Matrices and Determinants 2 Question 39
43. Let $\Delta_{a}=\left|\begin{array}{ccc}a-1 & n & 6 \\ (a-1)^{2} & 2 n^{2} & 4 n-2 \\ (a-1)^{3} & 3 n^{3} & 3 n^{2}-3 n\end{array}\right|$
Show that $\sum_{a=1}^{n} \Delta_{a}=c \in$ constant.
$(1989,5 M)$
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Solution:
- Given, $\Delta_{a}=\left|\begin{array}{ccc}a-1 & n & 6 \\ (a-1)^{2} & 2 n^{2} & 4 n-2 \\ (a-1)^{3} & 3 n^{3} & 3 n^{2}-3 n\end{array}\right|$
$ \therefore \sum_{a=1}^{n} \Delta_{a}=$ $ \begin{vmatrix} \sum_{a=1}^{n}(a-1) & n & 6 \\ \sum_{a=1}^{n}(a-1)^{2} & 2 n^{2} & 4 n-2 \\ \sum_{a=1}^{n}(a-1)^{3} & 3 n^{3} & 3 n^{2}-3 n \end{vmatrix} $
$ \begin{aligned} & =\left|\begin{array}{ccc} \frac{n(n-1)}{2} & n & 6 \\ \frac{n(n-1)(2 n-1)}{6} & 2 n^{2} & 4 n-2 \\ \frac{n^{2}(n-1)^{2}}{4} & 3 n^{3} & 3 n^{2}-3 n \end{array}\right| \\ & =\frac{n^{2}(n-1)}{2}\left|\begin{array}{ccc} 1 & 1 & 6 \\ \frac{\left(2 n^{3}-1\right)}{3} & 2 n & 4 n-2 \\ \frac{n\left(n^{-1)}\right.}{2} & 3 n^{2} & 3 n^{2}-3 n \end{array}\right| \\ & =\frac{n^{3}(n-1)}{12}\left|\begin{array}{ccc} 1 & 1 & 6 \\ 2 n-1 & 6 n & 12 n-6 \\ n-1 & 6 n & 6 n-6 \end{array}\right| \end{aligned} $
Applying $C_{3} \rightarrow C_{3}-6 C_{1}$
$ =\frac{n^{3}(n-1)}{12}\left|\begin{array}{ccc} 1 & 1 & 0 \\ 2 n-1 & 6 n & 0 \\ n-1 & 6 n & 0 \end{array}\right|=0 $
$\Rightarrow \quad \sum_{a=1}^{n} \Delta_{a}=c$
$[c=0$, i.e. constant $]$
