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Let be such that
MATHEMATICS · Single correct · CBSE · Mathematics · Determinants · Determinants
Problem Let A = [ a i j ] 3 × 3 A=[a_{ij}]_{3\times3} A = [ a ij ] 3 × 3 be such that a i j = { 3 , when i = j 0 , i ≠ j a_{ij}=\begin{cases}3, & \text{when } i=j\\0, & i\ne j\end{cases} a ij = { 3 , 0 , when i = j i = j . Then { det ( adj ( adj A ) ) 5 } \left\{\frac{\det(\operatorname{adj}(\operatorname{adj} A))}{5}\right\} { 5 d e t ( adj ( adj A )) } equals, where { ⋅ } \{\cdot\} { ⋅ } denotes the fractional-part function:
(a) 2 5 \frac{2}{5} 5 2
(b) 1 5 \frac{1}{5} 5 1
(c) 2 3 \frac{2}{3} 3 2
(d) 1 3 \frac{1}{3} 3 1
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