Linear Dependence/Independence

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The set S = {(3,-2,4),(4,6,-4),(3,6,-2),(-13,2,-18)} is linearly dependent, as indicated by the presence of a zero column in the reduced row echelon form (rref) of the corresponding matrix. The rref shows that there are non-trivial solutions to the equation c1*v1 + c2*v2 + c3*v3 + c4*v4 = 0. Specifically, the equations derived from the rref are c1 + 5.44*c4 = 0, c2 + 7.76*c4 = 0, and c3 - 9.24*c4 = 0. This confirms that the vectors cannot be linearly independent since there are more vectors than the dimension of the space (R^3). Therefore, a dependency equation can be constructed from these relationships.
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Homework Statement


Is the set S = {(3,-2,4),(4,6,-4),(3,6,-2),(-13,2,-18)} linearly dependent? If so, give a dependency equation.


Homework Equations





The Attempt at a Solution



I first placed the set S into a matrix equation
[3 4 3 -13
-2 6 6 2
4 -4 -2 -18]

then put it into rref:

[1 0 0 5.44 0
0 1 0 7.76 0
0 0 1 -9.24 0]

Since the final column is all zeros, is the system linearly independent?
 
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Can a set of more than 3 vectors in R^3 be linearly independent?
 
If the row rank is less than the number of variables then there has to be a parametric solution. How would I write a dependency equation, then?
 
This matrix,
[3 4 3 -13]
[-2 6 6 2]
[4 -4 -2 -18]
represents the vector equation c1*v1 + c2*v2 + c3*v3 + c4*v4 = 0. As a matrix equation this is Ac = 0, where the columns of A are your four vectors, and c = <c1, c2, c3, c4>^T.

Assuming that your work is correct and that you ended with the next matrix (I removed the 5th column of 0s),

[1 0 0 5.44]
[0 1 0 7.76]
[0 0 1 -9.24]

this matrix says that c1 + 5.44*c4 = 0, c2 + 7.76*c4 = 0, and c3 - 9.24*c4 = 0.
 
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