A matrix ARR[ – 4…6, 3….8] is stored in the memory with each element requiring 4 bytes of storage…

Row range $-4$ to $6$ $\to$ number of rows $M = 6-(-4)+1 = 11$ Column range $3$ to $8$ $\to$ number of columns $N = 8-3+1 = 6$ Each element $W = 4$ bytes; Base address $=1430$; target $ARR[3][6]$ Row Major formula: $Address(ARR[i][j]) = Base + W \times [(i - L_r) \times N + (j - L_c)]$ Here $i=3, j=6, L_r=-4, L_c=3$ $= 1430 + 4 \times [(3-(-4)) \times 6 + (6-3)]$ $= 1430 + 4 \times [42 + 3]$ $= 1430 + 4 \times 45$ $= 1430 + 180$ Final answer: 1610