$-\frac{23}{104} = -0.221$$ Sum approx: $$-0.221 + 0.631 = 0.41$$ So, $$R_1 = \begin{bmatrix}1, 0, 0, 0.41\end{bmatrix}$$ $$R_4 \to R_4 - \frac{43}{8} R_3 = \begin{bmatrix}0, 0, 0, \frac{35}{8} - \frac{43}{8} \times \frac{785}{753}\end{bmatrix}$$ Calculate: $$\frac{43}{8} \times \frac{785}{753} = \frac{33755}{6024} \approx 5.6$$ $$\frac{35}{8} = 4.375$$ Difference: $$4.375 - 5.6 = -1.225$$ So, $$R_4 = \begin{bmatrix}0, 0, 0, -1.225\end{bmatrix}$$ Step 8: Make pivot in row 4, column 4 equal to 1 by dividing row 4 by -1.225: $$R_4 \to -\frac{1}{1.225} R_4 = \begin{bmatrix}0, 0, 0, 1\end{bmatrix}$$ Step 9: Eliminate fourth column entries in rows 1, 2, and 3: $$R_1 \to R_1 - 0.41 R_4 = \begin{bmatrix}1, 0, 0, 0\end{bmatrix}$$ $$R_2 \to R_2 - \frac{15}{8} R_4 = \begin{bmatrix}0, 1, \frac{7}{8}, 0\end{bmatrix}$$ $$R_3 \to R_3 - \frac{785}{753} R_4 = \begin{bmatrix}0, 0, 1, 0\end{bmatrix}$$ **Final RREF of C:** $$\begin{bmatrix}1 & 0 & 0 & 0 \\ 0 & 1 & \frac{7}{8} & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1\end{bmatrix}$$ --- **Summary:** - Matrix B reduced to RREF with pivots in columns 1, 2, 3, and 4. - Matrix C reduced to RREF with pivots in columns 1, 2, 3, and 4. These forms help analyze linear independence and solve systems involving these matrices.