Determinants Matrices Question 72
Question: If $ {\Delta _{r}}= \begin{vmatrix} r-1 & n & 6 \\ {{(r-1)}^{2}} & 2n^{2} & 4n-2 \\ {{(r-1)}^{2}} & 3n^{3} & 3n^{2}-3n \\ \end{vmatrix}, $ then $ \sum\limits _{r=1}^{n}{{\Delta _{r}}} $ is.
Options:
A) $ 0 $
B) $ 1 $
C) $ 3 $
$ -1 $
D)
Show Answer
Answer:
Correct Answer: A
Solution:
- [a] Since $ C_1 $ has variable terms and $ C_2 $ and $ C_3 $ are constant, summation runs on $ C_1 $ . Therefore, $ \sum\limits _{r=1}^{n}{{\Delta _{r}}}= \begin{vmatrix} \sum\limits_1^{n}{(r-1)} & n & 6 \\ \sum\limits_1^{n}{{{(r-1)}^{2}}} & 2n^{2} & 4n-2 \\ \sum\limits_1^{n}{{{(r-1)}^{3}}} & 3n^{2} & 3n^{2}-3n \\ \end{vmatrix} $
$ = \begin{vmatrix} \frac{1}{2}(n-1)n & n & 6 \\ \frac{1}{6}(n-1)n(2n-1) & 2n^{2} & 4n-2 \\ \frac{1}{4}{{(n-1)}^{2}}n^{2} & 3n^{3} & 3n^{2}-3n \\ \end{vmatrix} $ Taking $ \frac{1}{12}n(n-1) $ common from $ C_1 $ and n common from $ C_2 $ , we get $ \sum{{\Delta _{r}}=\frac{1}{12}n^{2}(n-1)\times \begin{vmatrix} 6 & 1 & 6 \\ 2(2n-1) & 2n & 2(2n-1) \\ 3n(n-1) & 3n^{2} & 3n(n-1) \\ \end{vmatrix}} $
$ =0[\because C_1andC_3areidentical] $
BETA
