Supplementary notes for Math 229
Dan Bates
Tuesday, Sept. 2
I ran short on time in class today and had to cut a couple of examples down to the bare
minimum. Here are the details on those two topics.
Nonnumerical matrices
You can apply the techniques that we are discussing (like Gaussian elimination to RREF
form) to nonnumerical matrices. In fact, I have heard that these are sometimes on the
tests for 229 (which I don’t get to write). Here is an example:




2−x
1
2
1
2−x
2
R ↔ R2 
 1
2−x
2  1
2−x
1
0 
−→
1
1
3−x
1
1
3−x
At this point, it is tempting to do R2 − (2 − x)R1 → R1 to make entry (2, 1) zero. In the case
of a numerical matrix, that would be fine. In this case, though, check out what it would do
to column 2, row 2...it would make it even worse (degree 2 instead of degree 1!). Instead, we
follow a slightly different route that will make the entries of row 2 all divisible by the same
polynomial. Luckily, the homework problem is a bit more clear.




1
2−x
2
1
2−x
2
R2 − R1 → R2 
R − R1 → R3 
1−x x−1
0  3
1−x x−1
0 
−→
−→
1
1
3−x
0
x−1 1−x
Now I can divide the second and third rows by x − 1 to make them nice numbers:

1 2−x
2
→ R2 
1 −1
0 
−→
0 x−1 1−x
1
R
x−1 2



1 2−x 2
→ R3 
1 −1
0 
−→
0
1
−1
1
R
x−1 3
1
WARNING: We multiplied by x−1
. If x = 1, this amounts to division by 0 – bad news!
SO, the solution that we get here will be valid for x 6= 1, and we will need to do the x = 1
case separately.




1 2−x
2
1 −3 + x −2
R2 − R1 → R2 
R ↔ R3 
1 −3 + x −2  2
1 2−x
2 
−→
−→
0
1
−1
0
1
−1


1 2−x
2
R3 − (x − 3)R2 → R3 
1 −3 + x −2 
−→
0
0
x−5

1 2−x
2
→ R3 
1 −3 + x −2 
−→
0
0
1
1
R
x−5 3

WARNING: We did it again! We now need to do the case of x = 5 separately, too.
At this point, the matrix is in REF (not RREF). It is really simple from here to make
RREF. I’ll leave it to you, but you should get the identity matrix (1s on the diagonal and
0s everywhere else).
Are we done? Of course not - nobody ever asks that if they are done. We still need to handle
the special cases of x = 1 and x = 5. Just plug those into the original matrix and work out
the RREF. I won’t type this up since we have been over RREF for numerical matrices, but
you should get (for x = 1 and x = 5, respectively):




1 1 2
1 0 −1
 0 0 0  and  0 1 −1  .
0 0 0
0 0 0
Generalized augmented matrices
This one is easier. If you need to solve the same system repeatedly (with different right
hand sides, a.k.a., constant vectors), you can just toss all of the constant vectors into the
augmented matrix.
For example, to solve
2x + y = 3
x−y =1
and
2x + y = 1
x−y =2
simultaneously, just form the following generalized augmented matrix and follow the steps
for RREF as you would for a (normal) augmented matrix.
2
1 3 1
R1 ↔ R2
1 −1 1 2
1 −1 1 2
−→
2
1 3 1
R2 − 2R1 ↔ R2
−→
2
1 −1 1
0
3 1 −3
R1 + R2 ↔ R1
−→
1
R
3 2
1 0
0 1
↔ R2
−→
4
3
1
3
1
−1
1 −1 1
2
1
0
1 3 −1
.
Thus, as I mentioned briefly in class, the solution to the first system of linear equations is
( 43 , 31 ) while the solution to the second system is (1, −1).