This problem requires us to determine the nature of solutions for a system of linear equations $(A - 2I)X = B$. The core concept lies in analyzing the determinant and rank of the coefficient matrix $(A - 2I)$.

Key Concepts for Systems of Linear Equations $CX = B$:

For a system of linear equations $CX = B$, where $C$ is a square matrix of order $n$:

1. Unique Solution: If $\det(C) \neq 0$, then the matrix $C$ is invertible, and a unique solution $X = C^{-1}B$ exists.
2. No Solution or Infinitely Many Solutions: If $\det(C) = 0$, then the matrix $C$ is singular. In this case, we need to check the consistency of the system using the concept of rank:
   • If $\text{rank}(C) = \text{rank}([C|B]) < n$ (where $[C|B]$ is the augmented matrix), the system is consistent and has infinitely many solutions. This occurs when the right-hand side vector $B$ can be expressed as a linear combination of the column vectors of $C$, and there are free variables.
   • If $\text{rank}(C) \neq \text{rank}([C|B])$, the system is inconsistent and has no solution. This occurs when the right-hand side vector $B$ cannot be expressed as a linear combination of the column vectors of $C$.

Our strategy will be:

1. First, determine the matrix $A$ using the given conditions.
2. Then, calculate the coefficient matrix $C = (A - 2I)$.
3. Calculate $\det(C)$.
4. If $\det(C) = 0$, we will analyze the system $(A - 2I)X = B$ directly to check for consistency and determine if there are no solutions or infinitely many solutions.

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Step 1: Determining the Matrix A

Let the 3x3 matrix $A$ be represented as: $A = \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix}$

We are given three conditions involving matrix $A$ and specific column vectors:

Condition 1: $A \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}$ Performing the matrix multiplication: $\begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix} \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} = \begin{pmatrix} a(1) + b(1) + c(0) \\ d(1) + e(1) + f(0) \\ g(1) + h(1) + i(0) \end{pmatrix} = \begin{pmatrix} a+b \\ d+e \\ g+h \end{pmatrix}$ Equating this to the given result: $\begin{pmatrix} a+b \\ d+e \\ g+h \end{pmatrix} = \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} \Rightarrow \quad \begin{cases} a+b = 1 & \text{(Eq. 1.1)} \\ d+e = 1 & \text{(Eq. 1.2)} \\ g+h = 0 & \text{(Eq. 1.3)} \end{cases}$

Condition 2: $A \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix} = \begin{pmatrix} -1 \\ 0 \\ 1 \end{pmatrix}$ Performing the matrix multiplication: $\begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix} \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix} = \begin{pmatrix} a(1) + b(0) + c(1) \\ d(1) + e(0) + f(1) \\ g(1) + h(0) + i(1) \end{pmatrix} = \begin{pmatrix} a+c \\ d+f \\ g+i \end{pmatrix}$ Equating this to the given result: