14-240/Tutorial-December 2 - Revision history

Bug: /* Theorem */

2014-12-09T16:55:00Z

Theorem

← Older revision		Revision as of 12:55, 9 December 2014
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	:::::::<math>~~(-1)~~det\begin{pmatrix}...\\A_{i + 1}~~\\A_i~~\\...\\A_j\\...\end{pmatrix} = det\begin{pmatrix}...~~\\A_i~~\\A_{i + 1}\\...\\A_j\\...\end{pmatrix} = (-1)det\begin{pmatrix}...\\A_{i + 1}~~\\A_i~~\\...\\A_j\\...\end{pmatrix} =</math>		:::::::<math>det\begin{pmatrix}...\\A_i\\A_{i + 1}\\...\\A_j\\...\end{pmatrix} = (-1)det\begin{pmatrix}...\\A_{i + 1}\\A_i\\...\\A_j\\...\end{pmatrix} = (-1)^{j - i}det\begin{pmatrix}...\\A_{i + 1}\\...\\A_j\\A_i\\...\end{pmatrix} =</math>



	:::::::<math>(-1)^{j - i}det\begin{pmatrix}...\\A_{i + 1}\\...~~\\A_j~~\\A_i\\...\end{pmatrix} = (-1)^{~~j - i}~~(~~-1)^{~~j - i - 1}det\begin{pmatrix}...\\A_j\\A_{i + 1}\\...\\A_i\\...\end{pmatrix} =</math>		:::::::<math>(-1)^{j - i}(-1)^{j - i - 1}det\begin{pmatrix}...\\A_j\\A_{i + 1}\\...\\A_i\\...\end{pmatrix} = (-1)^{2(j - i) - 1}det\begin{pmatrix}...\\A_j\\A_{i + 1}\\...\\A_i\\...\end{pmatrix} =</math>



	:::::::<math>~~(-1)^{2(j - i) - 1}det\begin{pmatrix}...\\A_j\\A_{i + 1}\\...\\A_i\\...\end{pmatrix} =~~ (-1)^{- 1}det\begin{pmatrix}...\\A_j\\A_{i + 1}\\...\\A_i\\...\end{pmatrix} = (-1)det\begin{pmatrix}...\\A_j\\A_{i + 1}\\...\\A_i\\...\end{pmatrix} =</math>		:::::::<math>(-1)^{- 1}det\begin{pmatrix}...\\A_j\\A_{i + 1}\\...\\A_i\\...\end{pmatrix} = (-1)det\begin{pmatrix}...\\A_j\\A_{i + 1}\\...\\A_i\\...\end{pmatrix} =</math>

Boyang.wu: /* Scanned Lecture Notes by Boyang.wu */

2014-12-08T18:17:32Z

Scanned Lecture Notes by Boyang.wu

← Older revision		Revision as of 14:17, 8 December 2014
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	==Nikita==		==Nikita==
	==Scanned ~~Lecture~~ Notes by [[User Boyang.wu\|Boyang.wu]]==		==Scanned Tutorial Notes by [[User Boyang.wu\|Boyang.wu]]==
	[[File:Tut.pdf]]		[[File:Tut.pdf]]

Boyang.wu: /* Nikita */

2014-12-08T18:17:01Z

Nikita

← Older revision		Revision as of 14:17, 8 December 2014
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	==Nikita==		==Nikita==
			==Scanned Lecture Notes by [[User Boyang.wu\|Boyang.wu]]==
			[[File:Tut.pdf]]

Bug: /* Boris */

2014-12-07T21:01:41Z

Boris

← Older revision		Revision as of 17:01, 7 December 2014
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	For the proof of the theorem, assume that <math>B</math> is the matrix <math>A</math> with ~~row~~ <math>i</math> of <math>A</math> interchanged ~~with row <math>j</math> of <math>A</math>~~ and ~~that~~ <math>i \neq j</math>. By Lemma 1, we have the following:		For the proof of the theorem, assume that <math>B</math> is the matrix <math>A</math> with rows <math>i, j</math> of <math>A</math> interchanged and <math>i \neq j</math>. By '''Lemma 1''', we have the following:


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	:::::::<math>(-1)det\begin{pmatrix}...\\A_(i + 1)\\A_i\\...\\A_j\\...\end{pmatrix} = det\begin{pmatrix}...\\A_i\\A_{i + 1}\\...\\A_j\\...\end{pmatrix} = (-1)det\begin{pmatrix}...\\A_{i + 1}\\A_i\\...\\A_j\\...\end{pmatrix} =</math>		:::::::<math>(-1)det\begin{pmatrix}...\\A_{i + 1}\\A_i\\...\\A_j\\...\end{pmatrix} = det\begin{pmatrix}...\\A_i\\A_{i + 1}\\...\\A_j\\...\end{pmatrix} = (-1)det\begin{pmatrix}...\\A_{i + 1}\\A_i\\...\\A_j\\...\end{pmatrix} =</math>

Bug at 20:56, 7 December 2014

2014-12-07T20:56:22Z

← Older revision		Revision as of 16:56, 7 December 2014
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	:::::::<math>~~det(A) + det(B) =~~ det\begin{pmatrix}...\\A_i\\A_{i + 1}\\...\end{pmatrix} + det\begin{pmatrix}...\\A_{i + 1}\\A_i\\...\end{pmatrix} =</math>		:::::::<math>det\begin{pmatrix}...\\A_i\\A_{i + 1}\\...\end{pmatrix} + det\begin{pmatrix}...\\A_{i + 1}\\A_i\\...\end{pmatrix} =</math>

Bug at 20:54, 7 December 2014

2014-12-07T20:54:50Z

← Older revision		Revision as of 16:54, 7 December 2014
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	:::::::<math>(-1)det\begin{pmatrix}...\\A_(i + 1)\\A_i\\...\\A_j\\...\end{pmatrix} = det\begin{pmatrix}...\\A_i\\A_{i + 1}\\...\\A_j\\...\end{pmatrix} = (-1)det\begin{pmatrix}...\\A_{i + 1}\\A_i\\...\\A_j~~\\...\end{pmatrix} = (-1)^{j - i}det\begin{pmatrix}...\\A_{i + 1}\\...\\A_j\\A_i~~\\...\end{pmatrix} =</math>		:::::::<math>(-1)det\begin{pmatrix}...\\A_(i + 1)\\A_i\\...\\A_j\\...\end{pmatrix} = det\begin{pmatrix}...\\A_i\\A_{i + 1}\\...\\A_j\\...\end{pmatrix} = (-1)det\begin{pmatrix}...\\A_{i + 1}\\A_i\\...\\A_j\\...\end{pmatrix} =</math>



	:::::::<math>(-1)^{j - i~~}(-1)^{j - i - 1~~}det\begin{pmatrix}...~~\\A_j~~\\A_{i + 1}\\...\\A_i\\...\end{pmatrix} = (-1)^{2(j - i~~) - 1~~}~~det\begin{pmatrix}...\\A_j\\A_{i + 1}\\...\\A_i\\...\end{pmatrix} =~~ (-1)^{- 1}det\begin{pmatrix}...\\A_j\\A_{i + 1}\\...\\A_i\\...\end{pmatrix} =</math>		:::::::<math>(-1)^{j - i}det\begin{pmatrix}...\\A_{i + 1}\\...\\A_j\\A_i\\...\end{pmatrix} = (-1)^{j - i}(-1)^{j - i - 1}det\begin{pmatrix}...\\A_j\\A_{i + 1}\\...\\A_i\\...\end{pmatrix} =</math>



	:::::::<math>(-1)det\begin{pmatrix}...\\A_j\\A_{i + 1}\\...\\A_i\\...\end{pmatrix} = -det(B)</math>.		:::::::<math>(-1)^{2(j - i) - 1}det\begin{pmatrix}...\\A_j\\A_{i + 1}\\...\\A_i\\...\end{pmatrix} = (-1)^{- 1}det\begin{pmatrix}...\\A_j\\A_{i + 1}\\...\\A_i\\...\end{pmatrix} = (-1)det\begin{pmatrix}...\\A_j\\A_{i + 1}\\...\\A_i\\...\end{pmatrix} =</math>



			:::::::<math>-det(B)</math>.

Bug: /* Theorem */

2014-12-07T20:51:18Z

Theorem

@@ Line 12: / Line 12: @@
-All we need to show is that <math>det(A) + det(B) = 0</math>.  Assume that <math>B</math> is the matrix <math>A</math> with row <math>i</math> of <math>A</math> interchanged with row <math>i + 1</math> of <math>A</math>.  Since the determinant of a matrix with two identical rows is <math>0</math>, then:
+All we need to show is that <math>det(A) + det(B) = 0</math>.  Assume that <math>B</math> is the matrix <math>A</math> with rows <math>i, i + 1</math> of <math>A</math> interchanged.  Since the determinant of a matrix with two identical rows is <math>0</math>, then:
@@ Line 36: / Line 36: @@
-Then <math>det(A) + det(B) = 0</math> and <math>det(A) = -det(B)</math>.
+Then <math>det(A) + det(B) = 0</math> and <math>det(A) = -det(B)</math>.  The proof of the lemma is complete.
 ==Nikita==

Bug: /* Lemma 1 */

2014-12-07T20:17:29Z

Lemma 1

← Older revision		Revision as of 16:17, 7 December 2014
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	:::::::<math>det(A) + det(B) = det\begin{pmatrix}...\\A_i\\A_{i + 1}\\...\end{pmatrix} + det\begin{pmatrix}...\\A_{i + 1}\\A_i\\...\end{pmatrix} =</math>		:::::::<math>det(A) + det(B) = det\begin{pmatrix}...\\A_i\\A_{i + 1}\\...\end{pmatrix} + det\begin{pmatrix}...\\A_{i + 1}\\A_i\\...\end{pmatrix} =</math>


	:::::::<math>det\begin{pmatrix}...\\A_i\\A_i\\...\end{pmatrix} + det\begin{pmatrix}...\\A_i\\A_{i + 1}\\...\end{pmatrix} + det\begin{pmatrix}...\\A_{i + 1}\\A_{i + 1}\\...\end{pmatrix} + det\begin{pmatrix}...\\A_{i + 1}\\A_i\\...\end{pmatrix}</math>.		:::::::<math>det\begin{pmatrix}...\\A_i\\A_i\\...\end{pmatrix} + det\begin{pmatrix}...\\A_i\\A_{i + 1}\\...\end{pmatrix} + det\begin{pmatrix}...\\A_{i + 1}\\A_{i + 1}\\...\end{pmatrix} + det\begin{pmatrix}...\\A_{i + 1}\\A_i\\...\end{pmatrix}</math>.
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	:::::::<math>det\begin{pmatrix}...\\A_i\\A_i\\...\end{pmatrix} + det\begin{pmatrix}...\\A_i\\A_{i + 1}\\...\end{pmatrix} + det\begin{pmatrix}...\\A_{i + 1}\\A_{i + 1}\\...\end{pmatrix} + det\begin{pmatrix}...\\A_{i + 1}\\A_i\\...\end{pmatrix} = </math>		:::::::<math>det\begin{pmatrix}...\\A_i\\A_i\\...\end{pmatrix} + det\begin{pmatrix}...\\A_i\\A_{i + 1}\\...\end{pmatrix} + det\begin{pmatrix}...\\A_{i + 1}\\A_{i + 1}\\...\end{pmatrix} + det\begin{pmatrix}...\\A_{i + 1}\\A_i\\...\end{pmatrix} = </math>


	:::::::<math>det\begin{pmatrix}...\\A_i\\A_i + A_{i + 1}\\...\end{pmatrix} + det\begin{pmatrix}...\\A_{i + 1}\\A_i + A_{i + 1}\\...\end{pmatrix} = det\begin{pmatrix}...\\A_i + A_{i + 1}\\A_i + A_{i + 1}\\...\end{pmatrix} = 0</math>.		:::::::<math>det\begin{pmatrix}...\\A_i\\A_i + A_{i + 1}\\...\end{pmatrix} + det\begin{pmatrix}...\\A_{i + 1}\\A_i + A_{i + 1}\\...\end{pmatrix} = det\begin{pmatrix}...\\A_i + A_{i + 1}\\A_i + A_{i + 1}\\...\end{pmatrix} = 0</math>.

Bug: /* Theorem */

2014-12-07T20:17:03Z

Theorem

← Older revision		Revision as of 16:17, 7 December 2014
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	====Theorem====		====Theorem====

	Let <math>A</math> be a <math>n \times n</math> matrix and <math>B</math> be the matrix <math>A</math> with two rows interchanged. Then <math>det(A) = -det(B)</math>.		Let <math>A</math> be a <math>n \times n</math> matrix and <math>B</math> be the matrix <math>A</math> with two rows interchanged. Then <math>det(A) = -det(B)</math>. Boris decided to prove the following lemma first:


	Boris decided to prove the following lemma first:


	=====Lemma 1=====		=====Lemma 1=====

Bug: /* Theorem */

2014-12-07T20:14:49Z

Theorem

← Older revision		Revision as of 16:14, 7 December 2014
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	Boris decided to prove the following lemma first:		Boris decided to prove the following lemma first:


	=====Lemma 1=====		=====Lemma 1=====