The number of 3 × 3 matrices A whose entries are either 0 or 1 and for which the system A $\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}1\\0\\0\end{bmatrix}$ has exactly two distinct solutions, is

0

Let A $= \begin{bmatrix}a & b & c\\p & q & r\\l& m & n\end{bmatrix}$. Then,

$A\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}1\\0\\0\end{bmatrix}$

$⇒ ax + by + cz = 1, px + qy + rz = 0, lx + my + nz = 0 $

This gives three planes which cannot intersect at two distinct points. So, there cannot be any 3 × 3 matrix for which the given system of equations has two distinct solutions.