User:Kevindunn

From Process Model Formulation and Solution: 3E4
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The commented lines in the MATLAB code show how the function is used. The results are: \(x =[1.3333, 3.1667,1.5000,-6.0000]\). Also, the \(A\) and \(b\) (denoted as \(C=[A\quad b]\) below) are updated as:

\[\begin{split}C^1= \left[ \begin{array}{ccccc} 1 & 1 & 1 & 1 & 0\\ 0 & -3 & -3 & -3 & 4\\ 0 & 3 & -5 & 0 & 2\\ 0 & -6 & -6 & -4 & -4 \end{array} \right]\end{split}\]
\[\begin{split}C^2= \left[ \begin{array}{ccccc} 1 & 1 & 1 & 1 & 0\\ 0 & -3 & -3 & -3 & 4\\ 0 & 0 & -8 & -3 & 6\\ 0 & 0 & 0 & 2 & -12 \end{array} \right]\end{split}\]
\[\begin{split}C^3= \left[ \begin{array}{ccccc} 1 & 1 & 1 & 1 & 0\\ 0 & -3 & -3 & -3 & 4\\ 0 & 0 & -8 & -3 & 6\\ 0 & 0 & 0 & 2 & -12 \end{array} \right]\end{split}\]

The final result for \(x = [1.333, 3.167, 1.5, -6]\).