Answer
See below
Work Step by Step
Let $B=\frac{1}{|A|}adj(A)$
Obtain:
$(BA)_{ij}$$=\sum^n_{k-1}b_{ik}a_{kj}\\=\sum^n_{k-1} \frac{1}{\det A}(adj (A))_{ik}a_{kj}\\=\frac{1}{\det A}\sum^n_{k-1}(adj (A))_{ik}a_{kj}\\=\frac{1}{\det A}\sum^n_{k-1} C_{kj}a_{kj}\\=\frac{1}{\det A}\delta_{ji}\det A\\=\delta_{ji}$
Consequently, $(BA)_{ij}=(I_n)_{ij}$
