Square Matrix with Duplicate Rows has Zero Determinant/Proof 1

Theorem
If two rows of a square matrix over a commutative ring $\left({R, +, \circ}\right)$ are the same, then its determinant is zero.

Proof
Proof by induction over $n$, the order of the square matrix.

Basis for the Induction
Let $n=2$, which is the smallest natural number for which a square matrix of order $n$ can have two identical rows.

Let $\mathbf A = \left[{a}\right]_2$ be a square matrix over $R$ with two identical rows.

Then, by definition of determinant:


 * $\det \left({\mathbf A}\right) = a_{11}a_{22} - a_{12}a_{21} = a_{11}a_{22}-a_{22}a_{11} = 0$

Induction Hypothesis
Assume for $n \in \N_{\ge 2}$ that any square matrices of order $n$ over $R$ with two identical rows has determinant equal to $0$.

Induction Step
Let $\mathbf A$ be a square matrix of order $n+1$ over $R$ with two identical rows.

Let $i_1, i_2 \in \left\{ {1, \ldots, n+1}\right\}$ be the indices of the identical rows, and let $i_1 < i_2$.

Let $i \in \left\{ {1, \ldots, n+1}\right\}$.

Let $\mathbf A \left({i ; 1}\right)$ denote the submatrix obtained from $\mathbf A$ by removing row $i$ and column $1$.

If $i \ne i_1$ and $i \ne i_2$, then $\mathbf A \left({i ; 1}\right)$ still contains two identical rows.

By the induction hypotesis, it follows that $\det \left({\mathbf A \left({i ; 1}\right) }\right) = 0$.

Now consider the matrices $\mathbf A \left({i_1 ; 1}\right)$ and $\mathbf A \left({i_2 ; 1}\right)$.

Let $r_j$ denote row $j$ of $\mathbf A \left({i_1 ; 1}\right)$.

If we perform the following sequence of $i_2 - i_1 -1$ elementary row operations on $\mathbf A \left({i_1 ; 1}\right)$:


 * $r_{i_1} \leftrightarrow r_{i_1 +1} \ ;  \ r_{i_1 + 1} \leftrightarrow r_{i_1 +2}  \ ;  \ \ldots  \  ;  \ r_{i_2 - 1} \leftrightarrow r_{i_2}$

we will transform $\mathbf A \left({i_1 ; 1}\right)$ into $\mathbf A \left({i_2 ; 1}\right)$.

From Determinant with Rows Transposed, it follows that $\det \left({\mathbf A \left({i_1 ; 1}\right) }\right) = \left({-1}\right)^{i_2 - i_1 - 1} \det \left({\mathbf A \left({i_2 ; 1}\right) }\right)$

Then: