Find A^−1 modulo 26.

Asked Oct 15 '17 at 05:31

Active Oct 15 '17 at 05:55

Viewed 656 times

|4 9 15 |

|15 17 6|

|24 0 17|

i know i will have to do something like this

|4 9 15 | * | 1 0 0 |

|15 17 6| * | 0 1 0 |

|24 0 17| * | 0 0 1 |

Also if the product of the multiplication equals more then 26 i will have to mod 26 it

THE Problem is 5C)

edited Oct 15 '17 at 05:55

asked Oct 15 '17 at 05:31

Saad Abdullah

Here I did one modulo 29. Modulo 26 you have the extra problem that you can only find modular inverses of integers coprime to 26. So you have to be a bit more selective and/or possibly do extra row operations to get pivot entries coprime to 26. Anyway, other than such trifles (=use a modular inverse of an entry rather than the usual inverse) the algorithm you learned in linear algebra still works. – Jyrki Lahtonen Oct 15 '17 at 05:43
But, I don't understand what the products in the second group mean. May be you meant to have the augmented matrix there? You have taken linear algebra, haven't you? – Jyrki Lahtonen Oct 15 '17 at 05:47
Hey i posted pics giving links – Saad Abdullah Oct 15 '17 at 05:54
1

Looks like an ok beginning to me (+1). Why did you stop there? You can easily get a $25$ in the second column. Remember that modulo $25\equiv-1\pmod{26}$ so you get a $-1$. And $-1$ is pretty much as good as $1$ for the purposes of row operations (you can later multiply a row by $-1$ because $-1$ has no common factors with $26$). You are working modulo $26$ so you can add/subtract ANY integer multiple of $26$. – Jyrki Lahtonen Oct 15 '17 at 06:07
Wow thanks for the info – Saad Abdullah Oct 15 '17 at 13:35

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