Solve this system by rewriting in row-echelon form $x+y+z=6$, $2x-y+z=3$, $3x-z=0$

Question

This is my very first problem in Linear Algebra and I guess I really need to brush up on my Algebra skills..I'm at a loss as to how to solve this equation

My reading said that there are basically 3 changes I can make to a system of equations so that the equation remains equivalent.

1.) Interchange two equations
2.) Multiply an equation by a nonzero constant
3.) Add a multiple of an equation to another equation

What confuses me about this is how many times can I do each of the above in a given system of equations? For example, can I do multiple interchanges?

So here is the problem:

$ \left\{ \begin{aligned} x+y+z&=6 \\ 2x-y+z&=3 \\ 3x-z&=0 \end{aligned} \right. $

Here is what I did (which I didn't complete because I had no idea if I was doing the right thing):

Add (-2) times the first equation to the second equation to produce a new second equation

$ \left\{ \begin{aligned} x+y+z&=6 \\ -3y-z&=9 \\ 3x-z&=0 \end{aligned} \right. $

And next add (-3) times first equation to third equation to produce a new third equation:

$ \left\{ \begin{aligned} x+y+z&=6 \\ -3y-z&=9 \\ -3y-4z&=-18 \end{aligned} \right. $

And next add (-1) times second to third equation to produce a new third equation:

$ \left\{ \begin{aligned} x+y+z&=6 \\ -3y-z&=9 \\ -3z&=-27 \end{aligned} \right. $

Any help as to my journey in Linear Algebra and how to solve this is greatly appreciated.

Answer 1

For the most part, conceptually, you have the right idea, and your choices along the way are spot on. However, you did make an arithmetic error on the very first step, when adding $-2$ times the first row to the second row:

Specifically, $-2\cdot 6 + 3 = -9$. Correct for that error, and you can end where you ended, using the value you have for $z$ to solve for $y$, and using those values, solve for $x$, or you can perform additional row operations to obtain reduced echelon form.

Answer 2

You can do as many row additions/subtractions as you want/need. At first glance, you appear completely correct (except a -9 in the second equation, as David Mitra points out).

Now, you'll find that $z = \frac{-9}{-3} = 3$, use that result to compute $y$ from the second equation, and use both results to compute $x$ from the first. This process is called "back-substitution."

In general, if you have $n$ equations and $n$ unknowns, you should be able to do row-reduction using a total of $\frac23 n^2$ multiplications and additions. So, if you're careful about how you do things, you should only need to do the row operations a few times on a 3-by-3 matrix!