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Determinants — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsDeterminants1 Mark
MCQ
If $\text{D}_\text{k}=\begin{vmatrix}1&\text{n}&\text{n}\\2\text{k}&\text{n}^2+\text{n}+2&\text{n}^2+\text{n}\\2\text{k}-1&\text{n}^2&\text{n}^2+\text{n}+2\end{vmatrix} $ and $\sum\limits_{\text{k}=1}^\text{n}\text{D}_\text{k}=48,$ then $n$ equals:
  • ✓
    $4$
  • B
    $6$
  • C
    $8$
  • D
    None of these.
✓

Answer

Correct option: A.
$4$
$\text{D}_\text{k}=\begin{vmatrix}1&\text{n}&\text{n}\\2\text{k}&\text{n}^2+\text{n}+2&\text{n}^2+\text{n}\\2\text{k}-1&\text{n}^2&\text{n}^2+\text{n}+2\end{vmatrix} $
$\text{D}_\text{k}=\begin{vmatrix}1&\text{n}&\text{n}\\1&\text{n}+2&-2\\2\text{k}-1&\text{n}^2&\text{n}^2+\text{n}+2\end{vmatrix}\ \text{Applying}\ \text{R}_2\rightarrow\text{R}_2-\text{R}_3$
$\text{D}_\text{k}=\begin{vmatrix}1&\text{n}&\text{n}\\0&2&-2-\text{n}\\2\text{k}-1&\text{n}^2&\text{n}^2+\text{n}+2\end{vmatrix}\ \text{Applying}\ \text{R}_2\rightarrow\text{R}_2-\text{R}_1$
Now,
$\sum\limits_{\text{k}=1}^\text{n}\text{D}_\text{k}=\begin{vmatrix}1&\text{n}&\text{n}\\0&2&-2-\text{n}\\1&\text{n}^2&\text{n}^2+\text{n}+2\end{vmatrix}+\begin{vmatrix}1&\text{n}&\text{n}\\0&2&-2-\text{n}\\3&\text{n}^2&\text{n}^2+\text{n}+2\end{vmatrix}+.......+\begin{vmatrix}1&\text{n}&\text{n}\\0&2&-2-\text{n}\\2\text{n}-1&\text{n}^2&\text{n}^2+\text{n}+2\end{vmatrix}$
$\sum\limits_{\text{k}=1}^\text{n}\text{D}_\text{k}=\Big[1\Big(2(\text{n}^2+\text{n}+2)+\text{n}^2(2+\text{n})\Big)+\text{n}\Big(0+1(-2-\text{n})\Big)+\text{n}\Big(0+2\Big)\Big]\\+\Big[1\Big(2(\text{n}^2+\text{n}+2)+\text{n}^2(2+\text{n})\Big)+\text{n}\Big(0+3(-2-\text{n})\Big)+\text{n}\Big(0+6\Big)\Big]+......\\+\Big[1\Big(2(\text{n}^2+\text{n}+2)+\text{n}^2(2+\text{n})\Big)+\text{n}\Big(0+(2\text{n}-1)(-2-\text{n})\Big)+\text{n}\Big(0+2(2\text{n}-1)\Big)\Big]$
$\sum\limits_{\text{k}=1}^\text{n}\text{D}_\text{k}=\Big[1\Big(2(\text{n}^2+\text{n}+2)+\text{n}^2(2+\text{n})\Big)+\text{n}\Big(-2-\text{n}\Big)-2\text{n}\Big]\Big(1+3+5+.....+\text{n}\Big)$
$\sum\limits_{\text{k}=1}^\text{n}\text{D}_\text{k}=\Big[1\Big(2(\text{n}^2+\text{n}+2)+\text{n}^2(2+\text{n})\Big)+\text{n}\Big(-2-\text{n}\Big)-2\text{n}\Big]\Big(\text{n}^2\Big)$
$\Rightarrow\ 2\text{n}^2+4\text{n}=48$
$\Rightarrow\ (\text{n}+6)(\text{n}-4)=0$
$\Rightarrow\ \text{n}=4$

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