Square Matrix with Duplicate Rows has Zero Determinant/Proof 1

Proof
The proof proceeds 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 $\mathbf A = \sqbrk a_2$ be a square matrix over $R$ with two identical rows.

Then, by definition of determinant:


 * $\map \det {\mathbf A} = a_{1 1} a_{2 2} - a_{1 2} a_{2 1} = a_{1 1} a_{2 2} - a_{2 2} a_{1 1} = 0$

This is the basis for the induction.

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

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

Let $\map {\mathbf A} {i; 1}$ 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 $\map {\mathbf A} {i; 1}$ still contains two identical rows.

By the induction hypothesis:
 * $\map \det {\map {\mathbf A} {i; 1} } = 0$

Now consider the matrices $\map {\mathbf A} {i_1; 1}$ and $\map {\mathbf A} {i_2; 1}$.

Let $r_j$ denote row $j$ of $\map {\mathbf A} {i_1; 1}$.

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


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

we will transform $\map {\mathbf A} {i_1; 1}$ into $\map {\mathbf A} {i_2; 1}$.

From Determinant with Rows Transposed:


 * $\map \det {\map {\mathbf A} {i_1; 1} } = \paren {-1}^{i_2 - i_1 - 1} \map \det {\map {\mathbf A} {i_2; 1} }$

Then: