Find Reduced Row Echelon Form [[a,b,1,0],[c,d,0,1]]

Problem

[[a,b,1,0],[c,d,0,1]]

Solution

1. Divide the first row by a to create a leading 1 in the first column, assuming a≠0

(R_1)→1/a*(R_1)

[[1,b/a,1/a,0],[c,d,0,1]]

2. Eliminate the c in the second row by subtracting c times the first row from the second row.

(R_2)→(R_2)−c*(R_1)

[[1,b/a,1/a,0],[0,d−(b*c)/a,−c/a,1]]

3. Simplify the term in the second row, second column using a common denominator.

d−(b*c)/a=(a*d−b*c)/a

[[1,b/a,1/a,0],[0,(a*d−b*c)/a,−c/a,1]]

4. Divide the second row by (a*d−b*c)/a to create a leading 1 in the second column, assuming a*d−b*c≠0

(R_2)→a/(a*d−b*c)*(R_2)

[[1,b/a,1/a,0],[0,1,−c/(a*d−b*c),a/(a*d−b*c)]]

5. Eliminate the b/a in the first row by subtracting b/a times the second row from the first row.

(R_1)→(R_1)−b/a*(R_2)

(R_1,3)=1/a−b/a*(−c/(a*d−b*c))=1/a+(b*c)/(a*(a*d−b*c))=(a*d−b*c+b*c)/(a*(a*d−b*c))=(a*d)/(a*(a*d−b*c))=d/(a*d−b*c)

(R_1,4)=0−b/a*(a/(a*d−b*c))=−b/(a*d−b*c)

Final Answer

rref*[[a,b,1,0],[c,d,0,1]]=[[1,0,d/(a*d−b*c),−b/(a*d−b*c)],[0,1,−c/(a*d−b*c),a/(a*d−b*c)]]

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