Departament of Economic, Financial, and Actuarial Mathematics
Workshop I
Mathematics I
Workshop I. Solutions
The RnVector Space. Linear Combinations
Exercise 1.
a)(0,−1,1)
b)(8,−3,0)
c)(−4,0,−3)
Exercise 2.
a) A line passing through the points (1,3) and (0,0)
b) A line passing through the points (1,−1) and (0,0) (also through (−3,3).
c) The whole 2-dimensional space, hence it is a plane.
Exercise 3.
a) A line passing through the points (−1,−1,−1) and (0,0,0) (also through (−4,−4,−4)).
b) A plane passing through the points (2,0,0) ,(1,2,2), and (0,0,0).
c) A plane passing through the points (2,2,2),(1,0,2), and (3,2,3) (also through (0,0,0)).
It is difficult to realize that the three points are in a plane that passes through the origin
graphically. However, we can conclude this taking into account that the rank of the matrix
built from the three vectors is 2.
d) A plane passing through the points (2,−1,3),(1,4,1), and (5,2,7) (also through (0,0,0)).
This is checked in Exercise 10.
Exercise 4. No. The rank of the matrix built from {~u1, ~u2}is different from the rank of the
matrix built from {~u1, ~u2, ~u}. This also means that the three vectors do not lie in the same
plane.
Exercise 5. Yes.
Exercise 6. Yes. The rank of the matrix built from {~u1, ~u2,~u3}is 2. The rank of the matrix built
from {~u1, ~u2, ~u3,~u}is also 2. Note that to compute the rank of the augmented matrix 4different
determinants are to be computed (3new determinants and the determinant of the coefficient
matrix).
Exercise 7. For k= 8.
Exercise 8. The rank of the matrix built from {~u1, ~u2}is 2(it is so for any value of k). ~u is a
linear combination of {~u1, ~u2}if the rank of the matrix built from the 3vectors is 2. Solving the
equation obtained equalizing the determinant of that matrix to zero we obtain that ~u is a linear
combination of {~u1, ~u2}for k= 1.
M. Álvarez Mozos 1 ADE - 2013/2014
