The system of simultaneous linear equations $kx+2y-z=1,$ $(\mathrm{k}-1)\mathrm{y}-2\mathrm{z}=2\text{ and }(\mathrm{k}+2)\mathrm{z}=3$have a unique solution if K equals :

The system of equations$kx+2y-z= 1,$

$(\mathrm{k}-1)\mathrm{y}-2\mathrm{z}=2,(\mathrm{k}+2)\mathrm{z}=3$ will have a unique solution if $∆\ne 0$

$\begin{array}{l}\therefore \left|\begin{array}{ccc}k& 2& -1\\ 0& k-1& -2\\ 0& 0& k+2\end{array}\right|\ne 0\\ \Rightarrow k(k-1)(k+2)\ne 0\\ \therefore k\ne 0,k\ne 1,k\ne -2\end{array}$

$\therefore$ From the choices, we have $k = - 1$