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GATE | GATE-CS-2016 (Set 2) | Question 48

  • Last Updated : 28 Jun, 2021

Let A1, A2, A3, and A4 be four matrices of dimensions 10 x 5, 5 x 20, 20 x 10, and 10 x 5, respectively. The minimum number of scalar multiplications required to find the product A1A2A3A4 using the basic matrix multiplication method is

(A) 1500
(B) 2000
(C) 500
(D) 100

Answer: (A)

Explanation: We have many ways to do matrix chain multiplication because matrix multiplication is associative. In other words, no matter how we parenthesize the product, the result of the matrix chain multiplication obtained will remain the same. Here we have four matrices A1, A2, A3, and A4, we would have:

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((A1A2)A3)A4 = ((A1(A2A3))A4) = (A1A2)(A3A4) = A1((A2A3)A4) = A1(A2(A3A4)).

However, the order in which we parenthesize the product affects the number of simple arithmetic operations needed to compute the product, or the efficiency. Here, A1 is a 10 × 5 matrix, A2 is a 5 x 20 matrix, and A3 is a 20 x 10 matrix, and A4 is 10 x 5.

If we multiply two matrices A and B of order l x m and m x n respectively,then the number of scalar multiplications in the multiplication of A and B will be lxmxn.


The number of scalar multiplications required in the following sequence of matrices will be :

A1((A2A3)A4) = (5 x 20 x 10) + (5 x 10 x 5) + (10 x 5 x 5) = 1000 + 250 + 250 = 1500.

All other parenthesized options will require number of multiplications more than 1500.

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