Q:

Finding non-trivial fixed points of a matrix

Based on my notes I know the following if matrix A has a nontrivial fixed point, then $det(I-A) = 0$.   
Suppose I have a matrix A such that Row 1 is [1 0] and Row 2 is [1 0] (apologies for the notation).   
I know this matrix has a non trivial fixed point based on the calculation of $det(I-A)$ being equal to 0.   
But, how do I the find the fixed point(s)?    
Recall:
Solutions to the matrix equation $Ax = x$, if any, are called fixed points of A.

A:

The things you called fixed points are called eigenvectors with eigenvalue $1$.
In this particular case do note that if $x=\begin{pmatrix} x_1 \\ x_2 \end{pmatrix}$ then $Ax=\begin{pmatrix} 1 & 0 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} =\begin{pmatrix} x_1 \\ x_1 \end{pmatrix}$ so the vectors of the form $\begin{pmatrix} x_1 \\ x_1 \end{pmatrix}$ are "fixed points" in this particular case. 
In the general case to find the fixed points you want to find a basis of $Ker(I-A)$ and the span of that basis is the set of "fixed points".