Subscribe $4.99/month

Un-lock Verified Step-by-Step Experts Answers.

Find the Source, Textbook, Solution Manual that you are looking for in 1 click.

Our Website is free to use.
To help us grow, you can support our team with a Small Tip.

All the data tables that you may search for.

Need Help? We got you covered.

For Arabic Users, find a teacher/tutor in your City or country in the Middle East.

Products

Find the Source, Textbook, Solution Manual that you are looking for in 1 click.

For Arabic Users, find a teacher/tutor in your City or country in the Middle East.

Need Help? We got you covered.

Chapter 10

Q. 10.2

Complete the problem initiated in Example 10.1 by generating the final solution with forward and back substitution.

Step-by-Step

Verified Solution

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}