Elegant way to prove that this vector set is linear independent. I'm trying to find an elegant way to prove that this vector set $$\begin{bmatrix}1\\ 1\\1\\0\end{bmatrix}, ~~\begin{bmatrix}1\\ 1\\0\\1\end{bmatrix},~~\begin{bmatrix}1\\ 0\\1\\1\end{bmatrix},~~\begin{bmatrix}0\\ 1\\1\\1\end{bmatrix}$$ is linearly independent. I tried to find the matrix determinant (Where the column vectors are these vectors) through co factors expansion. Also, I tried to reduce the matrix to find which vectors are linearly independent based on the pivots. Is there another way to prove such linear independence?

You can observe that, if
$$
M=\begin{bmatrix}
1 & 1 & 1 & 0 \\
1 & 1 & 0 & 1 \\
1 & 0 & 1 & 1 \\
0 & 1 & 1 & 1
\end{bmatrix}
$$
then
$$
M\begin{bmatrix}1\\1\\1\\1\end{bmatrix}=
3\begin{bmatrix}1\\1\\1\\1\end{bmatrix},
\qquad
M\begin{bmatrix}1\\-1\\-1\\1\end{bmatrix}=
-\begin{bmatrix}1\\-1\\-1\\1\end{bmatrix},
\qquad
M\begin{bmatrix}0\\-1\\1\\0\end{bmatrix}=
\begin{bmatrix}0\\-1\\1\\0\end{bmatrix},
\qquad
M\begin{bmatrix}-1\\0\\0\\1\end{bmatrix}=
\begin{bmatrix}-1\\0\\0\\1\end{bmatrix}
$$
The last two vectors are obviously linearly independent. Thus $M$ has eigenvalues $3$, $-1$ and $1$ (double), so it is invertible.
However, row reduction is much easier than cofactor expansion:
\begin{align}
\begin{bmatrix}
1 & 1 & 1 & 0 \\
1 & 1 & 0 & 1 \\
1 & 0 & 1 & 1 \\
0 & 1 & 1 & 1
\end{bmatrix}
&\to
\begin{bmatrix}
1 & 1 & 1 & 0 \\
0 & 0 & -1 & 1 \\
0 & -1 & 0 & 1 \\
0 & 1 & 1 & 1
\end{bmatrix}
&& \begin{aligned}R_2&\gets R_2-R_1\\ R_3&\gets R_3-R_1\end{aligned}
\\
&\to
\begin{bmatrix}
1 & 1 & 1 & 0 \\
0 & -1 & 0 & 1 \\
0 & 0 & -1 & 1 \\
0 & 1 & 1 & 1
\end{bmatrix}
&&R_2\leftrightarrow R_3
\\
&\to
\begin{bmatrix}
1 & 1 & 1 & 0 \\
0 & -1 & 0 & 1 \\
0 & 0 & -1 & 1 \\
0 & 0 & 1 & 2
\end{bmatrix}
&&R_4\gets R_4+R_2
\\
&\to
\begin{bmatrix}
1 & 1 & 1 & 0 \\
0 & -1 & 0 & 1 \\
0 & 0 & -1 & 1 \\
0 & 0 & 0 & 3
\end{bmatrix}
&&R_4\gets R_4+R_3
\end{align}
so the determinant is $1\cdot(-1)\cdot(-1)\cdot 3=3$.
