14-240/Tutorial-December 2

DBN: Classes: 2014-15: 14-240 / Navigation Welcome to Math 240! (additions to this web site no longer count towards good deed points) # Week of... Notes and Links 1 Sep 8 About This Class, What is this class about? (PDF, HTML), Monday, Wednesday 2 Sep 15 HW1, Monday, Wednesday, TheComplexField.pdf,HW1_solutions.pdf 3 Sep 22 HW2, Class Photo, Monday, Wednesday, HW2_solutions.pdf 4 Sep 29 HW3, Wednesday, Tutorial, HW3_solutions.pdf 5 Oct 6 HW4, Monday, Wednesday, Tutorial, HW4_solutions.pdf 6 Oct 13 No Monday class (Thanksgiving), Wednesday, Tutorial 7 Oct 20 HW5, Term Test at tutorials on Tuesday, Wednesday 8 Oct 27 HW6, Monday, Why LinAlg?, Wednesday, Tutorial 9 Nov 3 Monday is the last day to drop this class, HW7, Monday, Wednesday, Tutorial 10 Nov 10 HW8, Monday, Tutorial 11 Nov 17 Monday-Tuesday is UofT November break 12 Nov 24 HW9 13 Dec 1 Wednesday is a "makeup Monday"! End-of-Course Schedule, Tutorial F Dec 8 The Final Exam Register of Good Deeds Add your name / see who's in!

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:

$det(A)=$

$(-1)det{\begin{pmatrix}...\\A_{i+1}\\A_{i}\\...\\A_{j}\\...\end{pmatrix}}=det{\begin{pmatrix}...\\A_{i}\\A_{i+1}\\...\\A_{j}\\...\end{pmatrix}}=(-1)det{\begin{pmatrix}...\\A_{i+1}\\A_{i}\\...\\A_{j}\\...\end{pmatrix}}=$

$(-1)^{j-i}det{\begin{pmatrix}...\\A_{i+1}\\...\\A_{j}\\A_{i}\\...\end{pmatrix}}=(-1)^{j-i}(-1)^{j-i-1}det{\begin{pmatrix}...\\A_{j}\\A_{i+1}\\...\\A_{i}\\...\end{pmatrix}}=$

$(-1)^{2(j-i)-1}det{\begin{pmatrix}...\\A_{j}\\A_{i+1}\\...\\A_{i}\\...\end{pmatrix}}=(-1)^{-1}det{\begin{pmatrix}...\\A_{j}\\A_{i+1}\\...\\A_{i}\\...\end{pmatrix}}=(-1)det{\begin{pmatrix}...\\A_{j}\\A_{i+1}\\...\\A_{i}\\...\end{pmatrix}}=$

$-det(B)$.

Then the proof of the theorem is complete.

Scanned Lecture Notes by Boyang.wu

File:Tut.pdf