14-240/Tutorial-December 2

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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{ (-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]\displaystyle{ (-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]\displaystyle{ (-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]\displaystyle{ -det(B) }[/math].

Then the proof of the theorem is complete.