Let be such that

MATHEMATICS · Single correct · CBSE · Mathematics · Determinants · Determinants

Problem

Let A=[aij]3×3A=[a_{ij}]_{3\times3} be such that aij={3,when i=j0,i≠ja_{ij}=\begin{cases}3, & \text{when } i=j\\0, & i\ne j\end{cases}. Then {det⁡(adj⁡(adj⁡A))5}\left\{\frac{\det(\operatorname{adj}(\operatorname{adj} A))}{5}\right\} equals, where {⋅}\{\cdot\} denotes the fractional-part function: (a) 25\frac{2}{5} (b) 15\frac{1}{5} (c) 23\frac{2}{3} (d) 13\frac{1}{3}

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