TITLE: Rank of a matrix
QUESTION [11 upvotes]: If a $3 \times 3$ matrix has determinant zero, then is it possible that its rank could be $3$? I think it only could be $2$ or less. I am right or wrong? Please explain.

REPLY [17 votes]: Just think about this image: A matrix can be viewed as a set of $n$ vectors of $n$ dimension (imagine the case of $n=3$). Thus the determinant is actually the volumn of the polyhedron spanned by these vectors (example: $(1,0,0)^T,(0,1,0)^T$ and $(0,0,1)^T$ form a unit cube), where the sign (plus or minus) indicates the orientation (right-handed system -- plus, and left-handed system otherwise).
Now you can understand many properties of determinants: Exchanging two columns (or rows) will flip the sign of determinant -- that's because the orientation has changed. Or, if one column (or row) becomes all zeroes, the determinant will also be reduced to zero -- that's because the polyhedron shrinks to a sheet, and thus occupies no space.
As for the rank, the rank of a matrix actually tells you what is the minimum dimension of the space holding all the vectors of the matrix. Take $n=3$ as an example again: If $r(A)=3$, it means, in order to contain all the column vectors of $A$, a space of at least 3 dimension is a must. As a result, they span a nonzero volumn. If $r(A)=2$, it means the vectors are in fact lying in a plane, or $r(A)=1$, lying within a line, either of which holds a zero volumn.