If $A =[a_{ij}]_{4×4}$ such that $a_{ij}=\left\{\begin{matrix}2,&i=j\\0,&i≠j\end{matrix}\right.$, then $\left\{\frac{1}{7}det(adj(adj\,A))\right\}$ is, where {-} denotes the fractional part function.
Answer & explanation
Correct answer: option 1
If A is a square matrix of order, then
$|adj (adj\, A)|=|A|^{(n-1)^2}$
We have,
$A =[a_{ij}]_{4×4}=\begin{bmatrix}2&0&0&0\\0&2&0&0\\0&0&2&0\\0&0&0&2\end{bmatrix}=2I_4$
$∴|A|=2^4=16$
So, $|adj (adj\, A)| = 16^9 = 2^{36} =(2^3)^{12} =(7+1)^{12}$
$⇒\frac{1}{7}|adj (adj\, A)| =\frac{1}{7}(1+7)^{12}$
$=\frac{1}{7}({^{12}C}_0+{^{12}C}_1(7)+{^{12}C}_2(7)^2+.....+{^{12}C}_{12}(7)^{12})$
$⇒\frac{1}{7}|adj (adj\, A)| =\frac{1}{7}$ + A natural number
$⇒\left\{\frac{1}{7}|(adj(adj\,A)|\right\}=\frac{1}{7}$