Question

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

Answer 1

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)`