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Let $A\in M_{m\times n}(\Bbb R)$ and let $b_0\in \mathbb R^m$. Suppose that the system of equations $Ax=b_0$ has a unique solution. Which of the following is true?

  1. $Ax=b$ has a solution for every $b \in \mathbb R^m$.

  2. if $Ax=b$ has a solution then it is unique.

  3. $Ax=0$ has a unique solution.

  4. $A$ has rank $m$.

My proceed: Let $A'$ be the augmented matrix of $A$. then the system of equation is consistent iff $r(A)=r(A')$. so the option $1$ is not always true. But I cannot understand how to proceed for other three options, please anyone help me. Thanks in advance.

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We consider the case $b_0=0$ first. So we can calssify true statements in this case.

Case 1 : $ b_0=0$ : Then kernel is trivial. (3) is true. And then $$ {\rm rank}(A) ={\rm dim.\ domain}(A)-{\rm dim.\ kernel}(A)=n$$

So if $n<m$ then $(4)$ is false. In further $A({\bf R}^n)$ is a proper subspace of dimension $n$ in ${\bf R}^m$. So $(1)$ is false.

If $Ax=b$ has two solutions $x_i$ then $$0=b-b= Ax_1-Ax_2= A(x_1-x_2) \Rightarrow x_1=x_2 $$ Contradiction. So $(2)$ is true.

So in this case $(2),\ (3)$ are true.

Case 2 : $ b_0\neq 0$ : If $Ax=0$ has nontrivial solution $x_0$, then $$ A(x+x_0)=Ax=b_0$$ Hence $x,\ x+x_0$ are solutions. Contradiction. So $Ax=0$ has unique solution. $(3)$ is true.

Assume that $Ax=b\neq 0$ have solutions $x_i$. Then $$ A(x_1-x_2)=b-b=0$$

So the equation $Ax=0$ has nontrivial solution. This is not true by the previous. So $(2)$ is true.

Hence by Case 1, 2, ultimately, $(2),\ (3)$ are true.

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The given condition $AX=b_{0}$ has a unique solution means kernel is trivial. So when the system $AX=b$ has a solution it is unique solution.So option 2 and 3 are true.

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