The question is such that, $A$ is an $m\times n$ matrix.
$x$ is an $n$ vector, $b$ is a $m$ vector. We want the condition that ensures the existence of a solution for $$Ax=b$$
The options had,
- $\det(A) \neq 0 $
- the columns of $A$ are linearly independent
- the rows of $A$ are linearly independent
So far from what I have studied from Anton's Linear Algebra, all these 3 options are equivalent and seem correct. I cant seem to figure out difference between them and also of course the answer for this question. I see that we can reject the option $1.$ on the basis that $A$ is not a square matrix and thus its determinant does not exist. But I can't figure out anything beyond this.
They are equivalent for square matrices.
For non-square matrices #1 makes no sense and #2 and #3 are different (in fact it is impossible for #2 and #3 to both be true for a non-square matrix).
Regarding the correct condition for existence of a solution to $Ax=b$ for any $b$, this requires that row operations will not produce a row of zeros in the un-augmented $A$. Otherwise, if $b$ is chosen appropriately, the augmented $A$ will reduce to have a row of the form $0,0,\dots,0,c$ where $c \neq 0$, which encodes the inconsistent equation $0=c$. This means the rows of $A$ must be linearly independent.
One can test the rows of $A$ for linear independence by doing elimination; one can test the columns for linear independence by doing elimination to $A^T$. Of course you really only have to do one or the other, considering the properties of the rank of a matrix.
Note that on the reverse side, the condition for uniqueness of a solution if one exists is that the columns are linearly independent. This is actually quite intuitive: given $x \neq y$ such that $Ax=b,Ay=b$ you have $A(x-y)=0$, and now you have a nontrivial linear combination of columns of $A$ adding up to zero.