14-240/Tutorial-December 2: Difference between revisions
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==Boris== |
==Boris== |
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====Theorem==== |
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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: |
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=====Lemma 1===== |
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Let <math>A</math> be a <math>n \times n</math> matrix and <math>B</math> be the matrix <math>A</math> with two '''adjacent''' rows interchanged. Then <math>det(A) = -det(B)</math>. |
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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 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: |
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:::::::<math>det(A) + det(B) =</math> |
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:::::::<math>det\begin{pmatrix}...\\A_i\\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>. |
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Since the determinant is linear in each row, then we continue where we left off: |
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:::::::<math>det(A) + det(B) =</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> |
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:::::::<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>. |
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Then <math>det(A) + det(B) = 0</math> and <math>det(A) = -det(B)</math>. The proof of the lemma is complete. |
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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>det(A) =</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}(-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)^{- 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>-det(B)</math>. |
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Then the proof of the theorem is complete. |
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==Nikita== |
==Nikita== |
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==Scanned Tutorial Notes by [[User Boyang.wu|Boyang.wu]]== |
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[[File:Tut.pdf]] |
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Latest revision as of 11:55, 9 December 2014
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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 rows [math]\displaystyle{ i, i + 1 }[/math] of [math]\displaystyle{ A }[/math] interchanged. 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\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]. The proof of the lemma is complete.
For the proof of the theorem, assume that [math]\displaystyle{ B }[/math] is the matrix [math]\displaystyle{ A }[/math] with rows [math]\displaystyle{ i, j }[/math] of [math]\displaystyle{ A }[/math] interchanged and [math]\displaystyle{ i \neq j }[/math]. By Lemma 1, we have the following:
- [math]\displaystyle{ det(A) = }[/math]
- [math]\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} = }[/math]
- [math]\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} = }[/math]
- [math]\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} = }[/math]
- [math]\displaystyle{ -det(B) }[/math].
Then the proof of the theorem is complete.
