How can I show that this set is linearly independent? Given the set:
$\{(1,β,1),(β,1,0),(0,1,β)\}$
For which values of $β \in \mathbb R$ is the set a basis for $\mathbb R^3$?

I decided to write the set as a matrix:
\begin{bmatrix}1&β&1\\β&1&0\\0&1&β\end{bmatrix}
Then through pivoting, I thought I could find out if it is linearly independent with $\dim=3$, which would make it a basis for $R^3$.
However, I'm stuck.

How can I proceed?
 A: You're on the right track. We know that a square matrix has linearly independent rows precisely when its determinant is non-zero. So, to check for linear independence, you can simply evaluate the determinant of your matrix and check when it's non-zero. The determinant is:
$$\det\begin{bmatrix}1&β&1\\β&1&0\\0&1&β\end{bmatrix}=\beta+\beta-\beta^3=\beta(2-\beta^2)$$
which is non-zero for $\beta\not\in\{-\sqrt{2},0,\sqrt{2}\}$.
A: The determinant of that matrix is $2\beta-\beta^3$, and therefore the vectors are linearly independent if and only if $\beta\notin\{0,\sqrt2,-\sqrt2\}$.
A: A set of three vectors, $v_1$, $v_2$, $v_3$ is "dependent" if and only if there exist three numbers, a, b, and c, not all 0, so that $av_1+ bv_2+ cv_3= 0$.  And, of course, they are "indpendent" if that is not true.
So, with $v_1= (1,\beta, 1)$, $v_2= (\beta, 1, 0)$, and $v_3= (0, 1,\beta)$ so we need to look at $a(1, \beta, 1)+ b(\beta, 1, 0)+ c(0, 1, \beta)= (a+ \beta b, \beta a+ b+ c, a+ \beta c)= (0, 0, 0)$.  That is $a+ \beta b= 0$, $\beta a+ b+ c= 0$, and $a+ \beta c= 0$.  The first of those equations says $b= -\frac{a}{\beta}$ and the third says $c= -\frac{a}{\beta}$.  Then $\beta a+ b+ c= \beta a- \frac{a}{\beta}- \frac{a}{\beta}= a\left(\beta- \frac{2}{\beta}\right)= 0$.
Either a= 0 which would mean b= c= 0 so the vectors are dependent, or $\beta- \frac{2}{\beta}= 0$.  $\beta= \frac{2}{\beta}$, $\beta^2= 2$, $\beta= \pm\sqrt{2}$.  Of course, dividing by $\beta$ requires that $\beta$ not be 0.  That has to be checked separately and it is easy to see that (1, 0, 1), (0, 1, 0), and (0, 1, 0), since the last two vectors are the same.  So the three vectors are "dependent" for $\beta$ equal to 0, $\sqrt{2}$, and $-\sqrt{2}$ and are "independent" for any other value of $\beta$.
A: The pivots will be accomplished via
Row Operations:
(1) Multiply/divide a row by a non-zero scalar.
(2) Add/subtract a scalar multiple of one row from another row.
(3) Exchange two rows.
$$\begin{bmatrix}1&β&1\\β&1&0\\0&1&β\end{bmatrix} \mapsto \begin{bmatrix}1&β&1\\0&1-β^2&-β\\0&1&β\end{bmatrix} \mapsto \begin{bmatrix}1&β&1\\0&1&β\\0&1-β^2&-β\end{bmatrix}\mapsto \begin{bmatrix}1&β&1\\0&1&β\\0&0&β(β^2-2)\end{bmatrix}$$
If (and only if) $β(β^2-2) \ne 0$ we can divide by it and arrive at the reduced row echelon form,
$$\begin{bmatrix}1&β&1\\0&1&β\\0&0&1\end{bmatrix}$$
to demonstrate independence.
