Question A.11: Consider the matrix A = |1 2 1 0 0 3 3 1 2 0 2 0 1 0 0 2| It......

Consider the matrix

\overline{A}= \left|\begin{matrix}1 &2 &1 &0 \\ 0& 3& 3& 1 \\ 2& 0& 2& 0 \\  1 &0 &0 &2 \end{matrix} \right|

It is required to convert it into LU form. This is the same matrix of Example A.10.

Step-by-Step
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Add an identity matrix, which will ultimately be converted into a U matrix and the \overline{A} matrix will be converted into an L matrix:

\left|\begin{matrix}1 &2 &1 &0 \\ 0& 3& 3& 1 \\ 2& 0& 2& 0 \\  1 &0 &0 &2 \end{matrix} \right|\left|\begin{matrix}1 &0 &0 &0 \\ 0& 1& 0& 0 \\ 0& 0& 1& 0 \\  0 &0 &0 &1 \end{matrix} \right|

First step, p = 1:
Table 1 and Table 2

The shaded columns and rows are converted into L and U matrix column and row and the elements of \overline{A} matrix are modified using Equation A.101, that is,
l_{ip} = a_{1}p  for ≥ p
u_{pj} = \frac{a_{pj}}{a_{pp}}  for j > p                                 (A.101)
a_{ij} = a_{1j} – l_{ip}u_{pj}  for i > p,  j > p

a_{32} = a_{32} – l_{31}u_{12}
= 0 – (2)(2) = -4
a_{33} = a_{33} – l_{31}u_{31}
= 2 – (2)(1) = 0

Second step, pivot column 2, p = 2:
Table 3 and Table 4

Third step, pivot column 3, p = 3:
Table 5 and Table 6

This is the same result as derived before in Example A.10.

Table 1
1
0 3 3 0
2 -4 0 0
1 -2 -1 2
Table 2
1 2 1 0
0 1 0 0
0 0 1 0
0 0 1 0
Table 3
1
0 3 3 0
2 -4 4 1.32
1 -2 1 2.66
Table 4
1 2 1 0
0 1 1 0.33
0 0 1 0
0 0 0 1
Table 5
1 0 0 0
0 3 0 0
2 -4 4 0
1 -2 1 2.33
Table 6
1 2 1 0
0 1 1 0.33
0 0 1 0.33
0 0 0 1

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