Probability that vector $x$ is orthogonal to the row of a non-singular $n \times n$ matrix Let $A$ be an $n \times n$ non-singular matrix with entries in $GF_q$, the finite field containing $q=p^a$ elements, where $p$ is prime and $a\geq 1$. Let $x=(x_1,\ldots, x_n)$ be a vector with non-zero entries in $GF_q$, where the entries are chosen randomly and independently according to the uniform distribution. 
I am trying to understand the following statement in a proof that I am reading:
$\textit{For every fixed row of $A$, the probability that $x$ is orthogonal to the row is at most $\frac{(q-1)^{n-1}}{(q-1)^n}=\frac{1}{q-1}$.}$
I understand that the possible number of vectors like $x$ is $(q-1)^n$, since there are $q-1$ possibilities for each entry. I do not understand why the denominator is $(q-1)^{n-1}$. 
Also, I understand that the expected number of zero entries in $Ax$ is at most $\frac{n}{q-1}$ for $q\geq n+2$. However, why is it that when $q=n+1$ and the expected number of zero entries in $Ax$ is precisely $1$ that the probability that $x$ is orthogonal to every row of $A$ is $\frac{1}{q-1}$?
Thank you for your help.
 A: Assume that the row of $A$ in question is $\vec{a}=(a_1,a_2,\ldots,a_n)$. Because $A$ is non-singular we have $a_i\neq0$ for at least one of the indices $i$. Without loss of generality we can assume that $a_1\neq0$.
Consider the orthogonality relation 
$$
a_1x_1+a_2x_2+\cdots+a_nx_n=0.\qquad(*)
$$
We can solve for $x_1$ from this
$$
x_1=-\frac1{a_1}(a_2x_2+a_3x_3+\cdots+a_nx_n).
$$
So orthogonality is equivalent to the first component $x_1$ having this precise value $x_1=z_0$ where
$$
z_0=-\frac1{a_1}(a_2x_2+a_3x_3+\cdots+a_nx_n).
$$
There are two cases. The first case, call it $A$, is that the random variables $x_2,x_3,\ldots,x_n$ happened to be chosen in such a way that $z_0=0$, then $(*)$ never holds because $x_1\neq0=z_0$. In this case the probability of orthogonality is equal to zero.
The complementary case $B$ is that $z_0\neq0$. As above $(*)$ holds if and only if $x_1=z_0$. This happens with probability $1/(q-1)$.
So the probability that $(*)$ holds is equal to
$$
p=0\cdot P(A)+\frac1{q-1} P(B)=\frac{P(B)}{q-1}\le \frac1{q-1}.
$$
