Question

How do you solve the system `3x-y+z=5`, `x+3y+3z=-6`, and `x+4y-2z=12`?

Answer 1

`x=3`, `y=1/2`, `z=-7/2`

Explanation:

Given:

`"(i)":color(white)(1/1)3x-y+z=5`
`"(ii)":color(white)(1/1)x+3y+3z=-6`
`"(iii)":color(white)(1/1)x+4y-2z=12`

Note that we can eliminate `z` by adding or subtracting suitable multiples of `"(i)"` to or from `"(ii)"` and `"(iii)"`:

`"(iv)" = "(ii)" - 3"(i)":color(white)(1/1)-8x+6y=-21`
`"(v)" = "(iii)" + 2"(i)":color(white)(1/1)7x+2y=22`

Then we can eliminate `y` by subtracting `3` times `"(v)"` from `"(iv)"`:

`"(vi)" = "(iv)"-3"(v)":color(white)(1/1)-29x = -87`

Dividing both sides of `"(vi)"` by `-29`, we find:

`x = 3`

Substituting this value of `x` in `"(v)"`, we find:

`7(color(blue)(3))+2y = 22`

Hence:

`y = 1/2`

Then from `"(i)"` we find:

`z = 5-3x+y = 5-3(color(blue)(3))+color(blue)(1/2) = -7/2`

`color(white)()`
Matrix formulation

We can represent the original equations as an augmented matrix:

`((3, -1, 1, |, 5), (1, 3, 3, |, -6), (1, 4, -2, |, 12))`

The operations we performed above correspond to various row operations:

Subtract `3xx"row1"` from `"row2"` and add `2xx"row1"` to `"row3"` to get:

`((3, -1, 1, |, 5), (-8, 6, 0, |, -21), (7, 2, 0, |, 22))`

Subtract `3xx"row3"` from `"row2"` to get:

`((3, -1, 1, |, 5), (-29, 0, 0, |, -87), (7, 2, 0, |, 22))`

Divide `"row2"` by `-29` to get:

`((3, -1, 1, |, 5), (1, 0, 0, |, 3), (7, 2, 0, |, 22))`

Subtract `7xx"row2"` from `"row3"` to get:

`((3, -1, 1, |, 5), (1, 0, 0, |, 3), (0, 2, 0, |, 1))`

Divide `"row3"` by `2` to get:

`((3, -1, 1, |, 5), (1, 0, 0, |, 3), (0, 1, 0, |, 1/2))`

Add `"row3"` to `"row1"` and subtract `3xx"row2"` to get:

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

Permute the rows, moving `"row1"` to the third row to get:

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

Now the left hand `3xx3` submatrix is an identity matrix, we can read off `x`, `y` and `z` from the right hand column.