If we take the size of each element as 1 bytes.
Then its actually asking for the address of A[4][2] in the lower triangular matrix.
In Row Major Order, A[i][j] = Base + Size( (i - lower1) *NNS + (j-lower2) ) //NNS = Natural No.Sum.
So, A[4][2] = 1000 + 1 ( ( 4-(-6) )*NNS + (2-(-6) ) )
= 1000 + (1+2+3+4+5+6+7+8+9+10) + 8
= 1000 + 55 + 8
= 1063