Difference between revisions of "User:Kevindunn"

From Process Model Formulation and Solution: 3E4
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* [mailto:kevin.dunn@connectmv.com kevin.dunn@connectmv.com] (alternate)
* [mailto:kevin.dunn@connectmv.com kevin.dunn@connectmv.com] (alternate)


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The commented lines in the MATLAB code show how the function is used. The results are: <span class="math">\(x =[1.3333, 3.1667,1.5000,-6.0000]\)</span>.
The commented lines in the MATLAB code show how the function is used. The results are: <span class="math">\(x =[1.3333, 3.1667,1.5000,-6.0000]\)</span>.
Also, the \(A\) and <span class="math">\(b\)</span> (denoted as <span class="math">\(C=[A\quad b]\)</span> below) are updated as:</p>
Also, the \(A\) and <span class="math">\(b\)</span> (denoted as <span class="math">\(C=[A\quad b]\)</span> below) are updated as:</p>
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Revision as of 07:30, 10 February 2017

Contact details

I don't have an office on campus. The best way to contact me is by email:


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 \(x = [1.333, 3.167, 1.5, -6]\).