Using determinants to find a unique solution I am supposed to find the values of $k$ for which the following system has a unique solution:
$$kx+y+z=1\\x+ky+z=1\\x+y+kz=1.$$
I came up with this:
$$\scriptsize\begin{bmatrix}k&1&1&1\\1&k&1&1\\1&1&k&1\end{bmatrix}\sim\begin{bmatrix}1&1&k&1\\1&k&1&1\\k&1&1&1\end{bmatrix}\sim\begin{bmatrix}1&1&k&1\\0&k-1&1-k&0\\0&1-k&1-k^2&1-k\end{bmatrix}\sim\begin{bmatrix}1&1&k&1\\0&k-1&1-k&0\\0&0&-k^2-k+2&1-k\end{bmatrix}.$$
Since $-k^2-k+2=-(k^2+k-2)=-(k+2)(k-1)$, the above system has a unique solution when $k\neq -2,1$. 
However, I was wondering if I could use determinants to come to the same conclusion. Is this possible? If so, how? 

If we let $x\in\mathbb R^3$ be a vector and name
$$E = \pmatrix{0&1&1\\1&0&1\\1&1&0}, e = \pmatrix{1\\1\\1}$$
Then your system is
$$(E + kI)x = e$$
This system is uniquely solvable iff $\det(E+kI) \ne 0$.
$$\det(E+kI) = k^3 + 1 + 1 - k - k - k = k^3 - 3k + 2 = (k-1)^2(k+2)$$
Thus we reach the same conclusion: $k\ne 1,-2$
A side note:
Notice that
$$\det(E+kI) = \det(E - (-k)I) = \chi_E(-k)$$
is the characteristic polynomial of $E$ evaluated at $-k$. Knowing the eigenvalues of $E$ will thus solve our problem.
