Question

The set of equations $x+y+3z=2,2x+y+z=4,x+2y+\alpha z=3$ has

• A. unique solution only for $\alpha =0$
• B. unique solution for $\alpha \ne 8$
• C. infinite number of solution of $\alpha =8$
• D. no solution of $\alpha =8$

Answer 1

The correct option is D no solution of $\alpha =8$

$\Delta =\left|\begin{array}{ccc}1& 1& 3\\ 2& 1& 1\\ 1& 2& \alpha \end{array}\right|=1(\alpha -2)-1(2\alpha -1)+3(4-1)=\alpha -2-2\alpha +1+9=-\alpha +8$
For unique solution $\Delta \ne 0\Rightarrow \alpha \ne 8$
For no solution $\Delta =0\Rightarrow \alpha =8$