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Question

Let k be a positve real number and let
A=⎢ ⎢2k12k2k2k12k2k2k1⎥ ⎥ and B=⎢ ⎢22k12k12k02kk2k0⎥ ⎥. If det (adj A)+det(adj B)=106, then [k] is equal to

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

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Solution

|A|=(2k+1)3,|B|=0
det (adj A)det (adj B) =(2k+1)^6 =106 k=92
[k]=4

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