(ii) A matrix N[11][8] is stored in the memory with each element requiring 2 bytes of storage. If the base address at N[2][3] is 2140, find the address of N[7][5] when the matrix is stored in Row Major Wise.

Solution

Address of N[7][5] (Row major) = B.A.+ W * [(I-Lr) * N + (J-Lc)]
= 2140 + 2[(7-2)*8 + (5-3)]
= 2140 + 2[5*8 + 2]
=2140 + 84
=2224