Question

The possible combination of values of $k,d\in R$ from the given set of options, for which the system of equations $x+dy+3z=0,kx+2y+2z=0,2x+3y+4z=0$ admits of non-trivial solution is

• A. $k=2$
• B. $d=5/2$
• C. $k=3$
• D. $d=5/4$

Answer 1

Answer: (A), (B)

The correct options are
A $d=5/2$
B $k=2$

$\left|\begin{array}{ccc}1& d& 3\\ k& 2& 2\\ 2& 3& 4\end{array}\right|=0$
$\Rightarrow 2-4d(k-1)+3(3k-4)=0$
$\Rightarrow 2-4dk+4d+9k-12=0$
$\Rightarrow -4dk+4d+9k-9-1=0$
$\Rightarrow -4d(k-1)+9(k-1)=1$
$\Rightarrow (9-4d)(k-1)=1$
The above equation will be satisfied when $(9-4d)=(k-1)=1$
So possible values:
$(k-1)=1\Rightarrow k=2$
and
$(9-4d)=1\Rightarrow d=\frac{5}{2}$