Do $3$ independent vectors retain independence if I add (subtract) their combinations of $2$? Suppose $w_1,w_2,w_3$ are independent vectors.
Does that mean $w_1-w_2,w_1-w_3,w_2-w_3$ and $w_1+w_2,w_1+w_3,w_2+w_3$ are independent vectors?

This is a similar question, but I didnt really understand the solution:
How to prove the sum of 2 linearly independent vectors is also linearly independent?
 A: 
Does that mean $w_1-w_2,w_1-w_3,w_2-w_3$ [...] are independent vectors?

Well we can play around a bit and add these vectors to each other. We find that
$$
(w_1 - w_2) + (w_2 - w_3) = w_1 - w_2 + w_2 - w_3 = (w_1 - w_3).
$$
So two of the vectors add up to get the third one! That must mean that they aren't linearly independent. Remember that if you can ever add up any linear combination of some of the vectors to get a remaining vector, then they aren't linearly independent (in fact, that is the definition of "linearly independent" in most textbooks/classes).

What about $w_1+w_2,w_1+w_3,w_2+w_3$ ?

This is a bit harder. We can play around with these: if we add up $(w_1 + w_2)$ and $(w_1 + w_3)$ we get $2 w_1 + w_2 + w_3$, which isn't the third vector. If we subtract them we get $(w_1 + w_2) - (w_1 + w_3) = w_1 - w_3$... that isn't the third vector either.
At this point, since we tried to add up a linear combination of some of the vectors to get the remaining one, and we didn't succeed, we are thinking it's probably impossible: we guess that they are probably linearly independent.
So how do we prove this? We can start with a linear combination of the three vectors and show that if it equals $0$, then all the coefficients must be zero (refer to your definition of "linearly independent"). Here, we have
$$
a (w_1 + w_2) + b (w_1 + w_3) + c (w_2 + w_3) = 0
$$
Now that expands to
$$
(a + b) w_1 + (a + c) w_2 + (b + c) w_3 = 0
$$
But we know that $w_1$, $w_2$, and $w_3$ are linearly independent! So by the definition that means that
$$
a + b = 0 \text{ and } a + c = 0 \text{ and } b + c = 0.
$$
Now solve this system of three equations to see what we find, and complete the proof.
A: The vectors $w_1,w_2,w_3$ are linearly independent if and only if the matrix $A$ whose columns are $w_i$ is non-singular i.e. $$A=\begin{bmatrix}w_1&w_2&w_3\end{bmatrix}\implies \det(A)\ne 0$$If we seek to build another triplet $v_1,v_2,v_3$ which are linearly independent then a similar matrix $B$ must be non-singular and a transformation matrix $T$ exists such that $$B=TA\implies \det(B)=\det(T)\cdot \det(A)$$since $\det(A),\det(B)\ne 0$ we must have $\det(T)\ne0$ which says that the transformation matrix must be non-singular.
In your question, the transformation matrices are $$T_1=\begin{bmatrix}1&1&0\\1&0&1\\0&1&1\end{bmatrix}$$and $$T_2=\begin{bmatrix}1&-1&0\\1&0&-1\\0&1&-1\end{bmatrix}$$ where $$\begin{align} \det(T_1)&=-2\\
\det(T_2)&=0\end{align}$$
