06-240/Classnotes For Tuesday November 21: Difference between revisions
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|<math>A_1=\begin{pmatrix}1 |
|<math>A_1=\begin{pmatrix}1</math> |
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|<math>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>A_1x=b</math> or <math>A_2x=b</math>. What does it tell us about the solutions? Let us start from the second system: |
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|<math>\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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|<math>x_1</math> |
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|<math>-4x_3</math> |
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|<math>-6x_5</math> |
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|<math>=</math> |
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|<math>b_1</math> |
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|<math>x_2</math> |
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|<math>+2x_3</math> |
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|<math>-2x_5</math> |
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|<math>=</math> |
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|<math>b_2</math> |
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|<math>x_4</math> |
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|<math>+2x_5</math> |
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|<math>=</math> |
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|<math>b_3</math> |
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|align=center|<math>0</math> |
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|<math>=</math> |
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|<math>b_4</math> |
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Well, quite clearly if <math>b_4\neq 0</math> this system has no solutions, but if <math>b_4=0</math> it has solutions no matter what <math>b_1</math>, <math>b_2</math> and <math>b_3</math> are. Finally, for any given values of <math>b_1</math>, <math>b_2</math> and <math>b_3</math> we can choose the values of <math>x_3</math> and <math>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>x_1=b_1+4x_3+6x_5</math>, <math>x_2=b_2-2x_3+2x_5</math> and <math>x_4=b_3-2x_5</math>. |
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What about the system corresponding to <math>A_1</math>? It is |
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|<math>\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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|<math>x_1</math> |
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|<math>+3x_2</math> |
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|<math>+2x_3</math> |
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|<math>+4x_4</math> |
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|<math>+2x_5</math> |
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|<math>=</math> |
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|<math>b_1</math> |
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|<math>x_2</math> |
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|<math>+2x_3</math> |
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|<math>+3x_4</math> |
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|<math>+4x_5</math> |
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|<math>=</math> |
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|<math>b_2</math> |
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|<math>x_4</math> |
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|<math>+2x_5</math> |
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|<math>=</math> |
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|<math>b_3</math> |
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|align=center|<math>0</math> |
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|<math>=</math> |
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|<math>b_4</math> |
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Here too we have solutions iff <math>b_4=0</math>, and if <math>b_4=0</math>, we have the freedom to choose the non-pivotal variables <math>x_3</math> and <math>x_5</math> as we please. But now the formulas for fixing the pivotal variables <math>x_1</math>, <math>x_2</math> and <math>x_4</math> in terms of the non-pivotal ones are a bit harder. |
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==Class notes== |
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[[Media:MAT Lect018.pdf|Scan of Week 11 Lecture 1 notes]] |
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[[Media:06-240-lec notes nov21st.pdf|Lecture_notes november21st]] |
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Revision as of 01:38, 28 May 2007
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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:
| Failed to parse (unknown function "\begin{pmatrix}"): {\displaystyle A_1=\begin{pmatrix}1} |
