Need help in solving a linear algebra ( System of Equations) quiz problem

Question

I am solving previous year quiz problem of my class and I am unable to solve this question in linear algebra.

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

1. $Ax = b$ has a solution for every $b \in \Bbb{R}^m$.
2. If $Ax = b$ has a solution then it is unique.
3. $Ax = 0$ has a unique solution.
4. $A$ has rank $m$.

( Multiple answers can be correct)

Answer is :

2, 3

I know the theorem that if $A$ is invertible then $Ax=0$ has only trivial solution and $A$ has rank $m$ and then $Ax=b$ has a solution for every $b \in \mathbb{R}^m$. But the problem is, how I can be sure if $A$ is invertible?

Kindly help.

Answer 1

Let $x_0$ be the unique solution of the equation $Ax = b_0$.

1)) For instance, consider Michael Hoppe’s example. Let $m=2$, $n=1$, $A=b_0=\begin{pmatrix} 1 \\ 0\end{pmatrix}$. It is easy to check that $x_0=1$. Since $A\Bbb R^n=\left\{\begin{pmatrix} \lambda \\ 0\end{pmatrix}:\lambda\in\Bbb R \right\}$, an equation $Ax=\begin{pmatrix} 1 \\ 1\end{pmatrix}$ has no solutions.

2)) Yes. Indeed, if $b\in\Bbb R^m$, $Ax=b$, and $Ax’=b$ then $A(x-x’+x_0)=Ax-Ax’+Ax_0=b-b+b_0$, so $x-x’+x_0=x_0$ and thus $x=x’$.

3)) We assume that $0$ from the question is a vector in $\Bbb R^m$. Then the answer is positive. Indeed, we have $A0=0$, and by (2), $0$ is a unique solution of the equation.

4)) The matrix $A$ from (1) has rank $1<2=m$.

Answer 2

Recall that the system $Ax=b$ is solvable iff $b$ is an element of the span of the column vectors of $A$. Now if this solution is unique the column vectors of $A$ are linearly independent. Does this clear up things?