For what value of the constants k does the system have (i) no solutions, (ii) infinitely many solutions, (iii) a unique solution?
$$ x − 2y + z = 7\\ x − 2y − kz = k\\ kx − 2y + kz = 7 $$
At first I converted it to augmented matrix: $$\left(\begin{array}{ccc|c} 1&-2&1&7\\ 1&-2&-k&k\\ k&-2&k&7 \end{array}\right)$$
Then I did some row operations and got this: $$\left(\begin{array}{ccc|c} 1&-2&1&7\\ 0&1&0&\frac{-7-7k}{-2-2k}\\ 0&0&1&\frac{k-7}{k-1} \end{array}\right)$$
I have no idea what to do now, I tried looking for any ways to solve this on the internet but couldn't find any. I only managed to figure out that with if division from 0 occurs on the right, then I have infinitely many solutions. (k = 1)
Answers for this exercise are: $$(i) k = −1\\ (ii) k = 1.\\ (iii) k = ±1.$$ I want to know what are the steps to solve this.
Label the equations (1) $x-2y+z=7$, (2) $x-2y-kz=k$, (3) $kx-2y+kz=7$.
If $k=-1$, then (1) and (2) contradict, so there are no solutions. If $k=1$ the last equation is the same as the fist, so there are infinitely many solutions. So assume $k\ne\pm1$.
Taking $k$(1)-(3) we have $(-2k+2)y=7k-7$. Since $k\ne1$, we can divide by $k-1$ to get $y=-\frac{7}{2}$. Taking (1)-(2) we have $(k+1)z=7-k$. Since $k\ne-1$ we can divide by $k+1$ to get $z=\frac{7-k}{k+1}$. Then (1) gives $x=-z=\frac{k-7}{k+1}$. So in this case we have found a unique solution.