Find Reduced Row Echelon Form [[3a,-4b,5],[-5a,6b,3]]

Problem

[[3*a,−4*b,5],[−5*a,6*b,3]]

Solution

1. Divide the first row by 3*a to create a leading 1 in the first column.

[[1,−(4*b)/(3*a),5/(3*a)],[−5*a,6*b,3]]

2. Eliminate the entry in the second row, first column by adding 5*a times the first row to the second row.

(R_2)⇒(R_2)+5*a*(R_1)

[[1,−(4*b)/(3*a),5/(3*a)],[0,6*b+5*a*(−(4*b)/(3*a)),3+5*a(5/(3*a))]]

3. Simplify the entries in the second row.

6*b−(20*b)/3=(18*b−20*b)/3=−(2*b)/3

3+25/3=(9+25)/3=34/3

[[1,−(4*b)/(3*a),5/(3*a)],[0,−(2*b)/3,34/3]]

4. Divide the second row by −(2*b)/3 to create a leading 1 in the second column.

(R_2)⇒−3/(2*b)*(R_2)

[[1,−(4*b)/(3*a),5/(3*a)],[0,1,−17/b]]

5. Eliminate the entry in the first row, second column by adding (4*b)/(3*a) times the second row to the first row.

(R_1)⇒(R_1)+(4*b)/(3*a)*(R_2)

[[1,0,5/(3*a)+(4*b)/(3*a)*(−17/b)]]

6. Simplify the final constant term in the first row.

5/(3*a)−(68*b)/(3*a*b)=5/(3*a)−68/(3*a)=−63/(3*a)=−21/a

Final Answer

rref*[[3*a,−4*b,5],[−5*a,6*b,3]]=[[1,0,−21/a],[0,1,−17/b]]

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