Matrices and Determinants 2 Question 38

42. Let the three digit numbers A28,3B9 and 62C, where A, B and C are integers between 0 and 9 , be divisible by a fixed integer k. Show that the determinant

|A3689C2B2| is divisible by k.

(1990,4 M)

Show Answer

Solution:

  1. We know, A28=A×100+2×10+8

3B9=3×100+B×10+9

and 62C=6×100+2×10+C

Since, A28,3B9 and 62C are divisible by k, therefore there exist positive integers m1,m2 and m3 such that, 100×A+10×2+8=m1k,100×3+10×B+9=m2k and 100×6+10×2+C=m3k

Δ=|A3689C2B2|

Applying R2100R1+10R3+R2

Δ=∣|A3100A+2×10+8100×3+10×B+92B|

|6100×6+10×2+C2|=|A36A283B962C2B2|

[from Eq. (i)]

=|A36m1km2km3k2B2|=k|A36m1m2m32B2|

Δ=mk

Hence, determinant is divisible by k.



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