Expert: Paul Klarreich - 11/6/2006

Question
Mr. Klarreich,
I have a real problem solving matrices in reduced row echelon form. I can apply an equation to a matrix easily enough, but when I begin to solve using row operations I find myself down the wrong path and I am spending far too much time on problems.
The problem I have been working has not been easy for me and I was wondering if you could shed some light on it for me.
R1: 2x+y=-4 R2: -2y+4z=0 R3: 3x-2z=-11
As I said before, I know how to apply it to a matrix, but I am unsure how to get solutions for x, y, and z. I also know how to use Cramer's rule, but my instructor is not allowing us to use it for this particular group of problems.
Thank you in advance,
Jonah


Answer
Questioner:  Jonah
Category:  Advanced Math
 
Subject:  Systems of Linear Equations: Matrices
Question:  Mr. Klarreich,
I have a real problem solving matrices in reduced row echelon form. I can apply an equation to a matrix easily enough, but when I begin to solve using row operations I find myself down the wrong path and I am spending far too much time on problems.
The problem I have been working has not been easy for me and I was wondering if you could shed some light on it for me.
R1: 2x+y=-4 R2: -2y+4z=0 R3: 3x-2z=-11
As I said before, I know how to apply it to a matrix, but I am unsure how to get solutions for x, y, and z. I also know how to use Cramer's rule, but my instructor is not allowing us to use it for this particular group of problems.
Thank you in advance,
Jonah
......................................
Hi, Jonah,

Here's your system:

R1: 2x+y=-4 R2: -2y+4z=0 R3: 3x-2z=-11

Let's take the variable symbols out.  If we're careful, we don't need them, and only careful people do math, right?

Here's the (augmented) matrix.
2    1    0    -4   R1
0   -2    4     0   R2
3    0   -2   -11   R3
 
If we can get it to look like this: (Form A)

1   ?    ?    ?
0   1    ?    ?
0   0    1    ?

OR to look like this:  (Form B)

1   0    0    ?
?   1    0    ?
?   ?    1    ?

Then we practically have the solution.  In fact, if we can get it to look like this: (Final form)

1   0    0    ?
0   1    0    ?
0   0    1    ?

Then we REALLY have the solution.  And if you have either Form A or B, it's easy to get to the final form.  So let's do some fiddling with it:

2    1    0    -4     R1
0   -2    4     0     R2
3    0   -2   -11     R3

When I look at that, I get the urge to use the -2(z) to kill the 4 above it.
Make  R2 = R2 + 2R3:

0   -2    4     0     R2
6    0   -4   -22     2R3
----------------------------
6   -2    0   -22, and we divide out a 2:
3   -1    0   -11  

so we now have:

2    1    0    -4     R1
3   -1    0   -11     R2
3    0   -2   -11     R3

Now I see a nice 'kill' -- use R1 = R1 + R2 to kill the 1(y) in the top row.  Getting the idea?  Look for easy ways to make a coefficient zero.

2    1    0    -4     R1
3   -1    0   -11     R2
----------------------------
5    0    0   -15, and we divide out a 5:
1    0    0   - 3

We have: (this is much more fun on a computer, where I can cut and paste.  On paper it's a lot of scribbling.)

1    0    0   - 3     R1
3   -1    0   -11     R2
3    0   -2   -11     R3

We're getting there -- Use  R2 = R2 - 3R1.

-3    0    0     9   -3R1
3   -1    0   -11     R2
-------------------------
0   -1    0    -2, and reverse the signs:
0    1    0     2


1    0    0   - 3     R1
0    1    0     2     R2
3    0   -2   -11     R3

Getting closer.  How about the same on R3:  R3 = R3 - 3R1
 
-3   0    0    +9   -3R1
3   0   -2   -11     R3
----------------------------
0   0   -2   -2, and get out a factor of -2:
0   0    1    1

That gives:

1    0    0   - 3     R1
0    1    0     2     R2
0    0    1     1     R3

AND THERE IS THE SOLUTION:  Just put back the symbols and the equations now say:

R1:  x = -3  
R2:  y = 2
R3:  z = 1

Happy kill... er,... eliminating.