Partial linear dependence let $S=\{s_1, s_2, s_3 \}$, if $s_1$ can be represented as a linear combination of $s_2$ and $s_3$, $s_2$ can be represented as a linear combination of $s_1$ and $s_3$ but $s_3$ can not be represented as a linear combination of $s_1$ or $s_2$ or $s_1$ and $s_2$, can we call $S$ a linearly dependent set? 
 A: Yes, the elements of $S$ are linearly dependent. To be linearly dependent means that there exist scalars $a, b, c$, not all zero such that
$$
as_1 + bs_2 + cs_3 = 0.
$$
This is true of your elements because we know
$$
s_1 = ms_2 + ns_3
$$
for some scalars $m, n$ and we can rearrange this equation to get $a = 1$, $b = -m$, $c = -n$. (The fact that we can't express $s_3$ as a linear combination of $s_1$ and $s_2$ implies that $n$ is going to have to be $0$, but this is not relevant to the definition of linear independence.)
A: You said that 
$$s_1=as_2+bs_3.$$
Therefore
$$s_1-as_2-bs_3=0$$
and your set is not linearly independent.
A: Warning!
By strictly adhering to notation, there is a special case where your $S=\{s_1, s_2, s_3\}$ with the given conditions  is linearly independent, namely if $s_1=s_2$ and $s_1, s_3$ are linearly independent.
This happens when one uses sets instead of families to talk about linear dependence and bases etc.
Concrete: In the vector space $\mathbb R^2$ let $s_1 = s_2 = (1, 0)$ and $s_3 = (0, 1)$. Then $s_1 = 1\cdot s_2+0\cdot s_3$ and $s_2 = 1\cdot s_1+0\cdot s_3$, whereas $s_3$ cannot be expressed as a linear combination of $s_1$ and $s_2$. The set $S=\{s_1, s_2, s_3\}$ has caridnality two and is linearly independent.
