Fundamentals of Matrix Algebra, 2011a

1.3 Elementary Row Operaons and Gaussian Eliminaon
so x 1 = 3.
This process of substung known values back into other equaons is called back
substuon. This process is essenally what happens when we perform the backward
steps of Gaussian eliminaon. We make note of this below as we finish out finding the
reduced row echelon form of our matrix.
5
R3 + R2 → R2
3
(knowing x 3 = 7 allows us
to find x 2 = 5)
1
R3 + R1 → R1
2
− 1 R 2 2 + R 1 → R 1
(knowing x 2 = 5 and x 3 = 7
allows us to find x 1 = 3)
⎡
1 1/2 −1/2
⎤
2
⎣ 0 1 0 5 ⎦
0 0 1 7
⎡
⎣ 1 0 0 3 ⎤
0 1 0 5 ⎦
0 0 1 7
We did our operaons slightly “out of order” in that we didn’t put the zeros above
our leading 1 in the third column in the same step, highlighng how back substuon
works. .
In all of our pracce, we’ve only encountered systems of linear equaons with exactly
one soluon. Is this always going to be the case? Could we ever have systems
with more than one soluon? If so, how many soluons could there be? Could we
have systems without a soluon? These are some of the quesons we’ll address in
the next secon.
Exercises 1.3
In Exercises 1 – 4, state whether or not the
given matrices are in reduced row echelon
form. If it is not, state why.
[ ] 1 0
1. (a)
0 1
[ ] 0 1
(b)
1 0
[ ] 1 0 0
2. (a)
0 0 1
[ ] 1 0 1
(b)
0 1 1
⎡
1 1
⎤
1
3. (a) ⎣ 0 1 1 ⎦
0 0 1
⎡
1 0
⎤
0
(b) ⎣ 0 1 0 ⎦
0 0 0
[ ] 1 1
(c)
1 1
[ ] 1 0 1
(d)
0 1 2
[ ] 0 0 0
(c)
1 0 0
[ ] 0 0 0
(d)
0 0 0
(c)
(d)
4. (a)
(b)
(c)
(d)
⎡
1 0
⎤
0
⎣ 0 0 1 ⎦
0 0 0
⎡
1 0 0
⎤
−5
⎣ 0 1 0 7 ⎦
0 0 1 3
⎡
2 0 0
⎤
2
⎣ 0 2 0 2 ⎦
0 0 2 2
⎡
0 1 0
⎤
0
⎣ 0 0 1 0 ⎦
0 0 0 0
⎡
0 0 1
⎤
−5
⎣ 0 0 0 0 ⎦
0 0 0 0
⎡
1 1 0 0 1
⎤
1
⎣ 0 0 1 0 1 1 ⎦
0 0 0 1 0 0
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