Suppose $\mathbf{Ax}=\mathbf b$ has no solution 
Suppose the matrix equation $\mathbf{Ax}=\mathbf b$ has no solution, where $\mathbf A$ is a 3×3 non-zero matrix of real numbers and $\mathbf b$ is a 3×1 vector of real numbers. Then
  a) the set of vectors $\mathbf x$ for which $\mathbf{Ax}=\mathbf 0$ is a plane
  b) the set of vectors $\mathbf x$ for which $\mathbf{Ax}=\mathbf 0$ is a line
  c) the rank of $\mathbf A$ is 3
  d) $\mathbf{Ax}=\mathbf 0$ has a non-zero solution

Since the matrix equation $\mathbf{Ax}=\mathbf 0$ has no solution, therefore, we can imply that $|\mathbf A|=0$. Therefore, option (c) is not the answer. I don't understand how to proceed after this. Can someone please explain?
 A: Because $\mathbf{Ax}=\mathbf b$ has no solution, the rank of $\mathbf A$ is less than 3. By the rank-nullity theorem, then, the nullity of $\mathbf A$ is at least one, so there must be a nonzero $\mathbf x$ with $\mathbf{Ax}=\mathbf 0$. (d) is correct.
(a) and (b) are not always true; they correspond to the rank of $\mathbf A$ being 1 and 2 respectively, when only the fact that it is less than 3 can be deduced from the given conditions.
A: Nice answer @Taxel.
Let me try to detail a bit: $\mathrm{rank} (A)  \le 2$.
Consider the row vectors in $A$:

*

*Let rank $A = 2$.
2 row vectors are independent. They span a plane. Vector $X$ is perpendicular to this plane to satisfy $AX = 0$.
(Matrix multiplication can be thought of taking the dot product of the 2 independent  row vectors of $A$ and column vector $X$.)


*let rank $A = 1$.
There is only one independent row vector in $A$. Can be thought of as a normal to a plane. The dot product of the  only independent row vector of A with the vector $X$ is $0$ , the vector $X$  is in this plane.
