$a[0...99,0...99]=a[100][100]$
$loc[80][90]:RMO$
$0^{th}\ row=100(0+x=100)$
$1^{st}\ row=99(1+x=100)$
$2^{nd}\ row=98(2+x=100)$
.
.
.
$79^{th}\ row=21(79+x=100)$
$Sum=100+99+98....+21=4840$
Now you are standing on $80^{th}$ row and it's first element will start from $80^{th}$ column but you need to go to $90^{th}$ column, hence add $11(80,81,82....90)$
$4840+11=4851$
But index starts from $0,$ so subtract $1$
$4851-1=4850$