[tex]\begin{cases}x+y+kz=9\\x+ky+z=2\\kx+y+z=5\end{cases}[/tex]
Consider the system in matrix form, with coefficient matrix
[tex]A=\begin{bmatrix}1&1&k\\1&k&1\\k&1&1\end{bmatrix}[/tex]
Then the system
[tex]A\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}9\\2\\5\end{bmatrix}[/tex]
has a unique solution if [tex]A[/tex] is invertible, which requires that [tex]\det A\neq0[/tex]. Compute the determinant: expanding along the first row gives
[tex]\det A=\begin{vmatrix}k&1\\1&1\end{vmatrix}-\begin{vmatrix}1&1\\k&1\end{vmatrix}+k\begin{vmatrix}1&k\\k&1\end{vmatrix}[/tex]
[tex]\det A=(k-1)-(1-k)+(1-k^2)=-1+2k-k^2=-(1-k)^2[/tex]
This is zero if [tex]k=1[/tex], and non-zero otherwise, hence there is no unique solution if [tex]k=1[/tex].
