# 14-240/Tutorial-December 2

## 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.