Answer
See answer below
Work Step by Step
We are given:
$\{(0,-1,0,k); (1,0,1,0); (0,1,1,0); (k,0,2,1)\}$ in $R^4$
$A=\begin{bmatrix}
0 & 1 & 0 & k\\
-1 & 0 & 1 & 0 \\
0 & 1 & 1 &2 \\
k & 0 & 0 & 1
\end{bmatrix}$
$\det A=\begin{vmatrix}
0 & 1 & 0 & k\\
-1 & 0 & 1 & 0 \\
0 & 1 & 1 &2 \\
k & 0 & 0 & 1
\end{vmatrix}=0-\begin{vmatrix}
-1 & 1 & 0\\
0 & 1 & 2 \\
k & 0 &1
\end{vmatrix}+0-k\begin{vmatrix}
-1 & 0 & 1 \\
0 & 1 & 1 \\
k & 0 & 0
\end{vmatrix}=-[(-1+0)-(0-2k)+0]-k[k(0-1)-0+0]=k^2-2k+1=(k-1)^2$
The set of vectors is linearly independent if and only if
$(k-1)^2 \ne0 \rightarrow k \ne 1$
Hence, the set of vectors $\{(0,-1,0,k); (1,0,1,0); (0,1,1,0); (k,0,2,1)\}$ is a basis for $R^4$ for all $k \in R$ \ $\{1\}$
