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

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Accepted Answer

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
[latex]\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}[/latex]
and that the forward-elimination phase of conventional Gauss elimination resulted in
[latex]\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}[/latex]
The forward-substitution phase is implemented by applying Eq. (10.8):

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

[latex]\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}[/latex] (10.3)
[latex]\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}[/latex]
which can be solved by back substitution (see Example 9.3 for details) for the final solution:
{x} =[latex]\begin{Bmatrix} 3 \ -2.5 \ 7.00003 \end{Bmatrix}[/latex]

Q. 10.2

Chapter 10

Q. 10.2

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