Question

If the system of linear equations
$2x+y-z=7$
$x-3y+2z=1$
$x+4y+\delta z=k,$ 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)

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