Linear Independence and Linear Dependence I have a question that asks if 3 $3\times 1$ vectors are linearly indepedent or linearly dependent. The answer to the question finds 3 scalars, multiplies them to each vector and sets it equal to $0$. After, it uses Gauss Jordan elimination to determine. However, I noticed that if I take the determinant of the resulting $3\times 3$ I get 0. 
If the determinant is 0 would this mean the set of vectors is linearly dependent, whereas if it does not equal 0 it's linearly indepdenent? 
 A: Given three $3 \times 1$ vectors $X,Y,Z$ where you want to find scalars $a,b,c \in F$ (the ground field, usually real numbers or complex numbers for example), such that $aX + bY + cZ = 0$ is equivalent to finding $Ax = 0$, where $A$ is the matrix with $X,Y,Z$ as its columns (in that order) and $x = \left( \begin{array}{c}
a \\
b \\
c \\ \end{array} \right)$, i.e. to find a vector in the nullspace. Now, a standard result is that the nullspace is nontrivial (which in the context of this problem, means there is a nontrivial sequence of scalars) if and only if the determinant is zero. So if you are getting that the determinant is zero, then this indicates that you should be able to find three numbers $a,b,c$ not all being zero, which by definition means that it is not linearly independent. This easily generalizes to higher dimensions than 3. 
In conclusion, you are correct in your assessment. The determinant of the natural square matrix via concatenation can be used to test whether a certain set of vectors are linearly independent. 
