$$x(k+2) + y(k-1) + z(k) = 2$$ $$y(k+2) + 2z(k) = 0$$ $$ z(k^2 + k -2) = k + 2$$
Determine the values of k for which the system has:
Exactly one solution, Infinitely many solutions, No solutions
I'm pretty sure there's exactly one solution when k != 1,-2. But I'm not sure about the other 2 conditions. I put this in augmented matrix form and realized that the system is inconsistent when k = 1. I also noticed that when k = -2, we have a row consisting entirely of 0's and so I was thinking this would mean infinitely many solutions, but my results didn't look right. I would love some help, thanks in advance.
Hint: Notice, the general solution by Crammer's Rule is given by
$$\frac{x}{\Delta_x}=\frac{y}{\Delta_y}=\frac{z}{\Delta_z}=\frac{1}{\Delta}$$
where, determinants are as follows $$\Delta_x=\left|\begin{array}\\ \color{red}{2}&(k-1)&k\\ \color{red}{0}&(k+2)& 2k\\ \color{red}{(k+2)}&0&(k^2+k-2) \end{array}\right|$$
$$\Delta_y=\left|\begin{array}\\ (k+2)&\color{red}{2}&k\\ 0&\color{red}{0}& 2k\\ 0&\color{red}{(k+2)}&(k^2+k-2) \end{array}\right|$$
$$\Delta_z=\left|\begin{array}\\ (k+2)&(k-1)&\color{red}{2}\\ 0&(k+2)& \color{red}{0}\\ 0&0&\color{red}{(k+2)} \end{array}\right|$$
$$\Delta=\left|\begin{array}\\ (k+2)&(k-1)&k\\ 0&(k+2)& 2k\\ 0&0&(k^2+k-2) \end{array}\right|=(k+2)^2(k^2+k-2)$$
1) For unique solution, we have $\Delta\neq 0$
2)For infinite solutions, we have $\Delta=0$, $\Delta_x=0$, $\Delta_y=0$, & $\Delta_z=0$
3)For no solution, we have $\Delta=0$ & at least one of $\Delta_x$, $\Delta_y$ & $\Delta_z$ is zero
I hope you can solve for $k$ by applying the above conditions.