3 and 4 .Determinants and Matrices
hard

Let $D _{ k }=\left|\begin{array}{ccc}1 & 2 k & 2 k -1 \\ n & n ^2+ n +2 & n ^2 \\ n & n ^2+ n & n ^2+ n +2\end{array}\right|$. If $\sum \limits_{ k =1}^n$ $D _{ k }=96$, then $n$ is equal to

A

$3$

B

$5$

C

$4$

D

$6$

(JEE MAIN-2023)

Solution

$D _{ k }=\left|\begin{array}{ccc}1 & 2 k & 2 k -1 \\ n & n ^2+ n +2 & n ^2 \\ n & n ^2+ n & n ^2+ n +2\end{array}\right|$

$\sum \limits_{ k =1}^{ n } D _{ k }=96 \Rightarrow$

$\left|\begin{array}{ccc}\sum \limits_{ k =1}^{ n } 1 & \sum_{2 k} & \sum(2 k-1) \\ n & n ^2+ n +2 & n ^2 \\ n & n ^2+ n & n ^2+ n +2\end{array}\right|=96$

$\Rightarrow\left|\begin{array}{ccc}n & n^2+n & n^2 \\ n & n^2+n+2 & n^2 \\ n & n^2+n & n^2+n+2\end{array}\right|=96$

$R _2 \rightarrow R _2- R _1$ and $R _3 \rightarrow R _3- R _1$

$\left|\begin{array}{ccc}n & n^2+n & n^2 \\ 0 & 2 & 0 \\ 0 & 0 & n+2\end{array}\right|=96$

$\Rightarrow n (2 n +4)=96 \Rightarrow n ( n +2)=48 \Rightarrow n =6$

Standard 12
Mathematics

Similar Questions

Start a Free Trial Now

Confusing about what to choose? Our team will schedule a demo shortly.