Find Floor Value of k from Adjoint Determinants Equation

To find the value of $[k]$, we first determine the determinants of matrices $A$ and $B$.

For any $n \times n$ matrix $M$, the determinant of its adjoint is given by:
$\det(\text{adj } M) = (\det M)^{n-1}$

For $3 \times 3$ matrices $A$ and $B$, this becomes:
$\det(\text{adj } A) = (\det A)^2 \quad \text{and} \quad \det(\text{adj } B) = (\det B)^2$

Step 1: Evaluation of $\det(B)$

Consider the matrix $B$:

$$B = \begin{bmatrix} 0 & 2k-1 & \sqrt{k} \\ 1-2k & 0 & 2\sqrt{k} \\ -\sqrt{k} & -2\sqrt{k} & 0 \end{bmatrix}$$

Taking the transpose of $B$:

$$B^T = \begin{bmatrix} 0 & 1-2k & -\sqrt{k} \\ 2k-1 & 0 & -2\sqrt{k} \\ \sqrt{k} & 2\sqrt{k} & 0 \end{bmatrix} = -B$$

Since $B^T = -B$, $B$ is a skew-symmetric matrix of order $3$ (odd order). Using the determinant properties:
$\det(B) = \det(B^T) = \det(-B) = (-1)^3 \det(B) = -\det(B)$
$\implies 2\det(B) = 0 \implies \det(B) = 0$

Therefore:
$\det(\text{adj } B) = (\det B)^2 = 0$

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Step 2: Evaluation of $\det(A)$

Consider the matrix $A$:

$$A = \begin{bmatrix} 2k-1 & 2\sqrt{k} & 2\sqrt{k} \\ 2\sqrt{k} & 1 & -2k \\ -2\sqrt{k} & 2k & -1 \end{bmatrix}$$

Expanding $\det(A)$ along the first row:
$\det(A) = (2k-1)\left[1(-1) - (-2k)(2k)\right] - 2\sqrt{k}\left[2\sqrt{k}(-1) - (-2k)(-2\sqrt{k})\right] + 2\sqrt{k}\left[2\sqrt{k}(2k) - 1(-2\sqrt{k})\right]$

Simplifying each term:

1. $(2k-1)(-1 + 4k^2) = (2k-1)(4k^2-1) = (2k-1)^2(2k+1)$
2. $-2\sqrt{k}\left[-2\sqrt{k} - 4k\sqrt{k}\right] = 4k(1+2k) = 4k(2k+1)$
3. $2\sqrt{k}\left[4k\sqrt{k} + 2\sqrt{k}\right] = 4k(2k+1)$

Adding all three terms together:
$\det(A) = (2k-1)^2(2k+1) + 4k(2k+1) + 4k(2k+1)$
$\det(A) = (2k+1) \left[ (2k-1)^2 + 8k \right]$
$\det(A) = (2k+1) \left[ 4k^2 - 4k + 1 + 8k \right]$
$\det(A) = (2k+1) \left[ 4k^2 + 4k + 1 \right]$
$\det(A) = (2k+1)(2k+1)^2 = (2k+1)^3$

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Step 3: Solving for $k$

We are given:
$\det(\text{adj } A) + \det(\text{adj } B) = 10^6$

Substituting $\det(\text{adj } A) = (\det A)^2$ and $\det(\text{adj } B) = 0$:
$\left((2k+1)^3\right)^2 + 0 = 10^6$
$(2k+1)^6 = 10^6$

Since $k > 0$, we have $2k+1 > 0$. Taking the $6^{\text{th}}$ root on both sides:
$2k+1 = 10$
$2k = 9 \implies k = 4.5$

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Step 4: Finding $[k]$

The floor function $[k]$ denotes the greatest integer less than or equal to $k$:
$[k] = [4.5] = 4$