Numerical Methods Course Notes Version 0.1 (UCSD Math 174, Fall ...
Numerical Methods Course Notes Version 0.1 (UCSD Math 174, Fall ...
Numerical Methods Course Notes Version 0.1 (UCSD Math 174, Fall ...
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32 CHAPTER 3. SOLVING LINEAR SYSTEMS<br />
3.3.1 An Example<br />
We consider solution of the following augmented form:<br />
⎛<br />
⎜<br />
⎝<br />
2 1 1 3 7<br />
4 4 0 7 11<br />
6 5 4 17 31<br />
2 −1 0 7 15<br />
⎞<br />
⎟<br />
⎠ (3.2)<br />
The naïve G.E. reduces this to<br />
⎛<br />
⎜<br />
⎝<br />
2 1 1 3 7<br />
0 2 −2 1 −3<br />
0 0 3 7 13<br />
0 0 0 12 18<br />
⎞<br />
⎟<br />
⎠<br />
We are going to run the naïve G.E., and see how it is a LU Factorization. Since this is the naïve<br />
version, we first pivot on the first row. Our multipliers are 2, 3, 1. We pivot to get<br />
⎛<br />
⎜<br />
⎝<br />
2 1 1 3 7<br />
0 2 −2 1 −3<br />
0 2 1 8 10<br />
0 −2 −1 4 8<br />
Careful inspection shows that we’ve merely multiplied A and b by a lower triangular matrix M 1 :<br />
M 1 =<br />
⎡<br />
⎢<br />
⎣<br />
1 0 0 0<br />
−2 1 0 0<br />
−3 0 1 0<br />
−1 0 0 1<br />
The entries in the first column are the negative e.r.o. multipliers for each row. Thus after the first<br />
pivot, it is like we are solving the system<br />
M 1 Ax = M 1 b.<br />
⎞<br />
⎟<br />
⎠<br />
⎤<br />
⎥<br />
⎦<br />
We pivot on the second row to get:<br />
⎛<br />
⎜<br />
⎝<br />
2 1 1 3 7<br />
0 2 −2 1 −3<br />
0 0 3 7 13<br />
0 0 −3 5 5<br />
⎞<br />
⎟<br />
⎠<br />
The multipliers are 1, −1. We can view this pivot as a multiplication by M 2 , with<br />
M 2 =<br />
⎡<br />
⎢<br />
⎣<br />
1 0 0 0<br />
0 1 0 0<br />
0 −1 1 0<br />
0 1 0 1<br />
⎤<br />
⎥<br />
⎦<br />
We are now solving<br />
M 2 M 1 Ax = M 2 M 1 b.