Matrices and Determinants 2 Question 39
«««< HEAD
43. Let $\Delta _a=\left(\begin{array}{lll}(a-1)^{2} & 2 n^{2} \quad 4 n-2\end{array}\right.$
$(a-1)^{3} \quad 3 n^{3} \quad 3 n^{2}-3 n$
####43. Let $\Delta _a=\left|\begin{array}{ccc}a-1 & n & 6 \\ (a-1)^{2} & 2 n^{2} & 4 n-2 \\ (a-1)^{3} & 3 n^{3} & 3 n^{2}-3 n\end{array}\right|$
3e0f7ab6f6a50373c3f2dbda6ca2533482a77bed
Show that $\sum _{a=1}^{n} \Delta _a=c \in$ constant.
$(1989,5 M)$
Show Answer
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 $]$
