© 2000−2019  P. Bogacki Solving a system of linear equations v. 1.25

 PROBLEM

Solve the following system of 3 linear equations in 3 unknowns:

 -3 x1 +1 x2 -2 x3 = -7 5 x1 +3 x2 -4 x3 = 2 1 x1 +2 x2 -3 x3 = -1

 SOLUTION

 Step 1: Transform the augmented matrix to the reduced row echelon form  (Hide details)

Row
Operation
1:

 -3 1 -2 -7 5 3 -4 2 1 2 -3 -1
multiply the 1st row by -1/3
 1 -1  3 2  3 7  3 5 3 -4 2 1 2 -3 -1
Row
Operation
2:

 1 -1  3 2  3 7  3 5 3 -4 2 1 2 -3 -1
add -5 times the 1st row to the 2nd row
 1 -1  3 2  3 7  3 0 14  3 -22  3 -29  3 1 2 -3 -1
Row
Operation
3:

 1 -1  3 2  3 7  3 0 14  3 -22  3 -29  3 1 2 -3 -1
add -1 times the 1st row to the 3rd row
 1 -1  3 2  3 7  3 0 14  3 -22  3 -29  3 0 7  3 -11  3 -10  3
Row
Operation
4:

 1 -1  3 2  3 7  3 0 14  3 -22  3 -29  3 0 7  3 -11  3 -10  3
multiply the 2nd row by 3/14
 1 -1  3 2  3 7  3 0 1 -11  7 -29  14 0 7  3 -11  3 -10  3
Row
Operation
5:

 1 -1  3 2  3 7  3 0 1 -11  7 -29  14 0 7  3 -11  3 -10  3
add -7/3 times the 2nd row to the 3rd row
 1 -1  3 2  3 7  3 0 1 -11  7 -29  14 0 0 0 3  2
Row
Operation
6:

 1 -1  3 2  3 7  3 0 1 -11  7 -29  14 0 0 0 3  2
multiply the 3rd row by 2/3
 1 -1  3 2  3 7  3 0 1 -11  7 -29  14 0 0 0 1
Row
Operation
7:

 1 -1  3 2  3 7  3 0 1 -11  7 -29  14 0 0 0 1
add 29/14 times the 3rd row to the 2nd row
 1 -1  3 2  3 7  3 0 1 -11  7 0 0 0 0 1
Row
Operation
8:

 1 -1  3 2  3 7  3 0 1 -11  7 0 0 0 0 1
add -7/3 times the 3rd row to the 1st row
 1 -1  3 2  3 0 0 1 -11  7 0 0 0 0 1
Row
Operation
9:

 1 -1  3 2  3 0 0 1 -11  7 0 0 0 0 1
add 1/3 times the 2nd row to the 1st row
 1 0 1  7 0 0 1 -11  7 0 0 0 0 1
 Step 2: Interpret the reduced row echelon form

The reduced row echelon form of the augmented matrix is

 1 0 1  7 0 0 1 -11  7 0 0 0 0 1

which corresponds to the system

 1 x1 +(1/7) x3 = 0 1 x2 +(-11/7) x3 = 0 0 = 1

Equation 3 cannot be solved, therefore, the system has no solution (i.e. the system is inconsistent).

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