Question

How do you solve `x-y+z=3` and `2y-z=1` and `-x+2y= -1` using matrices?

Answer 1

`x = 3`, `y = 1`, `z = 1`

Explanation:

Your linear equation can be written as the following matrix (first column representing the `x`, second column the `y` and third column the `z`):

`( (1, -1, 1, |, 3),(0, 2, -1, |, 1),(-1, 2, 0, |, -1) ) " " {: (I),(II),(III) :}`

To eliminate the `-1` in the first column, compute `I + III -> color(blue)(III)`:

`rArr ( (1, -1, 1, |, 3),(0, 2, -1, |, 1),(color(blue)(0), color(blue)(1), color(blue)(1), |, color(blue)(2)) ) " " {: (I),(II),(color(blue)(III)) :}`

To eliminate the `2` in the second column, compute `II -2 III -> color(red)(II)`:

`rArr ( (1, -1, 1, |, 3),(color(red)(0), color(red)(0), color(red)(-3), |,color(red)( -3)),(0, 1, 1, |, 2) ) " " {: (I),(color(red)(II)),(III) :}`

Compute `-1/3 II -> color(green)(II)`:

`rArr ( (1, -1, 1, |, 3),(color(green)(0), color(green)(0), color(green)(1), |, color(green)(1)),(0, 1, 1, |, 2) ) " " {: (I),(color(green)(II)),(III) :}`

Now, use the second row to eliminate the `1` in the third row, third column:
compute `III - II -> color(orange)(III)`:

`rArr ( (1, -1, 1, |, 3),(0, 0, 1, |, 1),(color(orange)(0), color(orange)(1), color(orange)(0), |, color(orange)(1)) ) " " {: (I),(II),(color(orange)(III)) :}`

We are almost done. Now, you need to use the third row to eliminate the `-1` in the first row, and the second row to eliminate the `1` in the first row, third column.

Let's do it step by step: first, `I + III -> color(violet)(I)`:

`rArr ( (color(violet)(1), color(violet)(0), color(violet)(1), |, color(violet)(4)),(0, 0, 1, |, 1),(0, 1, 0, |, 1) ) " " {: (color(violet)(I)),(II),(III) :}`

Last step: compute `I - II -> color(gray)(I)`:

`rArr ( (color(gray)(1), color(gray)(0), color(gray)(0), |, color(gray)(3)),(0, 0, 1, |, 1),(0, 1, 0, |, 1) ) " " {: (color(gray)(I)),(II),(III) :}`

If you would like, you can also swap `II` and `III` to have the perfect "step format" in your matrix.

The solution is `x = 3`, `y = 1` and `z = 1`.