Determining whether solution exists or not.Under a given condition

Question

Following Question has one or more correct answers.

Let $A ∈ \mathbb{M}_{m×n}(\mathbb{R})$ and let $b_0 \in R^m$

Suppose the system of equations $Ax = b_0$ has a unique solution. Which of the following statement(s) is/are true?

(A) $Ax = b$ has a solution for every $b \in R^m$.

(B) If $Ax = b$ has a solution then it is unique.

(C) $Ax = 0$ has a unique solution.

(D) $A$ has rank $m$.

What I Tried?

I will check every option one by one.

A)- Consider the following example

$7x+3y=10$

$4x+8y=12$

$14x+6y=20$

Here A=\begin{bmatrix} 7 & 3 \\ 4 & 8 \\ 14 & 6 \end{bmatrix}

and $b_0=$\begin{bmatrix} 10\\ 12 \\ 20 \end{bmatrix}

Here a unique solution exists.However what if I take $b=$\begin{bmatrix} 10\\ 12 \\ 19 \end{bmatrix} because third equation is the double of first one and the right hand side of this equation contradicts.

Hence A) is not always true.

B)- Dont Know

C)- Dont Know

D)- Same Example as given above.In which A=\begin{bmatrix} 7 & 3 \\ 4 & 8 \\ 14 & 6 \end{bmatrix} has rank $2 \neq 3=m$

Help me in getting the answer of B)- and C)- Please provide some linear algebra approach to solve A and D.

Answer 1

Hint:

If Ker($A$) is not the trivial subspace, then the equation $Ax=b$ can never have a unique solution.