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Week of...
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Notes and Links
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| 1
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Sep 11
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About, Tue, HW1, Putnam, Thu
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| 2
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Sep 18
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Tue, HW2, Thu
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| 3
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Sep 25
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Tue, HW3, Photo, Thu
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| 4
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Oct 2
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Tue, HW4, Thu
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| 5
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Oct 9
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Tue, HW5, Thu
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| 6
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Oct 16
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Why?, Iso, Tue, Thu
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| 7
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Oct 23
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Term Test, Thu (double)
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| 8
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Oct 30
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Tue, HW6, Thu
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| 9
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Nov 6
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Tue, HW7, Thu
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| 10
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Nov 13
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Tue, HW8, Thu
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| 11
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Nov 20
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Tue, HW9, Thu
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| 12
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Nov 27
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Tue, HW10, Thu
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| 13
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Dec 4
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On the final, Tue, Thu
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| F
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Dec 11
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Final: Dec 13 2-5PM at BN3, Exam Forum
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| Register of Good Deeds
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Add your name / see who's in!
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| edit the panel
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More about the Wongpak Matrices
In Talk:06-240/Classnotes_For_Tuesday_November_14, User:Wongpak asked something about row echelon form and reduced row echelon form, and gave the following matrices as specific examples:
| [math]\displaystyle{ A_1=\begin{pmatrix}1&3&2&4&2\\0&1&2&3&4\\0&0&0&1&2\\0&0&0&0&0 \end{pmatrix} }[/math]
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[math]\displaystyle{ A_2=\begin{pmatrix}1&0&-4&0&-6\\0&1&2&0&-2\\0&0&0&1&2\\0&0&0&0&0 \end{pmatrix} }[/math]
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So let us assume row reduction leads us to the systems [math]\displaystyle{ A_1x=b }[/math] or [math]\displaystyle{ A_2x=b }[/math]. What does it tell us about the solutions? Let us start from the second system:
| [math]\displaystyle{ \begin{pmatrix}1&0&-4&0&-6\\0&1&2&0&-2\\0&0&0&1&2\\0&0&0&0&0\end{pmatrix} \begin{pmatrix}x_1\\x_2\\x_3\\x_4\\x_5\end{pmatrix} = \begin{pmatrix}b_1\\b_2\\b_3\\b_4\end{pmatrix} }[/math]
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or
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| [math]\displaystyle{ x_1 }[/math]
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[math]\displaystyle{ -4x_3 }[/math]
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[math]\displaystyle{ -6x_5 }[/math]
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[math]\displaystyle{ = }[/math]
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[math]\displaystyle{ b_1 }[/math]
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[math]\displaystyle{ x_2 }[/math]
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[math]\displaystyle{ +2x_3 }[/math]
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[math]\displaystyle{ -2x_5 }[/math]
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[math]\displaystyle{ = }[/math]
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[math]\displaystyle{ b_2 }[/math]
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[math]\displaystyle{ x_4 }[/math]
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[math]\displaystyle{ +2x_5 }[/math]
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[math]\displaystyle{ = }[/math]
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[math]\displaystyle{ b_3 }[/math]
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[math]\displaystyle{ 0 }[/math]
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[math]\displaystyle{ = }[/math]
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[math]\displaystyle{ b_4 }[/math]
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Well, quite clearly if [math]\displaystyle{ b_4\neq 0 }[/math] this system has no solutions, but if [math]\displaystyle{ b_4=0 }[/math] it has solutions no matter what [math]\displaystyle{ b_1 }[/math], [math]\displaystyle{ b_2 }[/math] and [math]\displaystyle{ b_3 }[/math] are. Finally, for any given values of [math]\displaystyle{ b_1 }[/math], [math]\displaystyle{ b_2 }[/math] and [math]\displaystyle{ b_3 }[/math] we can choose the values of [math]\displaystyle{ x_3 }[/math] and [math]\displaystyle{ x_5 }[/math] (the variables corresponding the columns containing no pivots) as we please, and then get solutions by setting the "pivotal variables" in terms of the non-pivotal ones as follows: [math]\displaystyle{ x_1=b_1+4x_3+6x_5 }[/math], [math]\displaystyle{ x_2=b_2-2x_3+2x_5 }[/math] and [math]\displaystyle{ x_4=b_3-2x_5 }[/math].
What about the system corresponding to [math]\displaystyle{ A_1 }[/math]? It is
| [math]\displaystyle{ \begin{pmatrix}1&3&2&4&2\\0&1&2&3&4\\0&0&0&1&2\\0&0&0&0&0\end{pmatrix} \begin{pmatrix}x_1\\x_2\\x_3\\x_4\\x_5\end{pmatrix} = \begin{pmatrix}b_1\\b_2\\b_3\\b_4\end{pmatrix} }[/math]
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or
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| [math]\displaystyle{ x_1 }[/math]
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[math]\displaystyle{ +3x_2 }[/math]
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[math]\displaystyle{ +2x_3 }[/math]
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[math]\displaystyle{ +4x_4 }[/math]
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[math]\displaystyle{ +2x_5 }[/math]
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[math]\displaystyle{ = }[/math]
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[math]\displaystyle{ b_1 }[/math]
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[math]\displaystyle{ x_2 }[/math]
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[math]\displaystyle{ +2x_3 }[/math]
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[math]\displaystyle{ +3x_4 }[/math]
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[math]\displaystyle{ +4x_5 }[/math]
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[math]\displaystyle{ = }[/math]
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[math]\displaystyle{ b_2 }[/math]
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[math]\displaystyle{ x_4 }[/math]
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[math]\displaystyle{ +2x_5 }[/math]
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[math]\displaystyle{ = }[/math]
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[math]\displaystyle{ b_3 }[/math]
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[math]\displaystyle{ 0 }[/math]
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[math]\displaystyle{ = }[/math]
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[math]\displaystyle{ b_4 }[/math]
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Here too we have solutions iff [math]\displaystyle{ b_4=0 }[/math], and if [math]\displaystyle{ b_4=0 }[/math], we have the freedom to choose the non-pivotal variables [math]\displaystyle{ x_3 }[/math] and [math]\displaystyle{ x_5 }[/math] as we please. But now the formulas for fixing the pivotal variables [math]\displaystyle{ x_1 }[/math], [math]\displaystyle{ x_2 }[/math] and [math]\displaystyle{ x_4 }[/math] in terms of the non-pivotal ones are a bit harder.
Class notes
Scan of Week 11 Lecture 1 notes
Lecture_notes november21st
