Last Updated :
20 Nov, 2018
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) (A((BC)D))
(C) (AB)(CD)
(D) ((A(BC))D)
Answer: (A)
Explanation: Using Matrix Chain Multiplication :
- (A(B(CD))) will result into 124000 multiplication.
C50×60 and D60×20 = 60000 and resulting matrix = CD50×20
Now B40×50 and CD50×20 = 40000 and resulting matrix = BCD40×20
Now A30×40 and BCD40×20 = 24000 and resulting matrix = ABCD30×20
Total multiplication = 124000
- (A((BC)D)) will result into 192000 multiplication.
B40×50 and C50×60 = 120000 and resulting matrix = BC40×60
Now BC40×60 and D60×20 = 48000 and resulting matrix = BCD40×20
Now A30×40 and BCD40×20 = 24000 and resulting matrix = ABCD30×20
Total multiplication = 192000
- (AB)(CD) will result into 150000 multiplication
A30×40 and B40×50 = 60000 and resulting matrix = AB30×50
Now C50×60 and D60×20 = 60000 and resulting matrix = CD50×20
Now AB3×5 and CD5×2 = 30000 resulting matrix = ABCD30×20
Total multiplication = 150000.
- ((A(BC))D) will result into 228000 multiplication
B40×50 and C50×60 = 120000 and resulting matrix = BC40×60
Now A30×40 and BC40×60 = 72000 and resulting matrix = ABC30×60
Now ABC30×60 and D60×20 = 36000 and resulting matrix = ABCD30×20
Total multiplication = 228000
- (((AB)C)D) will result into 186000 multiplication
A30×40 and B40×50 = 60000 and resulting matrix = AB30×50
Now AB30×50 and C50×60 = 90000 and resulting matrix = ABC30×60
Now ABC30×60 and D60×20 = 36000 and resulting matrix = ABCD30×20
Total multiplication = 186000
Hence, minimuim multiplication = 124000 which will be resulted by (A(B(CD))).
So, option (A) is correct.
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