Determinants Matrices Question 66
Question: If $ A_1B_1C_1, $
$ A_2B_2C_2 $ and $ A_3B_3C_3 $ are three digit numbers, each of which is divisible by k, then $ \Delta = \begin{vmatrix} A_1 & B_1 & C_1 \\ A_2 & B_2 & C_2 \\ A_3 & B_3 & C_3 \\ \end{vmatrix} $ is
Options:
A) Divisible by k
B) Divisible by $ k^{2} $
C) Divisible by $ k^{3} $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
- [a] Since, $ A_1B_1C_1,A_2B_2C_2 $ and $ A_3B_3C_3 $ are divisible by k, therefore; $ 100A_1+10B_1+C_1=n_1k; $
$ 100A_2+10B_2+C_2 $
$ =n_2k;100A_3+10B_3+C_3=n_3k $ (where $ n_1,n_2,n_3 $ are integers) $ Now,\Delta = \begin{vmatrix} A_1 & B_1 & C_1 \\ A_2 & B_2 & C_2 \\ A_3 & B_3 & C_3 \\ \end{vmatrix} $
$ = \begin{vmatrix} A_1 & B_1 & 100A_1+10B_1+C_1 \\ A_2 & B_2 & 100A_2+10B_2+C_2 \\ A_3 & B_3 & 100A_3+10B_2+C_3 \\ \end{vmatrix} $
[Applying $ C_3\to C_3+10C_2+100C_1 $ ] $ = \begin{vmatrix} A_1 & B_1 & n_1k \\ A_2 & B_2 & n_2k \\ A_3 & B_3 & n_3k \\ \end{vmatrix}=k \begin{vmatrix} A_1 & B_1 & n_1 \\ A_2 & B_2 & n_2 \\ A_3 & B_3 & n_3 \\ \end{vmatrix}=k{\Delta_1} $
$ \Rightarrow \Delta $ is divisible by k [Since, elements of $ {\Delta_1} $ are integers
$ \therefore {\Delta_1} $ is an integer.]
BETA
