Question

If $3^n$ is a factor of the determinant $\left|\begin{array}{ccc}2& 2& 2\\ n_{C_1}& n+3_{C_1}& n+6_{C_1}\\ \frac{n_{C_2}}{2}& \frac{n+3_{C_2}}{2}& \frac{n+6_{C_2}}{2}\end{array}\right|$, then the maximum value of ‘n’ is:

• A. 3
• B. 4
• C. 2
• D. 5

Answer 1

The correct option is A 3

$LetD=\frac{2}{2}\left|\begin{array}{ccc}1& 1& 1\\ n_{C_1}& n+3_{C_1}& n+6_{C_1}\\ n_{C_2}& n+3_{C_2}& n+6_{C_2}\end{array}\right|=\left|\begin{array}{ccc}1& 1& 1\\ n& n+3& n+6\\ \frac{n(n-1)}{2}& \frac{(n+3)(n+2)}{2}& \frac{(n+6)(n+5)}{2}\end{array}\right|=\frac{1}{2}\left|\begin{array}{ccc}1& 1& 1\\ n& n+3& n+6\\ n^2-n& n^2+5n+6& n^2+11n+30\end{array}\right|C_2\to C_2-C_{1}and{C}_3-C_1=\frac{1}{2}\left|\begin{array}{ccc}1& 0& 0\\ n& 3& 6\\ n^2-n& 6n+6& 12n+30\end{array}\right|=\frac{1}{2}\left(3\right(12n+30)-6(6n+6\left)\right)=\frac{1}{2}(36n+90-36n-36)=\frac{1}{2}\left(54\right)=27=27=3^n$
$\Rightarrow$ Maximum value of 'n' will be '3'