# Hill 2-CipherWatch

Thread starter 4 years ago
#1
When using row reduction to obtain the inverse matrix on the left and side of the augmented coded and plain text matrices, and I want 1's across the leading diagonal to complete the inverse. Why is it that I can't just divide the whole row to get a 1, instead I have to multiply the row by the inverse of the matrix entry that I have to get a one in the Modulo that I am working with? I can do the whole encoding/decoding thing, I just don't quite understand why I have to multiply by the inverse, instead of just dividing the whole row by the entry value that I have.
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4 years ago
#2
(Original post by daveollie)
When using row reduction to obtain the inverse matrix on the left and side of the augmented coded and plain text matrices, and I want 1's across the leading diagonal to complete the inverse. Why is it that I can't just divide the whole row to get a 1, instead I have to multiply the row by the inverse of the matrix entry that I have to get a one in the Modulo that I am working with? I can do the whole encoding/decoding thing, I just don't quite understand why I have to multiply by the inverse, instead of just dividing the whole row by the entry value that I have.
I don't think there's any reason for not using division when working over a finite field, other than ease of computation.

Finding the inverse of the first element, and performing n-1 multiplications is a damn sight easier, than performing n-1 divisions mod(whatever).
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4 years ago
#3
(Original post by daveollie)
When using row reduction to obtain the inverse matrix on the left and side of the augmented coded and plain text matrices, and I want 1's across the leading diagonal to complete the inverse. Why is it that I can't just divide the whole row to get a 1, instead I have to multiply the row by the inverse of the matrix entry that I have to get a one in the Modulo that I am working with? I can do the whole encoding/decoding thing, I just don't quite understand why I have to multiply by the inverse, instead of just dividing the whole row by the entry value that I have.
Note that with modular arithmetic, the literal "dividing by integers" won't work because you'd end up with fractions, whereas in modular arithmetic the answer is always going to be a whole number. For example, with 3(mod 11), if you tried to divide by 5 literally you would get (3/5)(mod 11).

Hence why you multiply by the inverse -- e.g. for the above example, dividing by 5 is the same as multiplying by 9, so the proper answer should be (3 x 9)(mod 11), which is 5(mod 11).
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