Originally Posted by

**demode** Thanks, that was a typo. So, we must have that $\displaystyle 1- (p-1)(a \bmod\ p) \neq 0 \iff (p-1)(a \bmod\ p) \neq 1$. So I guess $\displaystyle (p-1)$ must be equal to $\displaystyle -1$ in $\displaystyle \mathbb{Z}_p$. Therefore $\displaystyle p-1 = -1 \neq a \bmod\ p$.

But how do I prove that (p-1)=-1?

!!!... As $\displaystyle p=0\!\!\pmod p$ , obviously $\displaystyle p-1=0-1=-1\!\!\pmod p$ ...!

Tonio

And for part **(b)** I think they are asking for the inverse of

$\displaystyle M_{10,p} = \left|\begin{array}{cc}1 & (10 \bmod\ p) \\ p-1 & 1\end{array}\right|$

Then the inverse would be of the form:

$\displaystyle \frac{1}{det(M_{10,p})} \left|\begin{array}{cc}1 & -(10 \bmod\ p) \\ 1-p & 1\end{array}\right|$

But what values do I use for $\displaystyle det(M_{10,p})$ and 10 mod p?