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

$\text{Given: } k > 0,\quad A = \begin{bmatrix} 2k-1 & 2\sqrt{k} & 2\sqrt{k} \\ 2\sqrt{k} & 1 & -2k \\ -2\sqrt{k} & 2k & -1 \end{bmatrix}$ $B = \begin{bmatrix} 0 & 2k-1 & \sqrt{k} \\ 1-2k & 0 & 2\sqrt{k} \\ -\sqrt{k} & -2\sqrt{k} & 0 \end{bmatrix}$ $\det(\operatorname{adj}A) + \det(\operatorname{adj}B) = 10^6$ $\text{To find: } [k] = \lfloor k \rfloor$ $1.\ \det(\operatorname{adj} M) = (\det M)^{n-1}$ $2.\ \det B,\ \text{then } \det A$ $3.\ \text{solve for } k,\ \text{then } [k]$ $\det(\operatorname{adj} M) = (\det M)^{n-1}$ $n = 3:\quad \det(\operatorname{adj} M) = (\det M)^2$ $B^{T} = -B \quad (\text{skew-symmetric})$ $\det B = 0 \quad (\text{odd order})$ $\det(\operatorname{adj}B) = 0 \ \Rightarrow\ (\det A)^2 = 10^6$ $\Rightarrow\ \det A = \pm 1000$ $\det A = (2k-1)\begin{vmatrix} 1 & -2k \\ 2k & -1 \end{vmatrix} - 2\sqrt{k}\begin{vmatrix} 2\sqrt{k} & -2k \\ -2\sqrt{k} & -1 \end{vmatrix} + 2\sqrt{k}\begin{vmatrix} 2\sqrt{k} & 1 \\ -2\sqrt{k} & 2k \end{vmatrix}$ $\det A = (2k-1)(4k^2-1) + 4k(1+2k) + 4k(1+2k)$ $\det A = (2k+1)\big[(2k-1)^2 + 8k\big] = (2k+1)(2k+1)^2$ $\det A = (2k+1)^3$ $(\det A)^2 = (2k+1)^6 = 10^6$ $2k+1 = 10 \ \ (\text{positive root})$ $2k = 9 \ \Rightarrow\ k = \dfrac{9}{2} = 4.5$ $[k] = \lfloor 4.5 \rfloor = 4$ $\boxed{[k] = 4}$