Determining Linear Independence and Linear Dependence? I understand that when solving for a linear dep/independent matrix, you can take the determinant of the matrix and if it is zero, then it is linearly dependent. However, how can I go about doing this for something like two $3\times 1$ vectors?
Example:
$$\begin{bmatrix} -6 \\ -1 \\ -7 \end{bmatrix}, \begin{bmatrix} -1 \\ -5 \\ 4\end{bmatrix} $$
The RREF yield infinitely many solutions, because R3 0=0. Does this tell me that the system is linearly independent if it has infinitely many solutions?

Define a matrix $A\in\mathbb{R}^{3\times 2}$ by setting $A:=\begin{bmatrix}-6 & -1 \\ -1 & -5 \\ -7 & 4 \end{bmatrix} $.
By performing the row operations on the matrix $A$, we can show that the null space $\mathcal{N}(A)=\left\{\begin{bmatrix}0 \\ 0\end{bmatrix}\right\}$.
Since the null space of $A$ contains only the zero vector, we have $Ax=0$ if and only if $x=\begin{bmatrix}0 \\ 0 \end{bmatrix}$.
But $Ax$ is the combinations of the column vectors of the matrix $A$, and thus the only combination of the vectors $v_1:=\begin{bmatrix}-6 \\ -1\\-7\end{bmatrix}$, $v_2:=\begin{bmatrix}-1 \\ -5\\4\end{bmatrix}$ which produces the zero vector $\begin{bmatrix}0 \\ 0\\0\end{bmatrix}$ is the trivial combination $(0,0)$.
In other words, we have $\displaystyle\sum_{1\le k\le 2}c_kv_k=0$ if and only if $c_k=0$ for all $1\le k\le 2$. Therefore by the definition of linear independence, the vectors $v_1,v_2$ are linearly independent.
