A matrix is stored in the memory with each element requiring 2 bytes of storage. If the base…

Row range $-1$ to $10$ $\to$ number of rows $M = 10-(-1)+1 = 12$ Column range $4$ to $13$ $\to$ number of columns $N = 13-4+1 = 10$ Each element $W = 2$ bytes; Base address $=1200$; target $M[2][7]$ Row Major formula: $Address(M[i][j]) = Base + W \times [(i - L_r) \times N + (j - L_c)]$ Here $i=2, j=7, L_r=-1, L_c=4$ $= 1200 + 2 \times [(2-(-1)) \times 10 + (7-4)]$ $= 1200 + 2 \times [30 + 3]$ $= 1200 + 66$ Final answer: 1266