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Chapter 10

Q. 10.3

Compute the LU factorization and find the solution for the same system analyzed in Example 9.4

\begin{bmatrix} 0.0003 &3.0000  \\ 1.0000 &  1.0000 \end{bmatrix} \begin{Bmatrix} x_{1} \\ x_{2}  \end{Bmatrix} = \begin{Bmatrix} 2.0001 \\ 1.0000  \end{Bmatrix}

Step-by-Step

Verified Solution

Before elimination, we set up the initial permutation matrix:
[P] = \begin{bmatrix} 1.0000 &0.0000 \\ 0.0000 &  1.0000 \end{bmatrix}
We immediately see that pivoting is necessary, so prior to elimination we switch the rows:
[A] =\begin{bmatrix} 1.0000 & 1.0000 \\ 0.0003 &  3.0000 \end{bmatrix}
At the same time, we keep track of the pivot by switching the rows of the permutation matrix:
[P] = \begin{bmatrix} 0.0000 & 1.0000 \\ 1.0000 &  0.0000 \end{bmatrix}

We then eliminate a_{21} by subtracting the factor l_{21} = a_{21}∕a_{11} = 0.0003∕1 = 0.0003 from the second row of A. In so doing, we compute that the new value of a_{2 2}^\prime = 3 − 0.0003(1) = 2.9997. Thus, the elimination step is complete with the result:
[U] = \begin{bmatrix} 1 & 1 \\ 0 &  2.9997 \end{bmatrix}              [L] = \begin{bmatrix} 1 & 0 \\ 0.0003 &  1 \end{bmatrix}
Before implementing forward substitution, the permutation matrix is used to reorder the right-hand-side vector to reflect the pivots as in
[P]{b} = \begin{bmatrix} 0.0000 & 1.0000 \\ 1.0000 &  0.0000 \end{bmatrix} \begin{Bmatrix} 2.0001 \\ 1.0000  \end{Bmatrix}= \begin{Bmatrix}  1\\2.0001 \end{Bmatrix}
Then, forward substitution is applied as in
\begin{bmatrix} 1 & 0  \\ 0.0003 &  1 \end{bmatrix} \begin{Bmatrix} d_{1} \\ d_{2}  \end{Bmatrix} = \begin{Bmatrix} 1 \\ 2.0001  \end{Bmatrix}
which can be solved for d_1 = 1 \text{ and }d_2 = 2.0001 − 0.0003(1) = 1.9998. At this point, the system is
\begin{bmatrix} 1 & 1  \\ 0 &  2.9997 \end{bmatrix} \begin{Bmatrix} x_{1} \\ x_{2}  \end{Bmatrix} = \begin{Bmatrix} 1 \\ 1.9998  \end{Bmatrix}
Applying back substitution gives the final result:
x_2 = \frac{1.9998}{2.9997} = 0.66667
x_1 =\frac{1 − 1(0.66667)}{1} = 0.33333