How do you solve using gaussian elimination or gauss-jordan elimination, #5x + y + 5z = 3 #, #4x − y + 5z = 13 #, #5x + 2y + 2z = 2#?

1 Answer
May 14, 2017

Answer:

#x=1 32/99#,
#y=-5 131/198#,
#z=9/22#

Explanation:

To solve the system of linear equations

#{(5x+y+5z=3),(4x-y+5z=13),(5x+2y+2z=2):}#

First convert it to the augmented matrix form

#Rightarrow [(5,1,5,|,3),(4,-1,5,|,13),(5,2,2,|,2)]#

The goal is to write this in row-echelon form, which is triangular form where the leading coefficient is 1. The three row operations we can use are:

  1. Switch rows
  2. Multiply a row by a constant
  3. Add a multiple of a row to another

Multiply the first row by #1/5#

#Rightarrow [(1,1/5,1,|,3/5),(4,-1,5,|,13),(5,2,2,|,2)]#

Multiply first row by -4 and add to second row

#Rightarrow [(1,1/5,1,|,3/5),(0,-9/5,1,|,53/5),(5,2,2,|,2)]#

Multiply first row by -5 and add to the third row

#Rightarrow [(1,1/5,1,|,3/5),(0,-9/5,1,|,53/5),(0,1,-3,|,-1)]#

Multiply row 2 by #-5//9#

#Rightarrow [(1,1/5,1,|,3/5),(0,1,-5/9,|,-53/9),(0,1,-3,|,-1)]#

Multiply row 2 by -1 and add to row 3

#Rightarrow [(1,1/5,1,|,3/5),(0,1,-5/9,|,-53/9),(0,0,-22/9,|,-1)]#

Multiply the last row by -9/22

#Rightarrow [(1,1/5,1,|,3/5),(0,1,-5/9,|,-53/9),(0,0,1,|,9/22)]#

Plugging these values back into formulas gives

#{(x+1/5y+z=3/5),(y-5/9z=-53/9),(z=9/22):}#

Using backwards substitution we get the remaining answers:
#x=131/99=1 32/99#,
#y=-1121/198=-5 131/198#,
#z=9/22#