Problem 1. [12 points] (a) Justify that the following list of vectors in R4 are linearly independent:








1
3
3
1
 1 
 2 
 −2 
 −1 




v4 = 
v3 = 
v2 = 
v1 = 
 1 
 −1 
 1 
 1 
1
−4
−2
−1
Answer: One way to justify this is to show that if we let A be the matrix A = [v1 v2 v3 v4 ] then the
equation Ax = 0 has only the trivial solution (since this corresponds to there being only the trivial linear
dependence relation). To do this we can row-reduce the augmented matrix [A|0], i.e.


1
3
3 1 0
 −1 −2 2 1 0 

;
 1
1 −1 1 0 
−1 −2 −4 1 0
if one row-reduces this we find pivots in each of the first four columns and thus there are no free variables.
Another solution is to do the “recursive” step by step method. The single vector v1 is linearly independent
since it’s nonzero, and then the list v1 , v2 is linearly independent since v2 is not a multiple of v1 . Then one
can check v3 is not a linear combination of v1 and v2 (a simple way to see that is that any linear combination
of v1 and v2 has the same 2nd and 4th entries, while v3 does not!), justifying that v1 , v2 , v3 is linearly
independent. Finally one can check v4 is not a linear combination as well, which is probably best done by
solving the system x1 v1 + x2 v2 + x3 v3 = v4 and seeing there is no solution.
(b) Is it possible to find a vector v5 ∈ R4 such that the list v1 , v2 , v3 , v4 , v5 remains linearly independent?
Why or why not?
Answer : No. This list would have five vectors in the 4-dimensional space R4 , and therefore must be
linearly dependent.
Problem 2. [8 points] For the following network flow diagram, write out the system of equations relating
the variables x1 , x2 , x3 , x4 and then solve it. (Your solution should tell which variables are free, and determine
all of the non-free variables in terms of them).
1
Answer: We get one equation for each node, giving the following system:
10 = x1 + x2
x1 + 20 = x3
x2 + 30 = x4
x3 + x4 = 60.
Solving this, we find we can take x4 to be free and we have x1 = 40 − x4 , x2 = x4 − 30, and x3 = 60 − x4 .
2