A more explicit way to do this would be by considering
\begin{align} v_1 = (\lambda,a,a),\\ v_2 = (a,\lambda,a), \\ v_3 = (a,a,\lambda),\end{align}
such that $a\in\mathbb{R}$ is fixed. (In this case, $a=-1/2$.)
We want to find $\lambda$ such that $v_1,v_2,v_3$ are LD. If these vectors are LD, then $\exists c,d \in\mathbb{R}: cv_1+dv_2=v_3$. That is:
\begin{align}\begin{cases}c\lambda+da=a \\ ca+d\lambda=a \\ (c+d)a=\lambda\end{cases}\implies \begin{cases}(c+d)(a+\lambda) = 2a \\ (c+d)a=\lambda\end{cases}\implies \begin{cases}(c+d+1)\lambda = 2a \\ (c+d)a=\lambda\end{cases}\end{align}
And we can deduce for $a,\lambda\neq 0$:
\begin{align}&(c+d)(c+d+1)\lambda a = 2\lambda a \\ \implies &(c+d)(c+d+1)= 2 \\ \implies &(c+d) = 1 \text{ or } -2.\end{align}
So, from $(c+d)a=\lambda$, we have $\lambda = -2a,\lambda = a$. In the case $a=-1/2$ we have, indeed, $\lambda=1,\lambda=-1/2$.
The determinant of the matrix is $a^2 + b^2$, not $a^2 - b^2$. This is non-zero if and only if at least one of $a$ and $b$ is non-zero.
This assumes that $a$ and $b$ are real; otherwise, the issue is more complicated. For example, the matrix
$$\left[\begin{array}{cc} i & -1 \\ 1 & i\end{array}\right]$$
is not invertible.
Best Answer
The key principle here is that a matrix is invertible if an only if its determinant is non-zero.
We apply this principle here as follows:
Set
$A(\lambda) = \begin{bmatrix} \lambda &-1 & 0 \\ -1 & \lambda & -1 \\ 0 & -1 & \lambda\end{bmatrix}; \tag{1}$
then we have
$\det A(\lambda) = \lambda^3 - 2\lambda; \tag{2}$
thus $\det A(\lambda) = 0$ precisely when
$0 = \lambda^3 - 2 \lambda = \lambda(\lambda^2 - 2); \tag{3}$
the roots of this equation are easily seen to be
$\lambda = 0, \pm \sqrt 2; \tag{4}$
$A(\lambda)$ is thus invertible for all other values of $\lambda$.
Hope this helps. Cheers,
and as always,
Fiat Lux!!!