Express the matrix [tex]I[/tex] like [tex](\delta_{ij})_{n\times n}[/tex], where [tex]\delta_{ij}=1[/tex] if [tex]i=j[/tex] and [tex]\delta_{ij}=0[/tex] if [tex]i\neq j[/tex]. ([tex]\delta_{ij}[/tex] is known as kronecker's delta)
In the same form we express the matrix [tex]A=(a_{ij})_{n\times n}[/tex].
The firs index indicate the row and the second the column.
By the multiplication of matrices we have [tex]AI=(c_{ij})_{n\times n}[/tex], where
[tex]c_{ij}=\sum_{k=1}^n a_{ik}\delta _{kj} = a_{ij}[/tex]
because only [tex]\delta_{jj}[/tex] is non-zero in the last sum.
therfore we have [tex]AI=(c_{ij})_{n\times n}=(a_{ij})_{n\times n}=A[/tex].
In the same manner we have
[tex]IA=(d_{ij})_{n\times n}[/tex], where
[tex]d_{ij}=\sum_{k=1}^n \delta _{ik}a_{kj} = a_{ij}[/tex]
And so, [tex]IA=A[/tex]
