Question

How do you solve the system `x+4y-6z=8`, `2x-y+3z=-10`, and `3x-2y+3z=-18`?

Answer 1

The solution is `(x,y,z)=(-4,"  "4,"  "2/3)`.

Explanation:

Write the coefficients in an augmented matrix:
`[(1,4,-6,|,8),(2,-1,3,|,-10),(3,-2,3,|,-18)]`

Our goal is to change this into a matrix that is upper-right triangular; that is, where everything below the diagonal is a zero. We do this by adding linear combinations of the rows together to create new rows (that replace the current ones) and by multiplying rows by constants. Remember: this does not change the information implied by the augmented matrix, it just presents it in a new (and hopefully more beneficial) way.

(Brief notation recap: `R_1`, `R_2`, and `R_3` mean "row 1", "row 2", and "row 3" respectively.)

Let's start with the first cell in row 2. Its value is `2`, which is `2/1=2` times as much as the `1` above it. If we want the `2` to become a `0`, we need to subtract `2R_1 - R_2`:

`2times[(1,4,"-"6,|,8"    ")]`
`ul("    "-[(2,"-"1,3,|,"-"10)])`
`"=    "[(0,9,"-"15,|,26)]`

This new row represents a new valid equation (that is, `0x+9y-15z=26`) which can replace one of the two rows used in forming it. Choosing to keep row 1 the same, we replace row 2:

`[(1,4,-6,|,8),(color(red)0,color(red)(9),color(red)(-15),|,color(red)26),(3,-2,3,|,-18)]`

A similar linear combination is then used to replace row 3—that combination is `3R_1-R_3`, and the following shorthand is used to indicate what row operation has been done to form a new row:

`=>color(white)[(),(),(color(black)(3R_1-R_3))][(1,4,-6,|,8),(0,9,-15,|,26),(color(red)0,color(red)14,color(red)(-21),|,color(red)42)]`

All the coefficients in `R_3` are multiples of 7, so let's reduce:

`=>color(white)[(),(),(color(black)(1/7*R_3))][(1,4,-6,|,8),(0,9,-15,|,26),(0,2,-3,|,6)]`

`=>color(white)[(),(),(color(black)(9R_3-2R_2))][(1,4,-6,|,8),(0,9,-15,|,26),(0,0,3,|,2)]`

Our newest `R_3` says `3z=2`, which means that `z=2/3`. And since the equation directly above `R_3` only involves one additional unknown (that is, `y`), we can plug this solved value for `z` into it to solve for `y`:

`9y-15z=26`
`=>9y-15(2/3)=26`
`=>9y-10=26`
`=>9y=36`
`=>y=4`

With the matrix in the upper-triangular form, solving for variables is like falling dominoes. Knowing `z` gives us `y`, and knowing `z` & `y` gives us `x`:

`x+4y-6z=8`
`=>x+4(4)-6(2/3)=8`
`=>x+16-4=8`
`=>x="-"4`

Our final solution is `(x,y,z)=(-4,"  "4,"  "2/3)`.
(It is best to check your solution to make sure it works; I will save that as an exercise.)

Note:

Many teachers will encourage you to reduce each new row you create so that the leading coefficients are `1`. This is perfectly fine, and you'll get the same solutions. It just means your matrix will likely have fractions in it. I chose not to reduce my rows for the sake of making the matrices easier to read.

Answer 2

Here is an alternate approach that turns the coefficient matrix into the identity matrix, thus making the augmented column the solution.

Explanation:

`"                         "[(1,4,-6,|,8),(2,-1,3,|,-10),(3,-2,3,|,-18)]`

`=>color(white)[(),(color(black)(2R_1-R_2)),()][(1,4,-6,|,8),(0,9,-15,|,26),(3,-2,3,|,-18)]`

`=>"      "color(white)[(),(color(black)(1/9*R_2)),()][(1,4,-6,|,8),(0,1,-5/3,|,26/9),(3,-2,3,|,-18)]`

`=>color(white)[(),(),(color(black)(3R_1-R_3))][(1,4,-6,|,8),(0,1,-5/3,|,26/9),(0,14,-21,|,42)]`

`=>"      "color(white)[(),(),(color(black)(1/7*R_3))][(1,4,-6,|,8),(0,1,-5/3,|,26/9),(0,2,-3,|,6)]`

`=>color(white)[(),(),(color(black)(R_3-2R_2))][(1,4,-6,|,8),(0,1,-5/3,|,26/9),(0,0,1/3,|,2/9)]`

`=>"       "color(white)[(),(),(color(black)(3*R_3))][(1,4,-6,|,8),(0,1,-5/3,|,26/9),(0,0,1,|,2/3)]`

`=>color(white)[(),(color(black)(R_2+5/3R_1)),(3*R_3)][(1,4,-6,|,8),(0,1,0,|,4),(0,0,1,|,2/3)]`

`=>color(white)[(color(black)(R_1-4R_2)),(),(3*R_3)][(1,0,-6,|,-8),(0,1,0,|,4),(0,0,1,|,2/3)]`

`=>color(white)[(color(black)(R_1+6R_3)),(),(3*R_3)][(1,0,0,|,-4),(0,1,0,|,4),(0,0,1,|,2/3)]`

These equations now say

`x"            "="-"4`
`"      "y"      "=4`
`"            "z=2//3`

So our answer is as before.