14-240/Tutorial-December 2

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Boris

Theorem

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

Lemma 1

Let A be a n \times n matrix and B be the matrix A with two adjacent rows interchanged. Then det(A) = -det(B).


All we need to show is that det(A) + det(B) = 0. Assume that B is the matrix A with rows i, i + 1 of A interchanged. Since the determinant of a matrix with two identical rows is 0, then:


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}.


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


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.


Then det(A) + det(B) = 0 and det(A) = -det(B). The proof of the lemma is complete.


For the proof of the theorem, assume that B is the matrix A with rows i, j of A interchanged and i \neq j. By Lemma 1, we have the following:


det(A) =


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} =


(-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} =


(-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} =


-det(B).


Then the proof of the theorem is complete.

Nikita

Scanned Tutorial Notes by Boyang.wu

File:Tut.pdf