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To solve this problem, we are given that \( A \) is a \( 2 \times 3 \) matrix, and the system of equations is:

\[
A \cdot \begin{pmatrix} x \\ y \\ z \end{pmatrix} = A \cdot \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}
\]

The question asks why this system has infinite solutions.

### Key observations:

1. **Matrix \( A \)'s Dimensions**: Since \( A \) is a \( 2 \times 3 \) matrix, it maps a 3-dimensional vector space to a 2-dimensional vector space. This implies that \( A \) cannot be full-rank, meaning it will have a non-trivial null space.

2. **Null Space of Matrix \( A \)**: The matrix \( A \), being a non-square matrix with more variables (3) than equations (2), implies that there are likely free variables. In particular, the null space of \( A \) will contain at least one non-zero vector, meaning that there are vectors \( v \neq 0 \) such that \( A \cdot v = 0 \).

3. **System Setup**: The system

   \[
   A \cdot \begin{pmatrix} x \\ y \\ z \end{pmatrix} = A \cdot \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}
   \]
   
   implies that the two vectors \( \begin{pmatrix} x \\ y \\ z \end{pmatrix} \) and \( \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} \) are mapped to the same vector in the 2-dimensional space (via the transformation matrix \( A \)).

4. **Conclusion**: Since the null space of \( A \) is non-trivial (i.e., there exists a vector \( v \neq 0 \) such that \( A \cdot v = 0 \)), the solution to this system can be written as:

   \[
   \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} + \alpha \cdot v
   \]
   
   for some scalar \( \alpha \in \mathbb{R} \), where \( v \) is a vector in the null space of \( A \). This shows that there are infinitely many solutions, since \( \alpha \) can take any real value.

### Summary:
The system has infinitely many solutions because the matrix \( A \), being a \( 2 \times 3 \) matrix, has a non-trivial null space. Therefore, any solution to the system can be written as the sum of a particular solution and any scalar multiple of a vector in the null space of \( A \), resulting in infinitely many solutions.

Would you like more details on this explanation?

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### Related Questions:
1. What is the rank-nullity theorem, and how does it apply to this situation?
2. How can we find the null space of a matrix?
3. Can a \( 2 \times 3 \) matrix ever have a unique solution to a system like this?
4. What happens if we change the dimensions of the matrix \( A \) (e.g., to \( 3 \times 3 \))?
5. How does this concept extend to systems with more or fewer equations than unknowns?

### Tip:
In systems of linear equations where the number of unknowns exceeds the number of equations, it is likely that the system will either have no solution or infinitely many solutions, depending on whether the system is consistent.

Si A es una matriz de 2 × 3, entonces el sistema de ecuaciones A·(x, y, z)^T = A·(1, 2, 3)^T tiene infinitas soluciones. ¿Por qué?

Descubre por qué un sistema de ecuaciones con una matriz de 2x3 tiene infinitas soluciones. Este problema explora conceptos clave de álgebra lineal, como los espacios nulos y las transformaciones de matrices.

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