Square Matrix with Duplicate Rows has Zero Determinant/Proof 2

Theorem

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

Proof

Suppose that $\forall x \in R: x + x = 0 \implies x = 0$.

From Determinant with Rows Transposed, if you exchange two rows of a square matrix, the sign of its determinant changes.

If you exchange two identical rows of a square matrix, then the sign of its determinant changes from $D$, say, to $-D$.

But the matrix stays the same.

So $D = -D$ and so $D = 0$.

$\blacksquare$

Sources

• 1998: Richard Kaye and Robert Wilson: Linear Algebra ... (previous) ... (next): Part $\text I$: Matrices and vector spaces: $1$ Matrices: $1.6$ Determinant and trace: Proposition $1.9$