Question

If the system of linear equations
$2x+y-z=3$
$x-y-z=\alpha$
$3x+3y+\beta z=3$
has infinitely many solutions, then $(\alpha +\beta -\alpha \beta )$ is equal to

Answer 1

For infinite solutions :
$\Delta =\left|\begin{array}{ccc}2& 1& -1\\ 1& -1& -1\\ 3& 3& \beta \end{array}\right|=0$
$\Rightarrow 2(-\beta +3)-1(\beta +3)-1(3+3)=0$
$\Rightarrow -2\beta +6-\beta -3-6=0$
$\therefore \beta =-1$
and
${\Delta }_1=\left|\begin{array}{ccc}3& 1& -1\\ \alpha & -1& -1\\ 3& 3& -1\end{array}\right|=0$
$\Rightarrow 3(1+3)-1(-\alpha +3)-1(3\alpha +3)=0$
$\Rightarrow 12+\alpha -3-3\alpha -3=0$
$\therefore \alpha =3$
Hence, $\alpha +\beta -\alpha \beta =5$