Find Reduced Row Echelon Form [[x,cos(3x),sin(4x)],[-2,1,3],[1,1,2]]

Problem

[[x,cos(3*x),sin(4*x)],[−2,1,3],[1,1,2]]

Solution

1. Swap rows to place a leading 1 in the first row. We swap (R_1) and (R_3)

[[1,1,2],[−2,1,3],[x,cos(3*x),sin(4*x)]]

2. Eliminate the first entry of the second row by performing (R_2)→(R_2)+2*(R_1)

[[1,1,2],[0,3,7],[x,cos(3*x),sin(4*x)]]

3. Eliminate the first entry of the third row by performing (R_3)→(R_3)−x*(R_1)

[[1,1,2],[0,3,7],[0,cos(3*x)−x,sin(4*x)−2*x]]

4. Normalize the second row by performing (R_2)→1/3*(R_2)

[[1,1,2],[0,1,7/3],[0,cos(3*x)−x,sin(4*x)−2*x]]

5. Eliminate the second entry of the first row by performing (R_1)→(R_1)−(R_2)

[[1,0,−1/3],[0,1,7/3],[0,cos(3*x)−x,sin(4*x)−2*x]]

6. Eliminate the second entry of the third row by performing (R_3)→(R_3)−(cos(3*x)−x)*(R_2)

[[1,0,−1/3],[0,1,7/3],[0,0,sin(4*x)−2*x−7/3*(cos(3*x)−x)]]

7. Simplify the expression in the third row, third column.

sin(4*x)−2*x−7/3*cos(3*x)+7/3*x=sin(4*x)−7/3*cos(3*x)+1/3*x

8. Normalize the third row by dividing by the non-zero expression (assuming it is not zero) to get a leading 1.

[[1,0,−1/3],[0,1,7/3],[0,0,1]]

9. Eliminate the third column entries in the first and second rows using (R_1)→(R_1)+1/3*(R_3) and (R_2)→(R_2)−7/3*(R_3)

[[1,0,0],[0,1,0],[0,0,1]]

Final Answer

rref*[[x,cos(3*x),sin(4*x)],[−2,1,3],[1,1,2]]=[[1,0,0],[0,1,0],[0,0,1]]

Want more problems? Check here!