The answer is a.
Given, x = [x1, x2, ....xn]T
Since, x is an n-tuple non-zero vector, that is, x is a non-singular matrix of order n, so it rank should be n.
i.e, I (x) = n
The vector (xT) also have rank (n) because transpose of any matrix does not altered its rank.
Then, matrix [v = x T'] must have the rank n, i. e., I(v) = n, because resultant of the multiplication of two same rank matrices also has the same rank as the rank of multiplicative matrix.