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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).