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Matrices and Determinants question

2010 · Shift 2 · Q32
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  5. /2010 · Shift 2 · Q32

Matrices and Determinants question

2010 · Shift 2 · Q32

JEE AdvancedMathematicsMatrices and DeterminantsNumerical+4 / −1
Let kkk be a positive real number and let A=[2k−12k2k2k1−2k−2k2k−1] and B=[02k−1k1−2k02k−k−2k0].\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}\right] \text { and } \\\\ \mathbf{B} & =\left[\begin{array}{ccc} 0 & 2 k-1 & \sqrt{k} \\ 1-2 k & 0 & 2 \sqrt{k} \\ -\sqrt{k} & -2 \sqrt{k} & 0 \end{array}\right] . \end{aligned}AB​=​2k−12k​−2k​​2k​12k​2k​−2k−1​​ and =​01−2k−k​​2k−10−2k​​k​2k​0​​.​ If det⁡(adj⁡A)+det⁡(adj⁡B)=106\operatorname{det}(\operatorname{adj} A)+\operatorname{det}(\operatorname{adj} B)=10^6det(adjA)+det(adjB)=106, then [k][k][k] is equal to ‾\underline{\hspace{2cm}}​. [ Note : adj M denotes the adjoint of a square matrix M and [k][k][k] denotes the largest integer less than or equal to kkk ].
Numerical answer
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