If these vectors are to be linearly independent, then there exists some not-all-zero choice for scalars [tex]c_1,c_2,c_3,c_4[/tex] such that
[tex]c_1(1,1,0,-1) + c_2(1,k,1,1) + c_3(5,1,k,1) + c_4(-1,1,1,k) = (0,0,0,0)[/tex]
We can recast this as a system of linear equations,
[tex]\begin{bmatrix}1&1&5&-1\\1&k&1&1\\0&1&k&1\\-1&1&1&k\end{bmatrix}\begin{bmatrix}c_1\\c_2\\c_3\\c_4\end{bmatrix} = \begin{bmatrix}0\\0\\0\\0\end{bmatrix}[/tex]
This system has a solution if the coefficient matrix on the left side is not singular. So we attempt to find k such that it *is* singular, i.e. the determinant is zero:
• A cofactor expansion along the first column gives
[tex]\begin{vmatrix}1&1&5&-1\\1&k&1&1\\0&1&k&1\\-1&1&1&k\end{vmatrix} = \begin{vmatrix}k&1&1\\1&k&1\\1&1&k\end{vmatrix} - \begin{vmatrix}1&5&-1\\1&k&1\\1&1&k\end{vmatrix} + \begin{vmatrix}1&5&-1\\k&1&1\\1&k&1\end{vmatrix}[/tex]
• Cofactor expansions along the first columns of the remaining 3x3 matrices give
[tex]\begin{vmatrix}k&1&1\\1&k&1\\1&1&k\end{vmatrix} = k \begin{vmatrix}k&1\\1&k\end{vmatrix} - \begin{vmatrix}1&1\\1&k\end{vmatrix} + \begin{vmatrix}1&1\\k&1\end{vmatrix} = k^3-3k+2[/tex]
[tex]\begin{vmatrix}1&5&-1\\1&k&1\\1&1&k\end{vmatrix} = \begin{vmatrix}k&1\\1&k\end{vmatrix} - \begin{vmatrix}5&-1\\1&k\end{vmatrix} + \begin{vmatrix}5&-1\\k&1\end{vmatrix} = k^2-4k+3[/tex]
[tex]\begin{vmatrix}1&5&-1\\k&1&1\\1&k&1\end{vmatrix} = \begin{vmatrix}1&1\\k&1\end{vmatrix} - \begin{vmatrix}5&-1\\k&1\end{vmatrix} + \begin{vmatrix}5&-1\\1&1\end{vmatrix} = -k^2 - 6k + 7[/tex]
It follows that
[tex]\begin{vmatrix}1&1&5&-1\\1&k&1&1\\0&1&k&1\\-1&1&1&k\end{vmatrix} = (k^3-3k+2) - (k^2-4k+3) + (-k^2 - 6k + 7) \\\\ \begin{vmatrix}\cdots\end{vmatrix} = k^3 - 2k^2 - 5k + 6 = (k-3)(k-1)(k+2)[/tex]
which makes the coefficient matrix singular if k = 3, k = 1, or k = -2.
Then the four vectors are linearly independent for
[tex]\left\{ k\in\mathbb R \mid k\not\in\{-2,1,3\}\right\}[/tex]
