Question
Mathematics Question on Determinants
If three-digit numbers A28,3B9 and 62C , where A,B and C are integers between 0 and 9 , are divisible by a fixed integer k , then the determinant A 8 239B6C2 is
A
divisible by k
B
divisible by k2
C
divisible by 2k
D
None of these
Answer
divisible by k
Explanation
Solution
Given, A28, 3B9 and 62C are divisible by k
∴A28=k
⇒100A+20+8=k:3B9=k
⇒300+10B+9=k and 62C=k
⇒600+20+C=k
Let Δ=A 8 239B6C2
=1001×101100A 8 20300910B600C20
Applying R1→R1+R2+R3
=10001
100A+20+8 8 20300+10B+9910B600+20+CC20
=10001k 8 20k910BkC20
=100k1 8 201910B1C20
Hence, it is always divisible by k