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

##### 2 Answers

#### Answer:

The solution is

#### Explanation:

Write the coefficients in an augmented matrix:

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:

Let's start with the first cell in row 2. Its value is

This new row represents a new valid equation (that is,

A similar linear combination is then used to replace row 3—that combination is

All the coefficients in

Our newest

With the matrix in the upper-triangular form, solving for variables is like falling dominoes. Knowing

Our final solution is

(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

#### Answer:

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

#### Explanation:

These equations now say

So our answer is as before.