© 2000−2019  P. BogackiSolving a system of linear equationsv. 1.25 

 PROBLEM

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

1 x1-3 x2 +1 x3=1 
2 x1-1 x2-2 x3=-1 
2 x1-6 x2 +2 x3=2 

 SOLUTION

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

Row
Operation
1:  
 1   -3   1    1 
 2   -1   -2   -1 
 2   -6   2   2 
add -2 times the 1st row to the 2nd row
 1   -3   1    1 
 0   5   -4   -3 
 2   -6   2   2 
Row
Operation
2:  
 1   -3   1    1 
 0   5   -4   -3 
 2   -6   2   2 
add -2 times the 1st row to the 3rd row
 1   -3   1    1 
 0   5   -4   -3 
 0   0   0   0 
Row
Operation
3:  
 1   -3   1    1 
 0   5   -4   -3 
 0   0   0   0 
multiply the 2nd row by 1/5
 1   -3   1    1 
 0   1   -4 
 5 		  -3 
 5 
 0   0   0   0 
Row
Operation
4:  
 1   -3   1    1 
 0   1   -4 
 5 		  -3 
 5 
 0   0   0   0 
add 3 times the 2nd row to the 1st row
 1   0   -7 
 5 		   -4 
 5 
 0   1   -4 
 5 		  -3 
 5 
 0   0   0   0 
 Step 2: Interpret the reduced row echelon form

The reduced row echelon form of the augmented matrix is

 1   0   -7 
 5 		   -4 
 5 
 0   1   -4 
 5 		  -3 
 5 
 0   0   0   0 

which corresponds to the system

1 x1  +(-7/5) x3=(-4/5) 
 1 x2 +(-4/5) x3=(-3/5) 
  0=0

No equation of this system has a form zero = nonzero; Therefore, the system is consistent.

The leading entries in the matrix have been highlighted in yellow.

A leading entry on the (i,j) position indicates that the j-th unknown will be determined using the i-th equation.

Those columns in the coefficient part of the matrix that do not contain leading entries, correspond to unknowns that will be arbitrary. The system has infinitely many solutions:

x1= +(7/5) x3 +(-4/5)
x2= +(4/5) x3 +(-3/5)
x3=arbitrary
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