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JEE Advance 2010
Paper-2
Question
Let k be a positive real number and let $\mathrm{A}=\left[\begin{array}{ccc}2 k-1 & 2 \sqrt{k} & 2 \sqrt{k} \\ 2 \sqrt{k} & 1 & -2 k \\ -2 \sqrt{k} & 2 k & -1\end{array}\right]$ and $\mathrm{B}=\left[\begin{array}{ccc}0 & 2 k-1 & \sqrt{k} \\ 1-2 k & 0 & 2 \sqrt{k} \\ -\sqrt{k} & -2 \sqrt{k} & 0\end{array}\right]$. If det $(\operatorname{adj} \mathrm{A})+\operatorname{det}(\operatorname{adj} \mathrm{B})=10^6$, then $[\mathrm{k}]$ is equal to [Note : adj M denotes the adjoint of a square matrix M and $[\mathrm{k}]$ denotes the largest integer less than or equal to k$]$.
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Solution
$\begin{aligned} & |A|=(2 k+1)^3,|B|=0 \quad(\text { Since } B \text { is a skew-symmetric matrix of order } 3) \\ & \Rightarrow \operatorname{det}(\operatorname{adj} A)=|A|^{n-1}=\left((2 k+1)^3\right)^2=106 \Rightarrow 2 k+1=10 \Rightarrow 2 k=9 \\ & {[k]=4}\end{aligned}$
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