Question

Let the system of linear equations
$ x+y+ kz =2$
$ 2 x+3 y-z=1 $
$ 3 x+4 y +2 z= k$
have infinitely many solutions. Then the system
$ ( k +1) x+(2 k -1) y=7$
$ (2 k +1) x+( k +5) y=10$
has:

• A. infinitely many solutions
• B. unique solution satisfying $x-y=1$
• C. no solution
• D. unique solution satisfying $x+y=1$

Answer 1

Answer: (D)

$\begin{vmatrix} 1 & 1 & k \\ 2 & 3 & -1 \\ 3 & 4 & 2 \end{vmatrix}=0 $
$ \Rightarrow 1(10)-1(7)+ k (-1)-0$
$ \Rightarrow k =3$
For $k =3,2^{ md }$ system is
$4 x +5 y =7 ....$(1)
and $7 x+8 y=10....$(2)
Clearly, they have a unique solution
(2) $-(1) \Rightarrow 3 x +3 y =3$
$\Rightarrow x+y=1$