I have no clue where to go with this matrix:
1 0 k = 1
k 1 0 = 1
0 k 1 = 1
The problem says to find all values of k, if any, for which the system of linear equations has no solution.
x + kz = 1
kx + y = 1
ky + z = 1
take the determinant of $\displaystyle A = \begin{bmatrix} 1 & 0 & k \\ k & 1 & 0 \\ 0 & k & 1 \end{bmatrix} $ to get $\displaystyle Det(A) = k^3 + 1 $, we want the determinant equal to zero. (Why? Can you anwer this?) so $\displaystyle k = -1 $. so the matrix $\displaystyle Ax = b $ either has zero solutions or infinite solutions. Now how do we know which? Well note that $\displaystyle \begin{bmatrix} 1 & 0 \\ -1 & 1 \\ 0 & -1 \end{bmatrix} $ are linearly. independent (the 3rd vector is not independent) so clearly if $\displaystyle \begin{bmatrix} 1 & 0 \\ -1 & 1 \\ 0 & -1 \end{bmatrix} x = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}$ has a solution at all. $\displaystyle x = \begin{bmatrix} x_1 \\ -1 \end{bmatrix} $ which means $\displaystyle x_2 = -2 $, but clearly $\displaystyle \begin{bmatrix} 1 & 0 \\ -1 & 1 \\ 0 & -1 \end{bmatrix} \begin{bmatrix} -2 \\ -1 \end{bmatrix} = \begin{bmatrix} -2 \\ 1 \\ 1 \end{bmatrix} \not =\begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} $ so. for k = -1, no soln. exist.