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−x1 2
1 2 −x2
1 1 3 −x
R1↔R2
−→
1 2 −x2
2−x1 0
1 1 3 −x
At this point, it is tempting to do R2−(2 −x)R1→R1to 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.
R2−R1→R2
−→
1 2 −x2
1−x x −1 0
1 1 3 −x
R3−R1→R3
−→
1 2 −x2
1−x x −1 0
0x−1 1 −x
Now I can divide the second and third rows by x−1 to make them nice numbers:
1
x−1R2→R2
−→
1 2 −x2
1−1 0
0x−1 1 −x
1
x−1R3→R3
−→
1 2 −x2
1−1 0
0 1 −1
WARNING: We multiplied by 1
x−1. If x= 1, this amounts to division by 0 – bad news!
SO, the solution that we get here will be valid for x6= 1, and we will need to do the x= 1
case separately.
R2−R1→R2
−→
1 2 −x2
1−3 + x−2
0 1 −1
R2↔R3
−→
1−3 + x−2
1 2 −x2
0 1 −1
R3−(x−3)R2→R3
−→
1 2 −x2
1−3 + x−2
0 0 x−5
1
x−5R3→R3
−→
1 2 −x2
1−3 + x−2
0 0 1
WARNING: We did it again! We now need to do the case of x= 5 separately, too.
