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## Q. 10.3

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

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

## 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:
$x_2 = \frac{1.9998}{2.9997} = 0.66667$
$x_1 =\frac{1 − 1(0.66667)}{1} = 0.33333$