So, as expected, I wrote those vectors as the matrix: \begin{pmatrix}1&t&1\\ \:t&1&0\\ \:0&1&t\end{pmatrix}
Which can be simplified to: $$\begin{pmatrix}t&1&0\\ 1&t&1\\ 0&1&t\end{pmatrix} \sim \begin{pmatrix}t&1&0\\ 0&\frac{t^2-1}{t}&1\\ 0&1&t\end{pmatrix} \sim \begin{pmatrix}t&1&0\\ 0&\frac{t^2-1}{t}&1\\ 0&0&\frac{t^3-2t}{t^2-1}\end{pmatrix}$$
To my understanding, to obtain the values of $t$ I should see from which values the rows are all zero. So first I should solve the equation $\frac{t^3-2t}{t^2-1} = 0$. That result in $t \neq \pm 1$ $t = 0$ or $t = \pm \sqrt2$. But now I need to check all the other rows, so I divide $\frac{t^3-2t}{t^2-1}$ by $\frac{t^2-1}{t^3-2t}$, and subtract it from the second row, leading to
\begin{pmatrix}t&1&0\\ 0&\frac{t^2-1}{t}&0\\ 0&0& 1\end{pmatrix}
Solving the equation I end up with: $t \neq 0$ and $t = \pm1$. Repeating the same process as above:
\begin{pmatrix}t&0&0\\ 0&1&0\\ 0&0& 1\end{pmatrix}
Which gives me $t=0$.
Everything appeared to be good and clear, but when I decided to test those values in the actual vectores, it only yield true to $ \pm \sqrt{2} $ and $0$. So why was $\pm 1$ discarded? If it was from the existence condition of $\frac{t^3-2t}{t^2-1}$ why wasn't $0$ discarded? Was my method correct at all?
For some context, this exercise was given as a complement to chapter 12 of Apostol's Calulus 1.
Thanks in advance.
Why you have complicated the problem I don't know. It is very easy calculation according to me. You have to show the vectors $(t,1,0), (1,t,1), (0,1,t)$ are linearly dependent,i.e., the equation: $a(t,1,0)+b(1,t,1)+c(0,1,t)=0$ is true for some non-zero $a,b,c$. i.e., the equations: $at+b=0, a+bt+c=0, b+ct=0$ has non-zero solution. So $det\begin{pmatrix}t&1&0\\ \:1&t&1\\ \:0&1&t\end{pmatrix}=t(t^2-1)+1(0-t)=0\implies t=0,\pm \sqrt{2}$