Question

If the system of linear equations
$2 x+y-z=7$
$x-3 y+2 z=1 $
$x+4 y+\delta z=k, $ where $ \delta, k \in R$
has infinitely many solutions, then $\delta+ k$ is equal to:

• A. $-3$
• B. 3
• C. 6
• D. 9

Answer 1

Answer: (B)

$\begin{vmatrix} 2 & 1 & -1 \\ 1 & -3 & 2 \\ 1 & 4 & \delta\end{vmatrix}=0$
$\Rightarrow \delta=-3$
And $\begin{vmatrix} 7 & 1 & -1 \\1 & -3 & 2 \\K & 4 & -3\end{vmatrix}=0 $ $\Rightarrow K =6$
$\Rightarrow \delta+ K =3$
Alternate
$2 x+y-z=7....$(1)
$x-3 y+2 z=1.....$(2)
$x+4 y+\delta z=k .....$(3)
Equation (2) $+(3)$
We get $2 x + y +(2+\delta) z =1+ K ....$(4)
For infinitely solution
Form equation (1) and (4)
$2+\delta=-1 $
$\Rightarrow \delta=-3$
$1+ k =7 $
$\Rightarrow k =6 $
$\delta+ k =3$