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
$D=0,D_1=0,D_2=0,D_3=0$
$D=\left|\begin{array}{ccc}1& -2& 5\\ 2& -1& 1\\ 11& -7& p\end{array}\right|=0\Rightarrow \left(-p+7\right)+2\left(2p-11\right)+5\left(-14+11\right)=0$
$\Rightarrow -p+7+4p-22-15=0$
$\Rightarrow 3p-30=0$
$\Rightarrow p=10$
$D_1=\left|\begin{array}{ccc}3& -2& 5\\ 1& -1& 1\\ q& -7& 10\end{array}\right|=0\Rightarrow 3\left(-10+7\right)+2\left(10-q\right)+5\left(-7+q\right)=0$
$\Rightarrow -9+20-2q-35+5q=0$
$\Rightarrow -24+3q=0$
$q=8$
$D_2=\left|\begin{array}{ccc}1& 3& 5\\ 2& 1& 1\\ 11& q& 10\end{array}\right|=0$ $\Rightarrow \left(10-q\right)-3\left(9\right)+5\left(2q-11\right)$
$\Rightarrow 10+9q-27-55=0$
$\Rightarrow q=8$
$D_3=\left|\begin{array}{ccc}1& -2& 3\\ 2& -1& 1\\ 11& -7& q\end{array}\right|=0$ $\Rightarrow \left(-q+7\right)+2\left(2q-11\right)+3\left(-3\right)=0$
$\Rightarrow q=8$