To solve the problem, we need to find the value of \( k \) given the matrices \( A \) and \( B \) and the condition that \( \text{det(Adj(A))} + \text{det(Adj(B))} = 10^6 \).
### Step 1: Calculate the determinant of matrix \( A \)
The matrix \( A \) is given by:
\[
A = \begin{pmatrix}
2k - 1 & 2\sqrt{k} & 2\sqrt{k} \\
2\sqrt{k} & 1 & -2k \\
-2\sqrt{k} & 2k & -1
\end{pmatrix}
\]
To find \( \text{det(A)} \), we can use the formula for the determinant of a \( 3 \times 3 \) matrix:
\[
\text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg)
\]
Where \( a, b, c, d, e, f, g, h, i \) are the elements of the matrix \( A \).
Calculating the determinant step-by-step:
1. \( a = 2k - 1 \), \( b = 2\sqrt{k} \), \( c = 2\sqrt{k} \)
2. \( d = 2\sqrt{k} \), \( e = 1 \), \( f = -2k \)
3. \( g = -2\sqrt{k} \), \( h = 2k \), \( i = -1 \)
Now substituting into the determinant formula:
\[
\text{det}(A) = (2k - 1)((1)(-1) - (-2k)(2k)) - (2\sqrt{k})((2\sqrt{k})(-1) - (-2k)(-2\sqrt{k})) + (2\sqrt{k})((2\sqrt{k})(2k) - (1)(-2\sqrt{k}))
\]
Calculating each term:
1. First term: \( (2k - 1)(-1 + 4k^2) = (2k - 1)(4k^2 - 1) \)
2. Second term: \( -2\sqrt{k}(-2\sqrt{k} + 4k) = 2\sqrt{k}(2\sqrt{k} - 4k) \)
3. Third term: \( 2\sqrt{k}(4k\sqrt{k} + 2\sqrt{k}) = 2\sqrt{k}(6k\sqrt{k}) = 12k^2 \)
Combining these terms gives us:
\[
\text{det}(A) = (2k - 1)(4k^2 - 1) + 2\sqrt{k}(2\sqrt{k} - 4k) + 12k^2
\]
### Step 2: Calculate the determinant of matrix \( B \)
The matrix \( B \) is given by:
\[
B = \begin{pmatrix}
0 & 2k - 1 & \sqrt{k} \\
1 - 2k & 0 & 0 \\
-\sqrt{k} & -2\sqrt{k} & 0
\end{pmatrix}
\]
Since the first column has a zero, we can use the determinant property that if a column (or row) is all zeros, the determinant is zero. Thus:
\[
\text{det}(B) = 0
\]
### Step 3: Calculate the determinants of the adjoints
The determinant of the adjoint of a matrix is given by:
\[
\text{det(Adj(A))} = \text{det(A)}^{n-1}
\]
Where \( n \) is the order of the matrix. For a \( 3 \times 3 \) matrix, \( n = 3 \), so:
\[
\text{det(Adj(A))} = \text{det(A)}^{2}
\]
And since \( \text{det(B)} = 0 \):
\[
\text{det(Adj(B))} = 0
\]
### Step 4: Set up the equation
Now we can substitute these into the given equation:
\[
\text{det(Adj(A))} + \text{det(Adj(B))} = 10^6
\]
This simplifies to:
\[
\text{det(A)}^{2} + 0 = 10^6
\]
Thus:
\[
\text{det(A)}^{2} = 10^6
\]
Taking the square root gives:
\[
\text{det(A)} = 10^3 = 1000
\]
### Step 5: Solve for \( k \)
Now we need to solve \( \text{det(A)} = 1000 \). We previously derived the expression for \( \text{det(A)} \). We set it equal to 1000 and solve for \( k \):
\[
(2k + 1)^3 = 1000
\]
Taking the cube root:
\[
2k + 1 = 10
\]
Thus:
\[
2k = 9 \implies k = \frac{9}{2} = 4.5
\]
### Final Answer
The value of \( k \) is:
\[
\boxed{4.5}
\]
