Determining linear independence Determine whether the vectors in $W$ are linearly independent.
 $W=\{(1,0,1),(3,4,5),(6, 5,1),(7,9, 2)\}$ 
Is there some way to show that these are linearly dependent? It seems obvious because there are four of them in a space of $\mathbb{R}^3$ so at least a pair must be dependent. 
But how do I show this? I can't use the matrix determinant method since it won't be square thus non invertible. Is this enough of an argument?
 A: The most directly convincing way to show linear dependence is to express one of the vectors as a linear combination of the others.
The determinant of the matrix formed by your first three vectors is $-30$, so those vectors are linearly independent. Solving the equation
$$\begin{bmatrix}
 1 & 3 & 6 \\
 0 & 4 & 5 \\
 1 & 5 & 1 \\
\end{bmatrix}
\cdot
\begin{bmatrix}
 x \\
 y \\
 z \\
\end{bmatrix}
=
\begin{bmatrix}
 7 \\
 9 \\
 2 \\
\end{bmatrix}$$
leads us to this equation:
$$\begin{bmatrix}
 7 \\
 9 \\
 2 \\
\end{bmatrix}
=\frac{-13}5\begin{bmatrix}
 1 \\
 0 \\
 1 \\
\end{bmatrix}
+\frac 23\begin{bmatrix}
 3 \\
 4 \\
 5 \\
\end{bmatrix}
+\frac{19}{15}\begin{bmatrix}
 6 \\
 5 \\
 1 \\
\end{bmatrix}
$$
which proves linear dependence.
A: Or taking the determinant of the matrix used by Rory Daulton:
$$
\mbox{det} A = \mbox{det}(w_1, w_2, w_3) = (4-25) + (15-24) = -29 \ne 0
$$
so we have a unique solution $u = A^{-1} w_4$, where $u \ne 0$, otherwise $A u = 0 \ne w_4$. 
As $u$ consists of the coefficients for the linear combination this means linear dependence of $w_4$ in terms of $w_1$, $w_2$, $w_3$.
