Question

How do you solve using gaussian elimination or gauss-jordan elimination, `3x-2y-z=7`, `z=x+2y-5`, `-x+4y+2z=-4`?

Answer 1

`(x,y,z) = (2,1/2,-2)`

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Try setting up a matrix:

`[(3,-2,-1,|,7),(1,2,-1,|,5),(-1,4,2,|,-4)]`

Do note that your second equation has `z` on the wrong side compared to the other equations in the system, so you must add `(5-z)` to both sides of the second equation before you form this matrix.

To row-reduce this using gaussian elimination, simply row-reduce this using elementary row operations, i.e. any of the following in any combination:

• Scale a row, i.e. `cR_i`, where `c` is a constant.
• Add row `i` onto row `j` and store in row `j`, i.e. `R_i + R_j`

Hence, if I write `cR_i + R_j`, it means `cR_i` is added to each entry in row `R_j` and the result is stored into row `j`.

Our process in general will be to get the matrix down to a solvable form. Gauss-Jordan elimination would have gotten it down to reduced-row-echelon form, though that is not necessary to solve this system.

`stackrel(R_2 + R_3" ")(->)[(3,-2,-1,|,7),(1,2,-1,|,5),(0,6,1,|,1)]`

`stackrel(-1/3R_1 + R_2" ")(->)[(3,-2,-1,|,7),(0,8/3,-2/3,|,8/3),(0,6,1,|,1)]`

`stackrel(3/2R_2" ")(->)[(3,-2,-1,|,7),(0,4,-1,|,4),(0,6,1,|,1)]`

`stackrel(R_3 + R_2; 1/5R_2" ")(->)[(3,-2,-1,|,7),(0,2,0,|,1),(0,6,1,|,1)]`

Good enough; we can solve backwards now. The current system of equations looks like:

`3x - 2y - z = 7`
`0x + 2y + 0 = 1`
`0x + 6y + 1 = 1`

Thus, from the second equation we have:

`2y = 1 => color(green)(y = 1/2)`

`6y + z = 1`

`=> 6(1/2) + z = 1 => color(green)(z = -2)`

`3x - 2y - z = 7`

`=>3x - 2(1/2) - (-2) = 7`

`=> 3x = 6 => color(green)(x = 2)`

Hence, we got `color(blue)((x,y,z) = (2,1/2,-2))`.

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Let's check...

`3x - 2y - z`

`= 3(2) - 2(1/2) - (-2) = 6 - 1 + 2 = 7 color(blue)(sqrt"")`

`x + 2y - 5`

`= 2 + 2(1/2) - 5 = 2 + 1 - 5 = -2 color(blue)(sqrt"")`

`-x + 4y + 2z`

`= -(2) + 4(1/2) + 2(-2) = -2 + 2 - 4 = -4 color(blue)(sqrt"")`