How do you solve the system #x+4y-6z=8#, #2x-y+3z=-10#, and #3x-2y+3z=-18#?
The solution is
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
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.)
Many teachers will encourage you to reduce each new row you create so that the leading coefficients are
Here is an alternate approach that turns the coefficient matrix into the identity matrix, thus making the augmented column the solution.
These equations now say
So our answer is as before.