Is not full rank matrix invertible?  Problem 
$A$ is a $4 \times 4$ matrix. It is known that $\text{rank}(A)=3$. Is matrix A invertible ?
 Attempt to solve 
$\text{rank(A)}=3 \implies \det(A)=0$
which implies matrix is $\textbf{not}$ invertible. One dimension is lost during linear transformation if matrix is not full rank by definition. This implies determinant will be $0$ and that some information is lost in this linear transformation.
Is my intuition behind this correct ?
 A: Your intuition seems fine. How you arrive at that conclusion depends on what properties you have seen, and/or which ones you are allowed to use.
The following properties are equivalent for a square matrix $A$:


*

*$A$ has full rank

*$A$ is invertible

*the determinant of $A$ is non-zero


There are more, but the first two are sufficient to immediately draw the desired conclusion.
A: This is exactly right. You're better off mentioning the rank-nullity theorem: for a linear map $f:U \to V$ we have $$\mbox{rank}+\mbox{nullity} = \dim U$$
where the nullity is the dimension of the kernel, $\ker f$. 
A four-by-four matrix represents a linear map $f: U \to V$ where $\dim U = \dim V = 4$. If the rank is three then $3+\mbox{nullity}=4$, i.e. there is a one-dimensional kernel. That means the map is not injective and has no inverse.
A: If $A$ and $B$ are matrices for which $AB$ makes sense, then
$$
\operatorname{rank}(AB)\le\min\{\operatorname{rank}(A),\operatorname{rank}(B)\}
$$
In particular, for every $B$, $\operatorname{rank}(AB)\le\operatorname{rank}(A)=3$. Can now $AB=I$?
A: $rank(A) = 3 \Rightarrow det(A) = 0$
needs to be proven, it is right though.
basically, you can say that:
$rank(A) < dim(A) \Rightarrow det(A) = 0$
but it still needs to be proven.
an easy way to prove it is by showing that you will get a row of zeroes when trying to use raw reduction.
