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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.

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