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:

$det(A)+det(B)=$

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

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

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 row $i$ of $A$ interchanged with row $j$ of $A$ and that $i\neq j$. 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.