Let $k$ be a positive real number and let $A = \begin{bmatrix} 2k & 2\sqrt{k} & 2\sqrt{k} \\ 2\sqrt{k} & 1 & -2k \\ -2\sqrt{k} & 2k & -1 \end{bmatrix}$ and $B = \begin{bmatrix} 0 & 2k-1 & \sqrt{k} \\ 1-2k & 0 & 2\sqrt{k} \\ -\sqrt{k} & -2\sqrt{k} & 0 \end{bmatrix}$. If $\det(\text{adj }A) + \det(\text{adj }B) = 10^6$, then greatest integer of $k$ is equal to
Step-by-Step Solution
Key Concept: For an n×n matrix M, det(adj M) = (det M)^(n-1). Since B is skew-symmetric of odd order 3, det(B) = 0 making det(adj B) = 0. Therefore, det(adj A) = 10^6, which requires computing det(A) and using det(A)^2 = 10^6.
After row operations $C_2 \to C_2 - C_1$ and $C_3 \to C_3 - C_1$, and $R_3 \to R_3 - R_1$, obtain a matrix with a skew-symmetric block. Since $B$ is skew-symmetric of odd order, $\det(B) = 0$. Using $\det(\text{adj}A) + \det(\text{adj}B) = 10^6$, solve $(2k+1)^3 + 0 = 10^6$ to get $2k+1 = 10$, hence $k = 4.5$ and $|k| = 4$.
Correct Answer: 4