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

 PROBLEM

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

(1/2) x1 +(-3/5) x2 +(-1/5) x3=50 
(-1/5) x1 +(4/5) x2 +(-1/10) x3=30 
(-1/10) x1 +(-1/5) x2 +(7/10) x3=20 

 SOLUTION

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

 1 
 2 		  -3 
 5 		  -1 
 5 		   50 
 -1 
 5 		  4 
 5 		  -1 
 10 		  30 
 -1 
 10 		  -1 
 5 		  7 
 10 		  20 

can be transformed by a sequence of elementary row operations to the matrix

 1   0   0    11300 
 39 
 0   1   0   1600 
 13 
 0   0   1   4100 
 39 

 Step 2: Interpret the reduced row echelon form

The reduced row echelon form of the augmented matrix is

 1   0   0    11300 
 39 
 0   1   0   1600 
 13 
 0   0   1   4100 
 39 

which corresponds to the system

1 x1  =(11300/39) 
 1 x2 =(1600/13) 
  1 x3=(4100/39) 

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.

Since every column in the coefficient part of the matrix has a leading entry that means our system has a unique solution:

x1=11300/39
x2=1600/13
x3=4100/39
This concludes the solution of the problem. Do you want to
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