Answer:
n-1 ways
Step-by-step explanation:
since there are 2 ways to order 3 matrices:
[tex](M_1M_2)M_3[/tex] and [tex]M_1(M_2M_3)[/tex]
you can do it the same way for 4 matrices and notice a pattern:
[tex](M_1M_2)M_3M_4\\M_1(M_2M_3)M_4\\M_1M_2(M_3M_4)\\[/tex]
the pattern here is that the number of ways to order the evaluation of multiplying matrices is 1 less than the number of matrices.
for [tex]n[/tex] matrices the order is [tex]n-1[/tex]
hope this helps!
