Find all $\lambda \in\Bbb R$ such that $\vec{a},\vec{b},\vec{c}$ don't span $V^{3}(0)$.
$\vec{a}=\lambda\vec{i}+\vec{j}+\vec{k}$
$\vec{b}=\vec{i}+\lambda{j}+\vec{k}$
$\vec{c}=\vec{i}+\vec{j}+\lambda\vec{k}$
Usually when I was asked to show that some set of vectors span a vector space, I would write it down like this $\alpha\vec{a}+\beta\vec{b}+\gamma\vec{c}=(x,y,z)$ and then I would find $\alpha, \beta, \gamma$ in terms of $x,y,z$. So I thought I might start with that and in the end I would get some contradiction or something. However, it brought me nowhere and I was told I should find $\lambda$ for which these vectors are linearly dependent. I managed to do it, but I don't understand why I did that. Can someone explain?
The solutions I got are $\lambda_1=1, \lambda_2=-2$ by the way.
The vectors $\vec{a},\vec{b},\vec{c}$ don't span $V^3$ if and only if they are a set linearly dependent if and only if $$\det(\vec{a},\vec{b},\vec{c})=0\iff \begin{vmatrix}\lambda & 1 & 1\\ 1 & \lambda & 1 \\1 & 1 & \lambda\end{vmatrix}=0.$$