# Solve linear congruence

• Aug 21st 2010, 05:06 PM
calfever
Solve linear congruence
• Aug 22nd 2010, 04:19 AM
HallsofIvy
That is, of course, the same as
$\displaystyle \begin{bmatrix}5- 1 & 13 \\ 3 & 18- 1\end{bmatrix}\begin{bmatrix}a \\ b\end{bmatrix}$$\displaystyle = \begin{bmatrix}4 & 13 \\ 3 & 17\end{bmatrix}\begin{bmatrix}a \\ b\end{bmatrix}= \begin{bmatrix}0 \\ 0\end{bmatrix}$.

But now the determinant is 3 modulo 26 and 3 is not a "0 divisor" modulo 26 the only solution is the trivial solution, a= b= 0.

More specifically, we can write that as the two equations 4a+ 13b= 0 and 3a+ 17b= 0. If we multiply the first equation by 3 we have 12a+ 13b= 0. If we multiply the second equation by -4 we have -12a- 22b= 0. Adding the two equations, 17b= 0 and, again, since 17 is not a "0 divisor" modulo 26, the only solution is a= b= 0.
• Aug 22nd 2010, 04:53 AM
calfever
ok. i understand.

thanks a lot.
• Aug 23rd 2010, 03:50 AM
HallsofIvy
Do you see why '3 is not a "0 divisor" modulo 26' and '17 is not a "0 divisor" modulo 26' are important?