Does this system of linear equations have infinite solutions?

Question

$$x(k+2)+y(k−1)+z(k)=2$$ $$y(k+2)+2z=0$$ $$z(k^2+k−2)=k+2$$

Is there any value of k for which this system of linear equations would have infinite solutions? I mean, it seems as if it does when k = -2, but it also seems as if it has no solutions when k = -2. I'm looking at this from a row-echelon form augmented matrix point of view and I really don't know.

Answer 1

If $k=-2$, the system has rank $2$: $$\begin{cases}-3y-2z=2\\2z=0\end{cases}$$ hence the solutions are $\{(x,-\frac23,0)\mid x\in\mathbf R\}$. It does have an infinite set of solutions, the straight line $\;(0,-\frac23,0)+x(1,0,0)$.

If $k=1$, the system also has rank $2$: $$\begin{cases}3x+z=2\\3y+2z=0\\0=3\end{cases}$$ but the last line is not identically $0$, hence there is no solution.

On all other cases, the ystem has rank $3$ and has a unique solution.

Answer 2

hope you will understand this.