Let $k$ be a positive real number and let $$ \begin{aligned} A=\left[\begin{array}{ccc} 2 k-1 & 2 \s

Question

Let $k$ be a positive real number and let $$ \begin{aligned} A=\left[\begin{array}{ccc} 2 k-1 & 2 \sqrt{k} & 2 \sqrt{k} \ 2 \sqrt{k} & 1 & -2 k \ -2 \sqrt{k} & 2 k & -1 \end{array}ight] 	ext { and } \ B=\left[\begin{array}{ccc} 0 & 2 k-1 & \sqrt{k} \ 1-2 \sqrt{k} & 0 & 2 \sqrt{k} \ -\sqrt{k} & -2 \sqrt{k} & 0 \end{array}ight] . \end{aligned} $$ If $\operatorname{det}(\operatorname{adj} A)+\operatorname{det}(\operatorname{adj} B)=10^{6}$, then $[k]$ is equal to [Note : adj $M$ denotes the adjoint of a square matrix $M$ and $[k]$ denotes the largest integer less than or equal to $k$ ].

Professor PC Thomas Classes · Answer

(4) $: \operatorname{det} A=\left|\begin{array}{ccc}2 k-1 & 2 \sqrt{k} & 2 \sqrt{k} \ 2 \sqrt{k} & 1 & -2 k \ -2 \sqrt{k} & 2 k & -1\end{array}ight|$\ $=\left|\begin{array}{ccc}2 k-1 & 2 \sqrt{k} & 2 \sqrt{k} \ 2 \sqrt{k} & 1 & -2 k \ 0 & (2 k+1) & -(2 k+1)\end{array}ight|\left[R_{3} ightarrow R_{3}+R_{2}ight]$\ $=(2 k+1)\left|\begin{array}{ccc}2 k-1 & 2 \sqrt{k} & 2 \sqrt{k} \ 2 \sqrt{k} & 1 & -2 k \ 0 & 1 & -1\end{array}ight|$\ $=(2 k+1)\left|\begin{array}{ccc}2 k-1 & 4 \sqrt{k} & 2 \sqrt{k} \ 2 \sqrt{k} & 1-2 k & -2 k \ 0 & 0 & -1\end{array}ight|$\ $\left[C_{2} ightarrow C_{2}+C_{3}ight]$\ $=(2 k+1)^{3} \quad$ (By expansion)\ $\operatorname{det} B=0 \quad(\because B$ is skew symmetric matrix of odd order)\ $\operatorname{det}(\operatorname{adj} A)=(\operatorname{det} A)^{n-1}=\left\{(2 k+1)^{3}ight\}^{2}$\ $=(2 k+1)^{6}$\ Now $(2 k+1)^{6}=10^{6}$\ $\Rightarrow \quad 2 k+1=10$\ $\Rightarrow \quad 2 k=9$. Thus $[k]=4$\