Question

The system of simultaneous equations
$\begin{array}{ccc}kx+2y-z=1& & \\ (k-1)y-2z=2& & \\ (k+2)z=3& & \end{array}$
have a unique solution if k equals

• A. $-2$
• B. $-1$
• C. $0$
• D. $1$

Answer 1

The correct option is B $-1$

$kx+2y-z=1$

$(k-1)y-2z=2$

$(k+2)z=3$

$A=\left|\begin{array}{ccc}k& 2& -1\\ 0& k-1& -2\\ 0& 0& k+2\end{array}\right|,D=\left|\begin{array}{c}1\\ 2\\ 3\end{array}\right|$

It has unique solutions if Rank$A=$ Rank$\left(AD\right)=3$

and $k+2=1k+2\ne 0$

$k=-1$ and $k\ne -2$

Option B