Let k be a positive real number and let and . If det (adjA) + det(adjB) = 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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|A| = (2k + 1)3, |B| = 0 (Since B is a skew-symmetric matrix of order 3)
⇒ det(adjA) = |A|n–1 = ((2k + 1)3)2 = 106 ⇒ 2k + 1 = 10 ⇒ 2k = 9
[k] = 4.
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