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 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: |
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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} = 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: |
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<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} = 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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==Nikita== |
==Nikita== |
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Revision as of 15:08, 7 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 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) = det\begin{pmatrix}...\\A_i\\A_{i + 1}\\...\end{pmatrix} + det\begin{pmatrix}...\\A_{i + 1}\\A_i\\...\end{pmatrix} = 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:
[math]\displaystyle{ 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} = 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].
