Full Rook Matrix is Invertible

Theorem

A full rook matrix is invertible.

Proof

Let $\mathbf A$ be a full rook matrix.

By definition, $\mathbf A$ is an instance of a permutation matrix.

By Determinant of Permutation Matrix, it follows that $\det \mathbf A = \pm 1$.

By Matrix is Invertible iff Determinant has Multiplicative Inverse:

$\mathbf A$ is invertible.

$\blacksquare$

Sources

• 2008: David Joyner: Adventures in Group Theory (2nd ed.) ... (previous) ... (next): Chapter $2$: 'And you do addition?': $\S 2.2$: Functions on vectors: $\S 2.2.4$: Multiplication and inverses: Ponderable $2.2.2 \ \text{(a)}$