Question

Value of $\alpha$ for which equations
$\alpha x+3y=\alpha -3;12x+\alpha y=\alpha$ has unique solution.

• A. All value of $\alpha$ (except 4 and -4)
• B. All value of $\alpha$ (except 6 and -6)
• C. All value of $\alpha$ (except 2 and -2)
• D. All value of $\alpha$ (except 3 and -3)

Answer 1

Answer: (B)

All value of $\alpha$ (except 6 and -6)

For a unique solution, we must have $\frac{a_1}{a_2}\ne \frac{b_1}{b_2}$
i.e., $\frac{\alpha }{12}\ne \frac{3}{\alpha }\Rightarrow {\alpha }^2\ne 36\Rightarrow \alpha \ne 6or-6$
Hence, for all valus of $\alpha$ (except $6$ and $-6$).

The given pair of linear equations has a unique solution.