I am really struggling with the following problem since it has no numbers to find a determinate, I am not sure how to solve/prove this.
The question: Let a,b,c in R^3 be linearly independent. Show that {a+b+c, a-b-c, a} is linearly independent. Is the same true for {a+b, a+c, c-b}
So far I have tried putting the matrix in Ax=b form setting it equal to zero and simplifying the matrix by elimination, really I can't tell if I am overthinking, or underthinking the problem, but for whatever reason I can't make sense of it.
Thanks for any help :)
The key idea for this exercise is that the linear independence of vectors does not depend on the choice of basis. SInce $a$, $b$ and $c$ are linearly independent, you can take them to be a new basis of $\mathbb{R}^3$. If you now express the vectors $\{a+b+c,a-b-c,a\}$ in this new basis, you obtain the vectors: $$ \left\{\begin{bmatrix}1 \\ 1 \\ 1\end{bmatrix},\begin{bmatrix}1 \\ -1 \\ -1\end{bmatrix},\begin{bmatrix}1 \\ 0 \\ 0\end{bmatrix}\right\}, $$ for which you can check linear independence in the usual way. A similar approach can be used to check if $\{a+b,a+c,b-c\}$ are linearly independent or not.