 Solution

A family $x_1,\dots ,x_n$ is linearly independent if ${\sum }_{i=1}^n{a}_i{x}_i=0$ forces $a_1=\cdots =a_n=0$. For the three specified vectors, direct substitution givesThe coefficients are not all zero, so the vectors are linearly dependent.

For the implication question, using the usual interpretation that a basis contains the listed family as independent entries, the answers in order (i)–(x) areThe key distinctions are that the unrestricted zero combination always exists, whereas linear dependence requires a nontrivial combination; and dependence does not necessarily let the specified third vector be solved in terms of the first two. For example $x=y=e_1$, $z=e_2$ makes (iii) and (iv) false, while $x=e_1,y=e_2,z=e_3$ disproves (vi).

There is a set-versus-family qualification in (v). If “contains” is interpreted as literal set containment, repeated nonzero vectors need only occur once in the basis. Then $x=y=e_1,z=e_2$ is dependent as a family but all three belong to the basis $\{e_1,e_2\}$, so (v) becomes false. With three distinct vectors, or with family-containment as above, basis extension gives (v) true. Implication (x) remains true on either reading.