Does the determinant of a matrix made up of column vectors being non-zero imply that the vectors are independent? Let's say we have 3 vectors and we make up a matrix where we depict the vectors as the columns of the matrix. If we calculate the determinant of the matrix and we get a non-zero number, does that mean that the vectors are linearly independent? 
 A: Yes. If the columns are linearly dependent, then we can show that the determinant is $0$. So if the determinant is not $0$ then the columns are linearly independent.
If the columns of the matrix are $\vec{c}_1, \vec{c}_2, \ldots, \vec{c}_n$ and there are scalars $a_1, a_2, \ldots, a_n$ such that
$$a_1 \vec{c}_1 + a_2 \vec{c}_2 + \cdots + a_n \vec{c}_n = \vec{0}$$
then we see that the vector
$$\langle a_1, a_2, \ldots , a_n\rangle^T$$
is an eigenvector of the matrix with eigenvalue $0$. Since $0$ is an eigenvalue, the determinant is $0$.
A: 
Let's say we have 3 vectors

$$
\left(
\begin{matrix}
a\\
b\\
c
\end{matrix}
\right),
\left(
\begin{matrix}
d\\
e\\
f
\end{matrix}
\right),
\left(
\begin{matrix}
g\\
h\\
i
\end{matrix}
\right)
$$

and we make up a matrix where we depict the vectors as the columns of the matrix

$$
\left(
\begin{matrix}
a & d & g\\
b & e & h\\
c & f & i\\
\end{matrix}
\right)
$$

If we calculate the determinant of the matrix and we get a non-zero number

$$
\left|
\begin{matrix}
a & d & g\\
b & e & h\\
c & f & i\\
\end{matrix}
\right|
=aei+dhc+gbf-gec-dbi-ahf\neq 0
$$

does that mean that the vectors are linearly independent?

Yes.
Look at the 6 summands that make up the determinant. Note how each one is the product of one element of each of the original vectors and also including one element from each "row". Let me illustrate
$$
\left(
\begin{matrix}
\color{red}{a}\\
\color{green}{b}\\
\color{blue}{c}
\end{matrix}
\right),
\left(
\begin{matrix}
\color{red}{d}\\
\color{green}{e}\\
\color{blue}{f}
\end{matrix}
\right),
\left(
\begin{matrix}
\color{red}{g}\\
\color{green}{h}\\
\color{blue}{i}
\end{matrix}
\right)
$$
and
$$\color{red}{a}\color{green}{e}\color{blue}{i}+\color{red}{d}\color{green}{h}\color{blue}{c}+\color{red}{g}\color{green}{b}\color{blue}{f}-\color{red}{g}\color{green}{e}\color{blue}{c}-\color{red}{d}\color{green}{b}\color{blue}{i}-\color{red}{a}\color{green}{h}\color{blue}{f}\neq 0$$
There seems to be a pattern to this. If you allow yourself for a moment to only care about the colors and substitute each summand by $\color{red}{R}\color{green}{G}\color{blue}{B}$, the result is clearly 0
$$\color{red}{R}\color{green}{G}\color{blue}{B}+\color{red}{R}\color{green}{G}\color{blue}{B}+\color{red}{R}\color{green}{G}\color{blue}{B}-\color{red}{R}\color{green}{G}\color{blue}{B}-\color{red}{R}\color{green}{G}\color{blue}{B}-\color{red}{R}\color{green}{G}\color{blue}{B} = 3\color{red}{R}\color{green}{G}\color{blue}{B}-3\color{red}{R}\color{green}{G}\color{blue}{B}= 0$$
And that's exactly what happens when the vectors are linear dependent, which means all of them can be expressed as one vector multiplied with a scalar, solely for artistic reasons that vector shall be $(\color{red}{R},\color{green}{G},\color{blue}{B})^T$ which is equal to $(\color{red}{a},\color{green}{b},\color{blue}{c})^T$
$$
\left(
\begin{matrix}
\color{red}{a}\\
\color{green}{b}\\
\color{blue}{c}
\end{matrix}
\right)=\left(
\begin{matrix}
\color{red}{R}\\
\color{green}{G}\\
\color{blue}{B}
\end{matrix}
\right),
\left(
\begin{matrix}
\color{red}{d}\\
\color{green}{e}\\
\color{blue}{f}
\end{matrix}
\right) = x\left(
\begin{matrix}
\color{red}{R}\\
\color{green}{G}\\
\color{blue}{B}
\end{matrix}
\right),
\left(
\begin{matrix}
\color{red}{g}\\
\color{green}{h}\\
\color{blue}{i}
\end{matrix}
\right)=y\left(
\begin{matrix}
\color{red}{R}\\
\color{green}{G}\\
\color{blue}{B}
\end{matrix}
\right)
$$
Let the transformation begin
$$
\begin{array}{c|ccccccl}
& \color{red}{a}\color{green}{e}\color{blue}{i}&+\color{red}{d}\color{green}{h}\color{blue}{c}&+\color{red}{g}\color{green}{b}\color{blue}{f}&-\color{red}{g}\color{green}{e}\color{blue}{c}&-\color{red}{d}\color{green}{b}\color{blue}{i}&-\color{red}{a}\color{green}{h}\color{blue}{f}&\neq 0\\
\hline
\color{red}{a}=\color{red}{R}& \color{red}{R}\color{green}{e}\color{blue}{i}&+\color{red}{d}\color{green}{h}\color{blue}{c}&+\color{red}{g}\color{green}{b}\color{blue}{f}&-\color{red}{g}\color{green}{e}\color{blue}{c}&-\color{red}{d}\color{green}{b}\color{blue}{i}&-\color{red}{R}\color{green}{h}\color{blue}{f}&\\
\color{green}{b}=\color{green}{G}& \color{red}{R}\color{green}{e}\color{blue}{i}&+\color{red}{d}\color{green}{h}\color{blue}{c}&+\color{red}{g}\color{green}{G}\color{blue}{f}&-\color{red}{g}\color{green}{e}\color{blue}{c}&-\color{red}{d}\color{green}{G}\color{blue}{i}&-\color{red}{R}\color{green}{h}\color{blue}{f}&\\
\color{blue}{c}=\color{blue}{B}& \color{red}{R}\color{green}{e}\color{blue}{i}&+\color{red}{d}\color{green}{h}\color{blue}{B}&+\color{red}{g}\color{green}{G}\color{blue}{f}&-\color{red}{g}\color{green}{e}\color{blue}{B}&-\color{red}{d}\color{green}{G}\color{blue}{i}&-\color{red}{R}\color{green}{h}\color{blue}{f}&\\
\color{red}{d}=x\color{red}{R}& \color{red}{R}\color{green}{e}\color{blue}{i}&+x\color{red}{R}\color{green}{h}\color{blue}{B}&+\color{red}{g}\color{green}{G}\color{blue}{f}&-\color{red}{g}\color{green}{e}\color{blue}{B}&-x\color{red}{R}\color{green}{G}\color{blue}{i}&-\color{red}{R}\color{green}{h}\color{blue}{f}&\\
\color{green}{e}=x\color{green}{G}& \color{red}{R}x\color{green}{G}\color{blue}{i}&+x\color{red}{R}\color{green}{h}\color{blue}{B}&+\color{red}{g}\color{green}{G}\color{blue}{f}&-\color{red}{g}x\color{green}{G}\color{blue}{B}&-x\color{red}{R}\color{green}{G}\color{blue}{i}&-\color{red}{R}\color{green}{h}\color{blue}{f}&\\
\color{blue}{f}=x\color{blue}{B}& \color{red}{R}x\color{green}{G}\color{blue}{i}&+x\color{red}{R}\color{green}{h}\color{blue}{B}&+\color{red}{g}\color{green}{G}x\color{blue}{B}&-\color{red}{g}x\color{green}{G}\color{blue}{B}&-x\color{red}{R}\color{green}{G}\color{blue}{i}&-\color{red}{R}\color{green}{h}x\color{blue}{B}&\\
\color{red}{g}=y\color{red}{R}& \color{red}{R}x\color{green}{G}\color{blue}{i}&+x\color{red}{R}\color{green}{h}\color{blue}{B}&+y\color{red}{R}\color{green}{G}x\color{blue}{B}&-y\color{red}{R}x\color{green}{G}\color{blue}{B}&-x\color{red}{R}\color{green}{G}\color{blue}{i}&-\color{red}{R}\color{green}{h}x\color{blue}{B}&\\
\color{green}{h}=y\color{green}{G}& \color{red}{R}x\color{green}{G}\color{blue}{i}&+x\color{red}{R}y\color{green}{G}\color{blue}{B}&+y\color{red}{R}\color{green}{G}x\color{blue}{B}&-y\color{red}{R}x\color{green}{G}\color{blue}{B}&-x\color{red}{R}\color{green}{G}\color{blue}{i}&-\color{red}{R}y\color{green}{G}x\color{blue}{B}&\\
\color{blue}{i}=y\color{blue}{B}& \color{red}{R}x\color{green}{G}y\color{blue}{B}&+x\color{red}{R}y\color{green}{G}\color{blue}{B}&+y\color{red}{R}\color{green}{G}x\color{blue}{B}&-y\color{red}{R}x\color{green}{G}\color{blue}{B}&-x\color{red}{R}\color{green}{G}y\color{blue}{B}&-\color{red}{R}y\color{green}{G}x\color{blue}{B}&\\
\end{array}
$$
And that last line is (apart from the x&y's) exactly as the one above, which equalled zero
$$xy\color{red}{R}\color{green}{G}\color{blue}{B}+xy\color{red}{R}\color{green}{G}\color{blue}{B}+xy\color{red}{R}\color{green}{G}\color{blue}{B}-xy\color{red}{R}\color{green}{G}\color{blue}{B}-xy\color{red}{R}\color{green}{G}\color{blue}{B}-xy\color{red}{R}\color{green}{G}\color{blue}{B} = 3xy\color{red}{R}\color{green}{G}\color{blue}{B}-3xy\color{red}{R}\color{green}{G}\color{blue}{B}= 0$$
Intuitively speaking, the linear dependency "smears" all 3 distinguishable a, d and g into one homogenous blob R. The same holds true for the other two rows. The scalar factors cancel out as well due to the construction of the determinant.

tl;dr; got color blind;
Checking vectors if they are linear dependent and assembling them into a matrix and checking if the determinant is 0 is the same thing.
A: Yes. I'll give a direct proof that tells you what's happening here.
Suppose the contrary: that it is possible to have a matrix with non-zero determinant and linearly dependent vectors.
We know that adding a multiple of a column to another column does not change the determinant (Lemma 1).
Why is that? Recall the definition of the determinant: $\text{det}(A)=\sum \limits _{\sigma \in S_n} (\text{sgn}(\sigma)\prod\limits_{i=1}^n a_{i,\sigma{i}})$.
For simplicity of notation, we'll prove that adding a row to another row does not change the determinant. The same statement for columns follows trivially from $\text{det}(A)=\text{det}(A^T)$.
Let's now subject an arbitrary column, call it $j$, to the transformation $j \to j+\beta k$, where $k$ is another arbitrary column and $\beta$ is an arbitrary scalar. Then for any $i$, the element $a_{ji}$ becomes $a_{ji}+\beta \cdot a_{ki}$
We rewrite the determinant of the resulting matrix, call it $B$:
$\text{det}(B) = \sum \limits _{\sigma \in S_n} (\text{sgn}(\sigma)\prod\limits_{i=1, i \ne j}^n a_{i,\sigma{i}})\cdot (a_j\sigma j+\beta \cdot a_{k\sigma j})$
$= \text{det}(A)+\beta  \sum \limits _{\sigma \in S_n} (\text{sgn}(\sigma) a_{k\sigma j} a_{k\sigma k}\prod\limits_{i=1, i \ne j,k}^n a_{i,\sigma{i}})$
Now, for any $\sigma \in S_n$ there is a corresponding unique permutation $\tau \in S_n$, such as $\tau(i) = \sigma (i), \forall i \ne j,k, \tau(j)=\sigma(k)$ and $\tau(k)=\sigma(j)$: the composition of $\sigma$ with the $(j,k)$ transposition. Also, since the sign of a permutation is defined as $(-1)^m$, where m is the number of transpositions in the permutation, clearly $\text{sgn}(\tau)=-\text{sgn}(\sigma)$
You can surely see where this is going now:
$2 \text{det}(B) = 2 \text{det}(A)+2 \beta  \sum \limits _{\sigma \in S_n} (\text{sgn}(\sigma) a_{k\sigma j} a_{k\sigma k}\prod\limits_{i=1, i \ne j,k}^n a_{i,\sigma{i}})$
$= 2 \text{det} (A) + \beta  \sum \limits _{\sigma \in S_n} (\text{sgn}(\sigma) a_{k\sigma j} a_{k\sigma k}\prod\limits_{i=1, i \ne j,k}^n a_{i,\sigma{i}})+\beta  \sum \limits _{\tau \in S_n} (\text{sgn}(\tau) a_{k\tau j} a_{k\tau k}\prod\limits_{i=1, i \ne j,k}^n a_{i,\tau{i}})$
$= 2 \text{det} (A) + \beta  \sum \limits _{\sigma \in S_n} (\text{sgn}(\sigma) (a_{k\sigma j} a_{k\sigma k}-a_{k\sigma k} a_{k\sigma j})\prod\limits_{i=1, i \ne j,k}^n a_{i,\sigma{i}})$
$= 2 \text{det} (A)$
In conclusion, $\text{det} A = \text{det} B$, and Lemma 1 is proven. The rest trivially follows: since the columns are linearly dependent, there exists a linear combination of them that adds up to zero. We can form that linear combination in one column by adding scalar multiples of other columns into it, and as we established that procedure does not change the determinant. It follows that we can obtain a matrix with a zero column that has the same determinant as the original one, and obviously a matrix with a zero column/row will have a zero determinant (from the definition). The statement is now proven.
A: There is a much simpler interpretation of the determinant of a $3\times 3$ matrix. Say we have vectors $x,y,z\in\mathbb{R}^3$. Then the determinant of the matrix $M=(x,y,z)$ with coloumn as the given vectors is called the Scalar Triple product. Let's say the determinant is $D$, then $|D|$ can be interpreted as the volume of the paralellopiped whose adjascent sides are given by $x,y,z$.
Now, suppose $|D|=0$, then we can deduce that $x,y,z$ lie on the same plane, and therefore they are linearly dependent. Conversely, if $x,y,z$ are linearly dependent, they lie on the same plane and hence we can see that $D=0$.
To the best of my knowledge, this interpretation is valid only when the vectors are $3$ dimensional.
