Number of Binary Matrices with Exactly Two Solutions for Linear System

To determine the number of $3 \times 3$ matrices $A$ with entries in $\{0, 1\}$ for which the linear system

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

has exactly two distinct solutions, consider the general theory of systems of linear equations over the field of real numbers $\mathbb{R}$.

Let the system be denoted by:

$$AX = B, \quad \text{where } X = \begin{bmatrix} x \\ y \\ z \end{bmatrix} \in \mathbb{R}^3 \quad \text{and} \quad B = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}$$

Suppose there exist two distinct solutions, $X_1$ and $X_2$, such that:
$A X_1 = B \quad \text{and} \quad A X_2 = B, \quad \text{with } X_1 \neq X_2$

Now, consider a linear combination of these two solutions defined for any scalar $\lambda \in \mathbb{R}$:
$X_\lambda = \lambda X_1 + (1 - \lambda) X_2$

Applying the matrix $A$ to $X_\lambda$:
$A X_\lambda = A\big(\lambda X_1 + (1 - \lambda) X_2\big) = \lambda (A X_1) + (1 - \lambda)(A X_2)$

Since $A X_1 = B$ and $A X_2 = B$, we have:
$A X_\lambda = \lambda B + (1 - \lambda) B = B$

Thus, for every $\lambda \in \mathbb{R}$, the vector $X_\lambda$ is also a solution to the system $AX = B$.

Since $X_1 \neq X_2$, each distinct choice of $\lambda$ yields a distinct vector $X_\lambda$. Because the field of scalars is $\mathbb{R}$, which is infinite, the system must have infinitely many solutions if it has more than one solution.

Hence, a system of linear equations over $\mathbb{R}$ can have:

1. No solution (inconsistent system)
2. Exactly one unique solution
3. Infinitely many solutions

It is impossible for such a system to have exactly two distinct solutions.

Therefore, no such matrix $A$ exists, and the number of such matrices is $0$.

Correct Option: A