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Question

Mathematics Question on Determinants

If three-digit numbers A28,3B9A28, 3B9 and 62C62C , where A,BA, B and CC are integers between 00 and 99 , are divisible by a fixed integer kk , then the determinant A36 89C 2B2\begin{vmatrix}A&3&6\\\ 8&9&C\\\ 2&B&2\end{vmatrix} is

A

divisible by kk

B

divisible by k2k^2

C

divisible by 2k2k

D

None of these

Answer

divisible by kk

Explanation

Solution

Given, A28A28, 3B93B9 and 62C62C are divisible by kk
A28=k\therefore A28=k
100A+20+8=k:3B9=k\Rightarrow 100A+20+8=k : 3B9 = k
300+10B+9=k\Rightarrow 300+10B+9=k and 62C=k62C=k
600+20+C=k\Rightarrow 600+20+C=k
Let Δ=A36 89C 2B2\Delta=\left|\begin{matrix}A&3&6\\\ 8&9&C\\\ 2&B&2\end{matrix}\right|
=1100×110100A300600 89C 2010B20=\frac{1}{100}\times\frac{1}{10} \left|\begin{matrix}100 A&300&600\\\ 8&9&C\\\ 20&10 B&20\end{matrix}\right|
Applying R1R1+R2+R3R_{1} \rightarrow R_{1}+R_{2}+R_{3}
=11000=\frac{1}{1000}
100A+20+8300+10B+9600+20+C 89C 2010B20\left|\begin{matrix}100A+20+8&300+10B+9&600+20+C\\\ 8&9&C\\\ 20&10B&20\end{matrix}\right|
=11000kkk 89C 2010B20=\frac{1}{1000} \left|\begin{matrix}k&k&k\\\ 8&9&C\\\ 20&10B&20\end{matrix}\right|
=k100111 89C 2010B20=\frac{k}{100} \left|\begin{matrix}1&1&1\\\ 8&9&C\\\ 20&10B&20\end{matrix}\right|
Hence, it is always divisible by kk