Answer
True
Work Step by Step
We have the elements of $adj(A)$ are $adj(A)_{ij}=C_{ji}$
which can be written as: $A.adj(A)_{ij}=\sum a_{ik}adj(A)_{ij}=\sum a_{ik}C_{jk} $
and we also get the formula: $\sum a_{ik}C_{jk} =δ_{ij} \det(A)$
Hence, $A.adj(A)_{ij}=δ_{ij} \det(A)$
$\rightarrow A.adj(A)_{ij}=\det (A)I_n$
The statement is true
