# 14-240/Tutorial-December 2

(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

## Boris

#### Theorem

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

##### Lemma 1

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

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

${\displaystyle det(A)+det(B)=}$

${\displaystyle det{\begin{pmatrix}...\\A_{i}\\A_{i+1}\\...\end{pmatrix}}+det{\begin{pmatrix}...\\A_{i+1}\\A_{i}\\...\end{pmatrix}}=}$

${\displaystyle det{\begin{pmatrix}...\\A_{i}\\A_{i}\\...\end{pmatrix}}+det{\begin{pmatrix}...\\A_{i}\\A_{i+1}\\...\end{pmatrix}}+det{\begin{pmatrix}...\\A_{i+1}\\A_{i+1}\\...\end{pmatrix}}+det{\begin{pmatrix}...\\A_{i+1}\\A_{i}\\...\end{pmatrix}}}$.

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

${\displaystyle det(A)+det(B)=}$

${\displaystyle det{\begin{pmatrix}...\\A_{i}\\A_{i}\\...\end{pmatrix}}+det{\begin{pmatrix}...\\A_{i}\\A_{i+1}\\...\end{pmatrix}}+det{\begin{pmatrix}...\\A_{i+1}\\A_{i+1}\\...\end{pmatrix}}+det{\begin{pmatrix}...\\A_{i+1}\\A_{i}\\...\end{pmatrix}}=}$

${\displaystyle det{\begin{pmatrix}...\\A_{i}\\A_{i}+A_{i+1}\\...\end{pmatrix}}+det{\begin{pmatrix}...\\A_{i+1}\\A_{i}+A_{i+1}\\...\end{pmatrix}}=det{\begin{pmatrix}...\\A_{i}+A_{i+1}\\A_{i}+A_{i+1}\\...\end{pmatrix}}=0}$.

Then ${\displaystyle det(A)+det(B)=0}$ and ${\displaystyle det(A)=-det(B)}$. The proof of the lemma is complete.

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

${\displaystyle det(A)=}$

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

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

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

${\displaystyle -det(B)}$.

Then the proof of the theorem is complete.