Answer
See answer below
Work Step by Step
We are given:
$\{(1,1,0,2); (2,1,3,-1); (-1,1,1,-2); (2,-1,1,2)\}$ in $R^4$
We can see that $(-1)(1,-1,1)+2(2,5,-2)=(-1,1,-1)+(4,10,-4)=(3,11,-5)$
We obtain:
$A=\begin{bmatrix}
1 & 2 & -1 &2\\
1 & 1&1& -1 \\
0 & 3 & 1 & 1\\
2 & -1 & -2 & 2
\end{bmatrix}=\begin{bmatrix}
1 & 2 & -1 &2\\
0 & -1&2& -3 \\
0 & 3 & 1 & 1\\
2 & -1 & -2 & 2
\end{bmatrix}=\begin{bmatrix}
1 & 2 & -1 &2\\
0 & -1&2& -3 \\
0 & 3 & 1 & 1\\
0 & -5 & 0 & -2
\end{bmatrix}$
$\det (A)=\begin{vmatrix}
-1 & 2 & 3\\
3 & 1 & 1\\
-5 & 0 & -2
\end{vmatrix}-0+0-0=(-1)\begin{vmatrix}
1 & 1\\
0 & 2
\end{vmatrix}-2\begin{vmatrix}
3 & 1\\
-5 & -2
\end{vmatrix}+(-3)\begin{vmatrix}
3 & 1\\
-5 & 0
\end{vmatrix}=-1(-2-0)-2(-6+5)-3(0+5)=2+2-15=-11$
Since $\det A = -11 \ne 0$, the set $\{(1,-1,1),(2,5,-2),(3,11,-5)\}$ is linearly independent in $R^4$.
Since $dim R^4=4$, the set of vectors is a basis for $R^4$
