How do you solve using gaussian elimination or gauss-jordan elimination, #4x-3y+z=9#, #3x+2y-2z=4#, #x-y+3z=5#?

1 Answer
Feb 11, 2018

Answer:

The values x=2, y=0, and z=1 satisfy the equations
#4x-3y+z=9---------(1)#
#3x+2y-2z=4--------(2)#
#x-y+3z=5-----------(3)#

Explanation:

Gaussian elimination method:
#4x-3y+z=9---------(1)#
#3x+2y-2z=4--------(2)#
#x-y+3z=5-----------(3)#

Eliminate x from (2)
Equation (2) becomes
#(2)-3/4(1)#
#3-3/4(4)=3-3= 0 #
#2-3/4(-3)=2+9/4=2+2.25=4.25#
#-2-3/4(1)=-2-3/4=-2-0.75=-2.75#
#4-3/4(9)=4-6.75=-2.75#
Changing the coefficients
#0x+4.25y-2.75z=-2.75#
#4.25y-2.75z=-2.75#

Eliminate x from (3)
Equation (3) becomes
#(3)-1/4(1)#
#1-1/4(4)=1-1=0#
#-1-1/4(-3)=-1+0.75=-0.25#
#3-1/4(1)=3-0.25=2.75#
#5-1/4(9)=5-2.25=2.75#
Changing the coefficients
#0x-0.25y+2.75z=2.75#
#-0.25y+2.75z=2.75#
Thus the equations where x is eliminated are:
#4.25y-2.75z=-2.75--------(4)#
#-0.25y+2.75z=2.75--------(5)#

Eliminate y from (5)
#(3)-(-0.25)/4.25(4)#
#-0.25-(-0.25)/4.25(4.25)=-0.25+0.25=0#
#2.75-(-0.25)/4.25(-2.75)=2.75+0.162=2.912#
#2.75-(-0.25)/4.25(-2.75)=2.75+0.162=2.912#
Changing the coefficients
#0y+2.912z=2.912#
#2.912z=2.912#
Thus, the equation where y is eliminated is
#2.912z=2.912#-----------(6)
Solving for z,
#z=2.912/2.912=1#

#z=1#

Substituting the value z=1 in (4)
#4.25y-2.75(1)=-2.7 Simplifying5#
#4.25y-2.75=-2.75#
Transposing -2,75 from lhs to rhs
#4.25y=-2.75+2.75#
#4.25y=0#
Solving for y
#y=0/4.25#

#y=0#

Substituting the values y=0, and z=1 in (1)
#4x-3(0)y+(1)=9#
Simplifying
#4x-0+1=9#
#4x=9-1#
#4x=8#
Solving for x

#x=8/4#

#x=2#

Check:
#4(2)-3(0)+(1)=9---------(1)#
#3(2)+2(0)-2(1)=4--------(2)#
#(2)-(0)+3(1)=5-----------(3)#
#8-0+1=9#, lhs=rhs
#6+0-2=4#, lhs=rhs
#2-0+3=5#, lhs=rhs

Thus, the values x=2, y=0, and z=1 satisfy the equations
#4x-3y+z=9---------(1)#
#3x+2y-2z=4--------(2)#
#x-y+3z=5-----------(3)#