Question

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

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