For which values of x do the vectors $(x,0,x + 1),(1,x,1),(0,x,−x)$ form a basis for R3? For which values of x do the vectors $(x,0,x + 1),(1,x,1),(0,x,−x)$ form a basis for R3?
Is 1 a correct answer to this. The vectors would each be linearly independent, however this seems too simple and I'm guessing theres an error somewhere.
If 1 works then it would seem any real number would work since the location of the 0 in the first and last vector will always mean the three vectors are linearly independent, right?
 A: Guide:


*

*Put those vectors as rows of a matrix, compute the determinant.

*If they form a basis, the determinant would not take value zero, hence you just have to find the root of the charactheristic polynomial and avoid those root to make it a basis.

*Remark: Find all values rather than find one that works.
Edit:
$$\begin{bmatrix} 1 & x & 1 \\ 0 & x & - x \\ x & 0 & x+ 1\end{bmatrix}$$
Perform $R_3-xR_1$: 
$$\begin{bmatrix} 1 & x & 1 \\ 0 & x & - x \\ 0 & -x^2 &  1\end{bmatrix}$$
Perform $R_3+xR_2$:
$$\begin{bmatrix} 1 & x & 1 \\ 0 & x & - x \\ 0 & 0 &  1-x^2\end{bmatrix}$$
To make it nonsingular, each column must have a pivot point, hence we require $x \neq 0$ and $1-x^2 \neq 0$.
Hence the conditions that you need are $x \neq 0$ and $1-x^2 \neq 0$.
A: How did you find this value?
You can also "glue" the vectors together to form a matrix (dimension 3x3), and try to find its determinant. You will obtain a polynom in $x$ and the zeros of this polynom are the values for which the three vectors are linearly dependant.
