Question

The set of equation
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 solutions for $\alpha$ = 8
• D. No solution for $\alpha$ = 8

Answer 1

Answer: (B), (D)

The correct options are
B Unique solution for $\alpha \ne 8$
D No solution for $\alpha$ = 8

If $D=\left|\begin{array}{ccc}1& -1& 3\\ 2& -1& 1\\ 1& -2& \alpha \end{array}\right|$
$\Rightarrow (-\alpha +2)+1(2\alpha -1)+3(-4+1)=0$
$\Rightarrow \alpha =8$
$\Rightarrow$ Hence no solution for $\alpha =8$ and unique solution for $\alpha \ne 8$