Linear Independence of 3 Vectors in R^4

VinnyCee
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LINEAR ALGEBRA: 3 vecotrs in R^4 (with 6 variables) -- Are they linearly independent?

For which values of the constants a, b, c, d, e, anf f are the following vectors linearly independent? Justify your answer.

\left[\begin{array}{c}a\\0\\0\\0\end{array}\right],\,\,\left[\begin{array}{c}b\\c\\0\\0\end{array}\right],\,\,\left[\begin{array}{c}d\\e\\f\\0\end{array}\right]

I figure that one would setup an equation:

x\,\left[\begin{array}{c}a\\0\\0\\0\end{array}\right]\,\,+\,\,y\,\left[\begin{array}{c}b\\c\\0\\0\end{array}\right]\,\,+\,\,z\,\left[\begin{array}{c}d\\e\\f\\0\end{array}\right]\,\,=\,\,\left[\begin{array}{c}0\\0\\0\\0\end{array}\right]

x\,a\,\,+\,\,y\,b\,\,+\,\,z\,d\,\,=\,\,0
\,\,\,\,\,\,\,y\,c\,\,+\,\,x\,e\,\,=\,\,0
\,\,\,\,\,\,\,\,\,\,\,\,\,\,z\,f\,\,=\,\,0

How does one proceed?
 
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You want to find a rule for when x, y, and z are all zero. Start with zf=0. When is z not zero?

Then you can take a look at your second equation, which should be yc + ze ;)
 
Do you know how to solve a system of linear equations? I suggest you try investigating the determinant of the system. Further on, what is the definition of linear independence? What must x, y, and z equal?
 
radou said:
I suggest you try investigating the determinant of the system.
I suggest you do not since it is not a system of 4 vectors in R^4 but 3, so there is no determinantal way to proceed, unless you wish to add a fourth arbitrary vector in that is not in the span of the first three. But that is unnecessarily complicated, although not particularyly hard. However if you can see how to do that then you can see the answer anyway. You can simply do it by inspection.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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