Let A, B, C, and D be four matrices of dimensions 30 x 40, 40 x 50, 50 x 60, and 60 x 20, respectively. The order of matrix multiplication that has minimum number of scalar multiplications required to find the product ABCD using the basic matrix multiplication method is ?
(A) (A(B(CD)))
(B) (AB)(CD)
(C) ((A(BC))D)
(D) (A(BC)D)


Answer: (A)

Explanation: Using Matrix Chain Multiplication :

1. (A(B(CD))) will result into 124000 multiplication.
   C_50×60 and D_60×20 = 60000 and resulting matrix = CD_50×20
   Now B_40×50 and CD_50×20 = 40000 and resulting matrix = BCD_40×20
   Now A_30×40 and BCD_40×20 = 24000 and resulting matrix = ABCD_30×20
   Total multiplication = 124000
2. (A((BC)D)) will result into 192000 multiplication.
   B_40×50 and C_50×60 = 120000 and resulting matrix = BC_40×60
   Now BC_40×60 and D_60×20 = 48000 and resulting matrix = BCD_40×20
   Now A_30×40 and BCD_40×20 = 24000 and resulting matrix = ABCD_30×20
   Total multiplication = 192000
3. (AB)(CD) will result into 150000 multiplication
   A_30×40 and B_40×50 = 60000 and resulting matrix = AB_30×50
   Now C_50×60 and D_60×20 = 60000 and resulting matrix = CD_50×20
   Now AB_3×5 and CD_5×2 = 30000 resulting matrix = ABCD_30×20
   Total multiplication = 150000.
4. ((A(BC))D) will result into 228000 multiplication
   B_40×50 and C_50×60 = 120000 and resulting matrix = BC_40×60
   Now A_30×40 and BC_40×60 = 72000 and resulting matrix = ABC_30×60
   Now ABC_30×60 and D_60×20 = 36000 and resulting matrix = ABCD_30×20
   Total multiplication = 228000
5. (((AB)C)D) will result into 186000 multiplication
   A_30×40 and B_40×50 = 60000 and resulting matrix = AB_30×50
   Now AB_30×50 and C_50×60 = 90000 and resulting matrix = ABC_30×60
   Now ABC_30×60 and D_60×20 = 36000 and resulting matrix = ABCD_30×20
   Total multiplication = 186000

Hence, minimuim multiplication = 124000 which will be resulted by (A(B(CD))).

So, option (A) is correct.

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