14-240/Tutorial-December 2: Difference between revisions

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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 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:


:::::::<math>det(A) + det(B) =</math>




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Since the determinant is linear in each row, then:
Since the determinant is linear in each row, then we continue where we left off:


:::::::<math>det(A) + det(B) =</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(A) + det(B) = 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>


Then <math>det(A) + det(B) = 0</math> and <math>det(A) = -det(B)</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>.


==Nikita==
==Nikita==

Revision as of 15:13, 7 December 2014

DBN: Classes: 2014-15: 14-240 / Navigation
Welcome to Math 240!
(additions to this web site no longer count towards good deed points)
# Week of... Notes and Links
1 Sep 8 About This Class, What is this class about? (PDF, HTML), Monday, Wednesday
2 Sep 15 HW1, Monday, Wednesday, TheComplexField.pdf,HW1_solutions.pdf
3 Sep 22 HW2, Class Photo, Monday, Wednesday, HW2_solutions.pdf
4 Sep 29 HW3, Wednesday, Tutorial, HW3_solutions.pdf
5 Oct 6 HW4, Monday, Wednesday, Tutorial, HW4_solutions.pdf
6 Oct 13 No Monday class (Thanksgiving), Wednesday, Tutorial
7 Oct 20 HW5, Term Test at tutorials on Tuesday, Wednesday
8 Oct 27 HW6, Monday, Why LinAlg?, Wednesday, Tutorial
9 Nov 3 Monday is the last day to drop this class, HW7, Monday, Wednesday, Tutorial
10 Nov 10 HW8, Monday, Tutorial
11 Nov 17 Monday-Tuesday is UofT November break
12 Nov 24 HW9
13 Dec 1 Wednesday is a "makeup Monday"! End-of-Course Schedule, Tutorial
F Dec 8 The Final Exam
Register of Good Deeds
Class Photo
Add your name / see who's in!
14-240-Splash.png

Boris

Theorem

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

Lemma 1

Let [math]\displaystyle{ A }[/math] be a [math]\displaystyle{ n \times n }[/math] matrix and [math]\displaystyle{ B }[/math] be the matrix [math]\displaystyle{ A }[/math] with two adjacent rows interchanged. Then [math]\displaystyle{ det(A) = -det(B) }[/math].

All we need to show is that [math]\displaystyle{ det(A) + det(B) = 0 }[/math]. Assume that [math]\displaystyle{ B }[/math] is the matrix [math]\displaystyle{ A }[/math] with row [math]\displaystyle{ i }[/math] of [math]\displaystyle{ A }[/math] interchanged with row [math]\displaystyle{ i + 1 }[/math] of [math]\displaystyle{ A }[/math]. Since the determinant of a matrix with two identical rows is [math]\displaystyle{ 0 }[/math], then:


[math]\displaystyle{ det(A) + det(B) = }[/math]


[math]\displaystyle{ 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]\displaystyle{ 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].


Since the determinant is linear in each row, then we continue where we left off:


[math]\displaystyle{ det(A) + det(B) = }[/math]


[math]\displaystyle{ 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]\displaystyle{ 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].


Then [math]\displaystyle{ det(A) + det(B) = 0 }[/math] and [math]\displaystyle{ det(A) = -det(B) }[/math].

Nikita

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