Question statement : Suppose $u,v,w$ are linearly independent vectors. Prove that $S$ is linearly independent where :
$\qquad$ b) $S = \{u+v-3w, u+3v-w, v+w\}. $
My attempt at the problem :
\begin{gather*}
c_1(u+v-3w) + c_2 (u+3v-w) + c_3 (v+w) = 0\\
\implies c_1 u + c_1 v -3c_1 w \ + \\
\qquad \: c_2u -c_2v -c_2w \ +\\
\quad c_3 u +c_3 w = 0 \\
\implies
\begin{bmatrix}
1 & 1 & 0 \\
1 & 3 & 1 \\
-3 & -1 & 1
\end{bmatrix} = (\dots) =
\begin{bmatrix}
1 & 0 & \frac{-1}{2} \\
0 & 1 & \frac12 \\
0 & 0 & 0
\end{bmatrix}
\end{gather*}
Which is clearly dependent. I don't understand what i'm doing wrong ,could there be a mistake in the book ? Thanks for help in advance.
Indeed there seems to be a problem. The identity $$ c_1(u+v-3w)+c_2(u+3v-w)+c_3(v+w) = 0 $$ implies $$ (c_1+c_2)u+(c_1+3c_2+c_3)v+(-3c_1-c_2+c_3)w = 0, $$ which is slightly different from what you write. Now $u,v,w$ are linearly independent, so each coefficient equals zero, hence \begin{cases} c_1+c_2 = 0 \\ c_1+3c_2+c_3 = 0 \\ -3c_1-c_2+c_3 = 0. \end{cases} Thus $c_2 = -c_1$, $c_3 = 2c_1$, and the third equation is redundant. For example, choose $c_1=1,c_2=-1,c_3=2$ and you get \begin{align} c_1(u+v-3w)+\dots+c_3(v+w) & = u+v-3w-u-3v+w+2v+2w \\ & = (u-u)+(v-3v+2v)+(-3w+w+2w) \\ & = 0. \end{align}