Suppose A is a singular square matrix.
It is known that the $(i,j)$-entry of $A[adj(A)] = a_{i1}A_{j1} + a_{i2}A_{j2} + \cdots + a_{in}A_{jn}$ $$ \begin{cases} det(A) & \text{if $i = j$} \\[2ex] 0, & \text{if $i ≠ j$} \end{cases} $$
So $A[adj(A)] = det(A)I$
Since $A$ is singular, $det(A) = 0$
and hence $A[adj(A)] = 0I = 0$
While i understand that if $A$ is singular, $det(A) = 0$. I do not understand how does the above chain of $A[adj(A)] = a_{i1}A_{j1} + a_{i2}A_{j2} + \cdots + a_{in}A_{jn}$ leads to the conclusion that $A[adj(A)] = det(A)I$.
Please explain. Thanks :)