Question 10.1: Derive an LU factorization based on the Gauss elimination pe...
Derive an LU factorization based on the Gauss elimination performed previously in Example 9.3.
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
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.