Question

If the three digit numbers A28, 3B9 and 62C when A, B and C are integers between 0 and 9 which are divisible by λ, then ∆ = $\left| \begin{matrix} A & 3 & 6 \\ 8 & 9 & C \\ 2 & B & 2 \end{matrix} \right|$ is divisible by –

• A. λ
• B. λ²
• C. 2λ
• D. None

Answer 1

Answer: (A)

λ

Answer 2

A28 = 100A + 2 . 10 + 8. Apply R₂ + 10R₃ + 100R₁ etc.

∆ = $\left| \begin{matrix} A & 3 & 6 \\ \lambda k_{1} & \lambda k_{2} & \lambda k_{3} \\ 2 & B & 2 \end{matrix} \right|$ = λ∆₁

∴ ∆ is divisible by λ.