Question

Question #2cc5b

Answer 1

The solution is `((x),(y),(z))=((7/2),(7/2),(-13/2))`

Explanation:

I assumed that there is `z` in the second equation instead of `x`

Perform the Gauss Jordan elimination on the augmented matrix

`((3,2,1,|,11),(4,-5,-1,|,3),(3,1,4,|,-12))`

`R3larrR3-R1`

`((3,2,1,|,11),(4,-5,-1,|,3),(0,-1,3,|,-23))`

`R2larrR2-5R3`

`((3,2,1,|,11),(4,0,-16,|,118),(0,-1,3,|,-23))`

`R2larr(R2)/4`

`((3,2,1,|,11),(1,0,-4,|,59/2),(0,-1,3,|,-23))`

`R1larr(R1)/3`

`((1,2/3,1/3,|,11/3),(1,0,-4,|,59/2),(0,-1,3,|,-23))`

`R2larrR2-R1`

`((1,2/3,1/3,|,11/3),(0,-2/3,-13/4,|,155/6),(0,-1,3,|,-23))`

`R2larr(R2)*(-3/2)`

`((1,2/3,1/3,|,11/3),(0,1,39/6,|,-155/4),(0,-1,3,|,-23))`

`R3larr(R3)+R2`

`((1,2/3,1/3,|,11/3),(0,1,13/2,|,-155/4),(0,0,19/2,|,-247/4))`

`R3larr(R3)*(2/19)`

`((1,2/3,1/3,|,11/3),(0,1,13/2,|,-155/4),(0,0,1,|,-13/2))`

`R2larr(R2)-(13/2R3)`

`((1,2/3,1/3,|,11/3),(0,1,0,|,7/2),(0,0,1,|,-13/2))`

`R1larr(R1)-(1/3R3)`

`((1,2/3,0,|,35/6),(0,1,0,|,7/2),(0,0,1,|,-13/2))`

`R1larr(R1)-(2/3R2)`

`((1,0,0,|,7/2),(0,1,0,|,7/2),(0,0,1,|,-13/2))`

Answer 2

`x=7/2, y = 7/2, z = -13/2`

Explanation:

I'll assume the middle equation was a typo and meant to be `4x-5y-z=3`. If not, then the solution is at the bottom to the problem provided.

First, begin by setting up the augmented matrix using the coefficients and constants from the 3 equations, For convenience reasons, I will actually make the 3rd equation my topmost equation in the matrix

`[ (3,1,4,|,-12), (3,2,1,|,11), (4,-5,-1,|,3) ]`

Now, we use elementary row operations to try and reduce the left portion of the matrix to the identity matrix:

`[(1,0,0), (0,1,0), (0,0,1) ]`

For readability I'll refer to an operation on Row n as `R_n`:

`[ (3,1,4,|,-12), (3,2,1,|,11), (4,-5,-1,|,3) ] {: (R_1 div 3),(-R_1),(->) :}`

`[ (1,1/3,4/3,|,-4), (0,1,-3,|,23), (4,-5,-1,|,3) ] {: (),(->),(-4R_1) :}`

`[ (1,1/3,4/3,|,-4), (0,1,-3,|,23), (0,-19/3,-19/3,|,19) ] {: (),(->),(-3/19R_3) :}`

`[ (1,1/3,4/3,|,-4), (0,1,-3,|,23), (0,1,1,|,-3) ] {: (),(->),(-R_2) :}`

`[ (1,1/3,4/3,|,-4), (0,1,-3,|,23), (0,0,4,|,-26) ] {: (),(->),(R_3 div 4) :}`

`[ (1,1/3,4/3,|,-4), (0,1,-3,|,23), (0,0,1,|,-13/2) ] {: (-4/3R_3),(+3R_3),(->) :}`

`[ (1,1/3,0,|,14/3), (0,1,0,|,7/2), (0,0,1,|,-13/2) ] {: (-1/3R_2),(->),() :}`

`[ (1,0,0,|,7/2), (0,1,0,|,7/2), (0,0,1,|,-13/2) ]`

We can read the final answer in the right portion of the matrix:

`x=7/2, y = 7/2, z = -13/2`

Note If the original problem was not a typo, similar steps can be used to arrive at a different solution:

`x=301/66, y=47/22, z=-153/22`