Question

The simultaneous equations
$ k x+2 y-z=1 $
$(k-1) y-2 z=2 $
$(k+2) z=3$
have only one solution when

• A. $k=-2$
• B. $k=-1$
• C. $k=0$
• D. $k=1$

Answer 1

Answer: (B)

The system of given equations is
$k x+2 y-z=1 $
$(k-1) y-2 z=2 $
and $(k+2) z=3$
This system of equations has unique solution, if
$ \begin{vmatrix} k & 2 & -1 \\0 & k-1 & -2 \\0 & 0 & k+2\end{vmatrix} \neq 0 $
$\Rightarrow (k+2)(k)(k-1) \neq 0$
$\Rightarrow k \neq-2,0,1$
i.e., $k=-1$, is a required answer.