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## Q. 10.1

Derive an LU factorization based on the Gauss elimination performed previously in Example 9.3.

## Verified Solution

In Example 9.3, we used Gauss elimination to solve a set of linear algebraic equations that had the following coefficient matrix:
$[A] = \begin{bmatrix}3& −0.1& −0.2\\ 0.1& 7& −0.3\\ 0.3& −0.2& 10\end{bmatrix}$
After forward elimination, the following upper triangular matrix was obtained:
$[U ] = \begin{bmatrix}3&−0.1& −0.2\\ 0& 7.00333& −0.293333\\ 0& 0& 10.0120\end{bmatrix}$
The factors employed to obtain the upper triangular matrix can be assembled into a lower triangular matrix. The elements $a_{21} \text{ and } a_{31}$ were eliminated by using the factors
$f_{21} = \frac{0.1}{3} = 0.0333333$              $f_{31} = \frac{0.3}{3} = 0.1000000$
and the element $a_{32}$ was eliminated by using the factor
$f_{32} = \frac{−0.19}{7.00333} = −0.0271300$
Thus, the lower triangular matrix is
[L ] = $\begin{bmatrix}1 &0& 0\\ 0.0333333& 1& 0\\ 0.100000& −0.0271300& 1\end{bmatrix}$
Consequently, the LU factorization is

[A] = [L ] [U ] = $\begin{bmatrix}1 &0 &0\\ 0.0333333& 1& 0\\ 0.100000& −0.0271300& 1\end{bmatrix} \begin{bmatrix}3& −0.1& −0.2\\ 0& 7.00333& −0.293333\\ 0& 0& 10.0120\end{bmatrix}$

This result can be verified by performing the multiplication of [L][U] to give

[L ] [U ] = $\begin{bmatrix}3& −0.1& −0.2\\ 0.0999999& 7& −0.3\\ 0.3 &−0.2& 9.99996\end{bmatrix}$

where the minor discrepancies are due to roundoff.