Can't do this system problem

Question

Question: For the value(s) of $k$, if any, will the following system have (a) no solution, (b) a unique solution, (c) infinity many solutons: $$x+y+kz=1,\\x+ky+z=1,\\xk+y+z=-2.$$

Answer: for $k=1$ there is no solution? For $k=2$ this is a unique solution? I tried these but I can't really make some progress.

Answer 1

Let's calculate this determinant

$$\det\left( \begin{matrix} 1 & 1 & k \\ 1 & k & 1 \\ k & 1 & 1 \end{matrix}\right)=\det\left( \begin{matrix} k+2 & 1 & k \\ k+2 & k & 1 \\ k+2 & 1 & 1 \end{matrix}\right)=\left( \begin{matrix} k+2 & 1 & k \\ 0 & k-1 & 1-k \\ 0 & 0 & 1-k \end{matrix}\right)=-(k+2)(k-1)^2$$ hence

• If $k\ne1$ and $k\ne-2$ there's exactly one solution.
• If $k=1$ we see easily that the $3$ equations aren't compatible so there is no solution.
• If $k=-2$ we can see also easily that there's infinitely many solutions.

Answer 2

Hint: Just add. This gives $(k+1)(x+y+z) = 0$. If $k = -1$, then there are an infinite number of solutions.

Then if $k = 0$... and so on.

Answer 3

If $k=1$, you have equations which say $x+y+z=1$ and $x+y+z=-2$, which is an obvious contradiction, hence no solutions.

If $k=2$ we get the matrix of coefficients

$$\left( \begin{matrix} 1 & 1 & 2 \\ 1 & 2 & 1 \\ 2 & 1 & 1 \end{matrix}\right)$$

which is invertible, and hence there is a unique solution.