Let $A$ be an $m \times n$ real matrix and $b \in \Bbb R^m$ with $b \neq 0$ . Then

Question

1. The set of all real solutions of $Ax=b$ is a vector space.

2. If $u$ and $v$ are two solutions of $Ax=b$, then $\lambda u+ (1-\lambda)v$ is also a solution of $Ax=b$ $\forall \lambda \in \Bbb R$.

3. For any two solutions $u$ and $v$ of $Ax=b$, the linear combination $\lambda u+(1-\lambda)v$ is also a solution of $Ax=b$ only when $0 \le \lambda \le 1$

4. If rank of $A$ is $n$, then $Ax=b$ has at most one solution.

Option 2 is correct because $A(\lambda u +(1-\lambda)v)=\lambda Au+ Av -\lambda Av=b$. Hence option 3 is false.

Option 1 seems to be incorrect because $x \neq 0$ by hypothesis.

How can I check whether option 4 is true or not?

Am I right so far?

Answer 1

What you have done is correct. For option $4$:

Since $\operatorname{Rank} A=n\implies \operatorname{Nullity} A=0$ by Rank-Nullity Theorem.

Hence $Ax=0$ has only solution as $x=0$.Hence if $Ax=b$ has a solution then it is unique.