Question

If the system of equations

$x-2y+5z=3,$

$2x-y+z=1,$

and $11x-7y+pz=q$

has infinitely many solutions, then

• A. $p+q=2$
• B. $p+q=10$
• C. $p-q=2$
• D. $p-q=5$

Answer 1

Answer: (C)

$\Rightarrow $ for infinite solution
$D=0,D_{1}=0,D_{2}=0,D_{3}=0$
$D=\begin{vmatrix} 1 & -2 & 5 \\ 2 & -1 & 1 \\ 11 & -7 & p \end{vmatrix}=0\Rightarrow \left(- p + 7\right)+2\left(2 p - 11\right)+5\left(- 14 + 11\right)=0$
$\Rightarrow -p+7+4p-22-15=0$
$\Rightarrow 3p-30=0$
$\Rightarrow p=10$
$D_{1}=\begin{vmatrix} 3 & -2 & 5 \\ 1 & -1 & 1 \\ q & -7 & 10 \end{vmatrix}=0\Rightarrow 3\left(- 10 + 7\right)+2\left(10 - q\right)+5\left(- 7 + q\right)=0$
$\Rightarrow -9+20-2q-35+5q=0$
$\Rightarrow -24+3q=0$
$q=8$
$D_{2}=\begin{vmatrix} 1 & 3 & 5 \\ 2 & 1 & 1 \\ 11 & q & 10 \end{vmatrix}=0$ $\Rightarrow \left(10 - q\right)-3\left(9\right)+5\left(2 q - 11\right)$
$\Rightarrow 10+9q-27-55=0$
$\Rightarrow q=8$
$D_{3}=\begin{vmatrix} 1 & -2 & 3 \\ 2 & -1 & 1 \\ 11 & -7 & q \end{vmatrix}=0$ $\Rightarrow \left(- q + 7\right)+2\left(2 q - 11\right)+3\left(- 3\right)=0$
$\Rightarrow q=8$