Given $\alpha = \langle 1,1,1 \rangle, \beta = \langle 0,1,-1 \rangle, \gamma = \langle 2,0,1 \rangle$, show that each of $\alpha, \beta, \gamma$ is a linear combination of the other two. Show that it is impossible to find coefficients $x, y$ and $z$ such that $x \alpha + y \beta + z \gamma = \delta ^1$, where $\delta ^1 = \langle 1,0,0 \rangle$.
What does it mean to show that each $\alpha, \beta, \gamma$ is a linear combination of the other two? To me, that means there are scalars $x,y \in \mathbb{R}$ such that
$x\alpha + y\beta = \gamma $
$x\gamma + y\beta = \alpha$
$x\alpha + y\gamma = \beta$
Then I am struggling to show the above. Let's take $x\alpha + y\beta = \gamma$ for example.
\begin{align} x \langle 1, 1, 1 \rangle + y \langle 0,1,-1 \rangle &= \langle x, x, x \rangle + \langle 0, y, -y \rangle \\ &= \langle 2,0,1 \rangle \end{align}
$$\begin{vmatrix} 1 & 0 & 2 \\ 1 & 1 & 0 \\ 1 & -1 & 1 \end{vmatrix}$$
But then it doesn't make sense because in reduced row echelon form, we get
$$\begin{vmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{vmatrix}$$
So $x$ and $y$ have to equal $0$, but then the above system of linear equations wouldn't be correct.