06-240/Classnotes For Tuesday November 21 - Revision history

Drorbn: Reverted edit of 161.53.232.46, changed back to last version by Gokmen

2007-05-28T14:47:26Z

Reverted edit of 161.53.232.46, changed back to last version by Gokmen

@@ Line 7: / Line 7: @@
 {| align=center
 |-
-|<math>A_1=\begin{pmatrix}1</math>
+|<math>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>

161.53.232.46 at 06:38, 28 May 2007

2007-05-28T06:38:42Z

@@ Line 7: / Line 7: @@
 {| align=center
 |-
-|<math>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>
+|<math>A_1=\begin{pmatrix}1</math>

Gokmen at 09:30, 4 December 2006

2006-12-04T09:30:54Z

← Older revision		Revision as of 05:30, 4 December 2006
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	[[Media:MAT Lect018.pdf\|Scan of Week 11 Lecture 1 notes]]		[[Media:MAT Lect018.pdf\|Scan of Week 11 Lecture 1 notes]]

	[[Media:06-240-lec notes nov21st\|Lecture_notes november21st]]		[[Media:06-240-lec notes nov21st.pdf\|Lecture_notes november21st]]

Gokmen at 09:30, 4 December 2006

2006-12-04T09:30:17Z

← Older revision		Revision as of 05:30, 4 December 2006
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	[[Media:MAT Lect018.pdf\|Scan of Week 11 Lecture 1 notes]]		[[Media:MAT Lect018.pdf\|Scan of Week 11 Lecture 1 notes]]

			[[Media:06-240-lec notes nov21st\|Lecture_notes november21st]]

Alla: Uploaded Week 11 Lecture 1 notes

2006-11-26T17:25:50Z

Uploaded Week 11 Lecture 1 notes

← Older revision		Revision as of 13:25, 26 November 2006
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	==Class notes==		==Class notes==

			[[Media:MAT Lect018.pdf\|Scan of Week 11 Lecture 1 notes]]

Alla: added header

2006-11-26T17:24:15Z

added header

← Older revision		Revision as of 13:24, 26 November 2006
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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.		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.

			==Class notes==

Drorbn: /* More about the Wongpak Matrices */

2006-11-21T01:17:51Z

More about the Wongpak Matrices

← Older revision		Revision as of 21:17, 20 November 2006
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	==More about the [[User:Wongpak\|Wongpak]] Matrices==		==More about the [[User:Wongpak\|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:		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:

	{\| align=center		{\| align=center

Drorbn at 01:15, 21 November 2006

2006-11-21T01:15:57Z

New page

{{06-240/Navigation}}

==More about the [[User:Wongpak|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:

{| align=center
|-
|<math>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>
|width=10%|
|<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>
|}

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:

{| align=center
|- align=center
|<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>
|width=15%|or
|
{|
|-
|<math>x_1</math>
|
|<math>-4x_3</math>
|
|<math>-6x_5</math>
|<math>=</math>
|<math>b_1</math>
|-
|
|<math>x_2</math>
|<math>+2x_3</math>
|
|<math>-2x_5</math>
|<math>=</math>
|<math>b_2</math>
|-
|
|
|
|<math>x_4</math>
|<math>+2x_5</math>
|<math>=</math>
|<math>b_3</math>
|-
|
|
|
|
|align=center|<math>0</math>
|<math>=</math>
|<math>b_4</math>
|}
|}

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>.

What about the system corresponding to <math>A_1</math>? It is
{| align=center
|- align=center
|<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>
|width=15%|or
|
{|
|- align=center
|<math>x_1</math>
|<math>+3x_2</math>
|<math>+2x_3</math>
|<math>+4x_4</math>
|<math>+2x_5</math>
|<math>=</math>
|<math>b_1</math>
|- align=center
|
|<math>x_2</math>
|<math>+2x_3</math>
|<math>+3x_4</math>
|<math>+4x_5</math>
|<math>=</math>
|<math>b_2</math>
|- align=center
|
|
|
|<math>x_4</math>
|<math>+2x_5</math>
|<math>=</math>
|<math>b_3</math>
|-
|
|
|
|
|align=center|<math>0</math>
|<math>=</math>
|<math>b_4</math>
|}
|}

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.