Answer
The $7\times5$ matrix must have a pivot position in all five columns.
Work Step by Step
If one of the columns lacked a pivot position, then the matrix equation $\mathbf{A}\vec{x}=\vec{0}$ would have a free variable. But this equation has a free variable if and only if the corresponding vector equation $c_{1}\vec{v}_{1}+...+c_{5}\vec{v}_{5}=\vec{0}$ has a nontrivial solution, which by definition means that the vectors are linearly dependent. Since we are told the vectors (i.e., the columns of the matrix) are linearly independent, it must be that no column lacks a pivot position.
