Step 1: Introduction
This problem concerns basic feasible solutions (BFS) in linear algebra and linear programming. A basic solution to \(AX=b\) is found by setting some variables to zero and solving for the rest.
Typo Note: If A is \(m \times n\) and X is \(n \times 1\), then b must be \(m \times 1\). The question's \(n \times 1\) for b is a typo. We assume b is \(m \times 1\).
Step 2: Definitions
- The system has \(m\) equations and \(n\) variables.
- The rank of A is \(m\), implying \(m \le n\) and linearly independent equations.
- A basic solution:
1. Choose \(m\) variables as "basic variables".
2. Set the other \(n-m\) variables ("non-basic variables") to zero.
3. Solve the \(m \times m\) system for the basic variables.
- Valid selection of \(m\) basic variables requires the corresponding columns of A to be linearly independent, forming an invertible \(m \times m\) submatrix.
- A basic feasible solution (BFS) is a basic solution where all variables (basic ones) are non-negative.
Step 3: Explanation
The question asks for the maximum number of basic feasible solutions. The number of BFS depends on A and b's values. However, the theoretical maximum is limited by the total number of basic solutions.
A basic solution depends on which \(m\) variables are basic. Since rank(A) is \(m\), A has at least one set of \(m\) linearly independent columns. The maximum number of basic solutions is the number of ways to choose \(m\) columns from \(n\).
This is given by the binomial coefficient:
\[ \text{Maximum number of basic solutions} = \binom{n}{m} = {}^nC_m \]Since every BFS is a basic solution, the number of BFS cannot exceed the total number of basic solutions. Therefore, the maximum possible number of basic feasible solutions is \( {}^nC_m \).
Step 4: Answer
The maximum number of basic feasible solutions is \( {}^nC_m \).