= Solution
On fixed labelled <ground sets> of <cardinality> $1,2,3$, the numbers of <matroids> are respectively \b[$2,5,16$].
For one element, the element is either a <loop of a matroid> or a <nonloop>, giving two choices. For two elements, <rank of a matroid> zero and two each give one possibility. At <rank of a matroid> one, the <nonloops> form any nonempty <subset>, giving $2^2-1=3$ more possibilities: total $1+3+1=5$.
For three elements, classify by the total <rank of a matroid>. At <rank of a matroid> zero every element is a <matroid loop>; at <rank of a matroid> three every <subset> is a <matroid independent set>. At <rank of a matroid> one, choose any nonempty <subset> of <nonloops>; exactly the <empty set> and their <singletons> are <matroid independent sets>. This gives seven possibilities.
At <rank of a matroid> two, either exactly two elements are <nonloops>, giving three choices for the remaining <loop of a matroid>, or all three are <nonloops>. In the latter case, pairs that are <matroid dependent sets> define <parallel elements of a matroid>. There can be no such pair, giving the <uniform matroid> $U_{2,3}$, or exactly one such pair, giving three choices. There cannot be exactly two such pairs: if $\{a,b\}$ and $\{a,c\}$ were <matroid dependent sets> but $\{b,c\}$ a <matroid independent set>, augmentation from $\{a\}$ to $\{b,c\}$ would fail. If all three pairs were <matroid dependent sets>, the <rank of a matroid> would be one. Hence <rank of a matroid> two gives $3+1+3=7$ examples.
Every listed possibility satisfies augmentation: the only nontrivial comparison on three elements is from a <singleton> to a two-element <matroid independent set>, and with at most one two-element <matroid dependent set> at least one addition works. The total is therefore $1+7+7+1=16$.
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