Find Reduced Row Echelon Form

Problem

[[4,−3,1,−7],[0,7/4,15/4,23/4],[0,0,27/7,54/7]]

Solution

1. Normalize the third row by multiplying (R_3) by 7/27 to create a leading 1.

7/27*(R_3)→(R_3)

[[4,−3,1,−7],[0,7/4,15/4,23/4],[0,0,1,2]]

2. Eliminate the third column entries above the leading 1 in (R_3) by performing (R_2)−15/4*(R_3)→(R_2) and (R_1)−(R_3)→(R_1)

(R_2)−15/4*(R_3)→(R_2)

[[4,−3,1,−7],[0,7/4,0,−7/4],[0,0,1,2]]

(R_1)−(R_3)→(R_1)

[[4,−3,0,−9],[0,7/4,0,−7/4],[0,0,1,2]]

3. Normalize the second row by multiplying (R_2) by 4/7 to create a leading 1.

4/7*(R_2)→(R_2)

[[4,−3,0,−9],[0,1,0,−1],[0,0,1,2]]

4. Eliminate the second column entry above the leading 1 in (R_2) by performing (R_1)+3*(R_2)→(R_1)

(R_1)+3*(R_2)→(R_1)

[[4,0,0,−12],[0,1,0,−1],[0,0,1,2]]

5. Normalize the first row by multiplying (R_1) by 1/4 to create the final leading 1.

1/4*(R_1)→(R_1)

[[1,0,0,−3],[0,1,0,−1],[0,0,1,2]]

Final Answer

[[4,−3,1,−7],[0,7/4,15/4,23/4],[0,0,27/7,54/7]]=[[1,0,0,−3],[0,1,0,−1],[0,0,1,2]]

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