Question

Consider the linear system \( A x = b \), where \(A\) is an \(m \times n\) matrix, \(x\) is an \(n \times 1\) vector of unknowns and \(b\) is an \(m \times 1\) vector. Further, suppose there exists an \(m \times 1\) vector \(c\) such that the linear system \(A x = c\) has NO solution. Then, which of the following statements is/are necessarily TRUE?

• A. If \(m \le n\) and \(d\) is the first column of \(A\), then the linear system \(A x = d\) has a unique solution
• B. If \(m \ge n\), then \(\text{Rank}(A)<n\)
• C. \(\text{Rank}(A)<m\)
• D. If \(m>n\), then the linear system \(A x = 0\) has a solution other than \(x = 0\)

Hint:

If \(A x = c\) has no solution for some \(c\), it means \(c\) lies outside the column space of \(A\), implying the rank of \(A\) is less than the number of rows \(m\).

Answer 1

The Correct Option is C

Step 1: Analyze the given condition.
The statement says that there exists a vector \(c\) such that \(A x = c\) has no solution. This means that \(c\) does not belong to the column space (range) of \(A\).
Step 2: Interpret the implication.
Since not all vectors \(c \in \mathbb{R}^m\) can be represented as \(A x\), the column space of \(A\) is a proper subspace of \(\mathbb{R}^m\). Hence, \[ \text{Rank}(A)<m. \]
Step 3: Check other options.
(A) There is no reason that \(A x = d\) must have a unique solution since uniqueness requires full column rank (\(\text{Rank}(A) = n\)), which is not given.
(B) The case \(\text{Rank}(A)<n\) is not necessarily true; \(A\) could still have full column rank with \(\text{Rank}(A) = n<m\).
(D) Homogeneous system \(A x = 0\) always has \(x = 0\) as a solution, but having a nontrivial solution requires \(\text{Rank}(A)<n\), which is not guaranteed.
Thus, only (C) is necessarily true. Final Answer: \[ \boxed{\text{Rank}(A)<m} \]