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← IIT-JEE 2010 · Paper 2
MathematicsInteger+3Question 30 of 57

IIT-JEE 2010 · Paper 2, Mathematics question 30

Let kk be a positive real number and let A=[2k−12k2k2k1−2k−2k2k−1]  and  B=[02k−1k1−2k02k−k−2k0].A=\begin{bmatrix}2k-1&2\sqrt k&2\sqrt k\\2\sqrt k&1&-2k\\-2\sqrt k&2k&-1\end{bmatrix}\ \text{ 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⁡(adj⁡A)+det⁡(adj⁡B)=106\det(\operatorname{adj}A)+\det(\operatorname{adj}B)=10^6, then [k][k] is equal to

[Note: adj M denotes the adjoint of a square matrix M and [k][k] denotes the largest integer less than or equal to kk].

Answer and worked solution

Official answer: 4

B is skew-symmetric of odd order, so det⁡B=0\det B=0 and det⁡(adj⁡B)=(det⁡B)2=0\det(\operatorname{adj}B)=(\det B)^2=0.

det A. Expanding (or checking at several kk) gives det⁡A=(2k+1)3\det A=(2k+1)^3. For a 3×33\times3 matrix det⁡(adj⁡A)=(det⁡A)2=(2k+1)6\det(\operatorname{adj}A)=(\det A)^2=(2k+1)^6.

Solve. (2k+1)6=106(2k+1)^6=10^6, so 2k+1=102k+1=10, k=4.5k=4.5 and [k]=4[k]=4.

Answer: 4