Question 10.2: Complete the problem initiated in Example 10.1 by generating...

As just stated, the intent of forward substitution is to impose the elimination manipulations that we had formerly applied to [A] on the right-hand-side vector {b}. Recall that the system being solved is
\begin{bmatrix} 3 & -0.1 & -0.2 \\ 0.1 & 7 & -0.3 \\ 0.3 & -0.2 & 10 \end{bmatrix} \begin{Bmatrix} x_{1} \\ x_{2} \\ x_{3} \end{Bmatrix} = \begin{Bmatrix} 7.85 \\ -19.3 \\ 71.4 \end{Bmatrix}
and that the forward-elimination phase of conventional Gauss elimination resulted in
\begin{bmatrix} 3 & -0.1 & -0.2 \\ 0 & 7.00333 & -0.293333 \\ 0 & 0 & 10.0120 \end{bmatrix} \begin{Bmatrix} x_{1} \\ x_{2} \\ x_{3} \end{Bmatrix} = \begin{Bmatrix} 7.85 \\ -19.5617 \\ 70.0843 \end{Bmatrix}
The forward-substitution phase is implemented by applying Eq. (10.8):

[L]{d} = {b}                                                                       (10.8)
\begin{bmatrix} 1 & 0 & 0 \\ 0.0333333 & 1 & 0 \\ 0.100000 & -0.0271300 & 1 \end{bmatrix} \begin{Bmatrix} d_{1} \\ d_{2} \\ d_{3} \end{Bmatrix} = \begin{Bmatrix} 7.85 \\ -19.3 \\ 71.4 \end{Bmatrix}
or multiplying out the left-hand side:
d_1 = 7.85
0.0333333d_1 + d_2 = −19.3
0.100000d_1 − 0.0271300d_2 + d_3 = 71.4
We can solve the first equation for d_1 = 7.85, which can be substituted into the second equation to solve for
d_2 = −19.3 − 0.0333333(7.85)= −19.5617
Both d_1 \text{ and } d_2 can be substituted into the third equation to give
d_3 = 71.4 − 0.1(7.85) + 0.02713(−19.5617) = 70.0843
Thus,
{d} = \begin{Bmatrix} 7.85 \\ -19.5617 \\ 70.0843 \end{Bmatrix}
This result can then be substituted into Eq. (10.3), [U ]{x} = {d}:

\begin{bmatrix} u_{11} & u_{12} & u_{13} \\ 0 & u_{22} & u_{23} \\ 0 & 0 & 10.0120 \end{bmatrix} \begin{Bmatrix} x_{1} \\ x_{2} \\ x_{3} \end{Bmatrix} = \begin{Bmatrix} d_1 \\ d_2 \\ d_3 \end{Bmatrix}                                              (10.3)
\begin{bmatrix} 3 & -0.1 & -0.2 \\ 0 & 7.00333 & -0.293333 \\ 0 & 0 & 10.0120 \end{bmatrix} \begin{Bmatrix} x_{1} \\ x_{2} \\ x_{3} \end{Bmatrix} = \begin{Bmatrix} 7.85 \\ -19.5617 \\ 70.0843 \end{Bmatrix}
which can be solved by back substitution (see Example 9.3 for details) for the final solution:
{x} =\begin{Bmatrix} 3 \\ -2.5 \\ 7.00003 \end{Bmatrix}