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  4. If 3n is a factor of the determinant |1 &1&1 nC1&n+3C1&n+6C1 nC2&n+3C2&n+6C2|, then the maximum value of n is

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Q. If $3^n$ is a factor of the determinant $\begin{vmatrix}1 &1&1\\ ^{n}C_{1}&^{n+3}C_{1}&^{n+6}C_{1}\\ ^{n}C_{2}&^{n+3}C_{2}&^{n+6}C_{2}\end{vmatrix}$, then the maximum value of $n$ is

COMEDKCOMEDK 2006Determinants

Solution:

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