Question

If $3^n$ is a factor of the determinant $\left|\begin{array}{ccc}1& 1& 1\\ {}^n{C}_1& {}^{n+3}{C}_1& {}^{n+6}{C}_1\\ {}^n{C}_2& {}^{n+3}{C}_2& {}^{n+6}{C}_2\end{array}\right|$, then the maximum value of $n$ is

• A. 7
• B. 5
• C. 3
• D. 1

Answer 1

Answer: (C)

$\Delta =\left|\begin{array}{ccc}1& 1& 1\\ {}^n{C}_1& {}^{n+3}{C}_1& {}^{n+6}{C}_1\\ {}^n{C}_2& {}^{n+3}{C}_2& {}^{n+6}{C}_2\end{array}\right|$
$=\frac{1}{2}\left|\begin{array}{ccc}1& 1& 1\\ n& \left(n+3\right)& \left(n+6\right)\\ n\left(n-1\right)& \left(n+3\right)\left(n+2\right)& \left(n+6\right)\left(n+5\right)\end{array}\right|$
Applying $C_2\to C_2-C_1$ and $C_3\to C_3-C_1$, we get
$=\frac{1}{2}\left|\begin{array}{ccc}1& 0& 0\\ n& 3& 6\\ n^2-n& 6n+6& 12n+30\end{array}\right|$
$=27=3^3$
$\therefore n=3$