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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: |
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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> |
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:::::::<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> |
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:::::::<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> |
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:::::::<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> |
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:::::::<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> |
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:::::::<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> |
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==Nikita== |
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==Nikita== |
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==Scanned Tutorial Notes by [[User Boyang.wu|Boyang.wu]]== |
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[[File:Tut.pdf]] |
Latest revision as of 12:55, 9 December 2014
Welcome to Math 240! (additions to this web site no longer count towards good deed points)
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#
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Week of...
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Notes and Links
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1
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Sep 8
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About This Class, What is this class about? (PDF, HTML), Monday, Wednesday
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2
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Sep 15
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HW1, Monday, Wednesday, TheComplexField.pdf,HW1_solutions.pdf
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3
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Sep 22
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HW2, Class Photo, Monday, Wednesday, HW2_solutions.pdf
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4
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Sep 29
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HW3, Wednesday, Tutorial, HW3_solutions.pdf
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5
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Oct 6
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HW4, Monday, Wednesday, Tutorial, HW4_solutions.pdf
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6
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Oct 13
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No Monday class (Thanksgiving), Wednesday, Tutorial
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7
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Oct 20
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HW5, Term Test at tutorials on Tuesday, Wednesday
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8
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Oct 27
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HW6, Monday, Why LinAlg?, Wednesday, Tutorial
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9
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Nov 3
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Monday is the last day to drop this class, HW7, Monday, Wednesday, Tutorial
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10
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Nov 10
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HW8, Monday, Tutorial
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11
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Nov 17
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Monday-Tuesday is UofT November break
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12
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Nov 24
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HW9
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13
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Dec 1
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Wednesday is a "makeup Monday"! End-of-Course Schedule, Tutorial
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F
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Dec 8
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The Final Exam
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Register of Good Deeds
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![Class Photo](/images/thumb/6/6b/14-240-ClassPhoto.jpg/310px-14-240-ClassPhoto.jpg) Add your name / see who's in!
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Boris
Theorem
Let
be a
matrix and
be the matrix
with two rows interchanged. Then
. Boris decided to prove the following lemma first:
Lemma 1
Let
be a
matrix and
be the matrix
with two adjacent rows interchanged. Then
.
All we need to show is that
. Assume that
is the matrix
with rows
of
interchanged. Since the determinant of a matrix with two identical rows is
, then:
![{\displaystyle det(A)+det(B)=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6f6f221712ba4ef348447412c519a2ea9da0a699)
![{\displaystyle det{\begin{pmatrix}...\\A_{i}\\A_{i+1}\\...\end{pmatrix}}+det{\begin{pmatrix}...\\A_{i+1}\\A_{i}\\...\end{pmatrix}}=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7aabf4f10ea4f5cc4b2d954664e17266a8d3b251)
.
Since the determinant is linear in each row, then we continue where we left off:
![{\displaystyle det(A)+det(B)=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6f6f221712ba4ef348447412c519a2ea9da0a699)
![{\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}}=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0d78436acae9ef478a4fff58a8fb22b8a63a6190)
.
Then
and
. The proof of the lemma is complete.
For the proof of the theorem, assume that
is the matrix
with rows
of
interchanged and
. By Lemma 1, we have the following:
![{\displaystyle det(A)=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6a943e582cff343c93a9cbdd40005c52a7aa8ab8)
![{\displaystyle 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}}=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ff3cbc1127786fee5e402fa5ebe3310c77a89f9d)
![{\displaystyle (-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}}=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8c24050d4a3e2bfa26f1eaf9624e7dbdf9dae7e2)
![{\displaystyle (-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}}=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/285f47b9ecb3201400a5b9f75cf3b77661947cb1)
.
Then the proof of the theorem is complete.
Nikita
Scanned Tutorial Notes by Boyang.wu
File:Tut.pdf