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A full rook matrix is a permutation matrix, ie it shows an injective homomorphism from 𝔖n in GLn(𝔽) where 𝔽 is an arbitrary field.
a) To prove the ponderable by hand, let’s calculate det(M) where M is a full rook matrix. Imagine that only the i-th element of the first line is 1, the others are 0. If we develop the first line of det(M), det(M)=(−1)i+1×det(Mn−1), where Mn−1 is the submatrix extracted from M without the first line and the i-th column (both of them are filled with 0). Mn−1 is also a full rook matrix, and with an immediate recurrence, we can see that det(M)=±1; M is invertible.
b) Now, if M and N are rook matrices, imagine that we compute the first line of the M×N product. There is at most one 1 in the first line of M and in each column of N, therefore, and because there is at most one 1 in each line of N, there can not be more than one 1 in the first line of M×N. So, M×N is also a rook matrix.