Problem: Let A, B, C, D be real matrices such that A^T = BCD ; B^T = CDA ; C^T = DAB and D^T = ABC for the matrix M = ABCD, then find M^2016?

Solution:
Let's start with the given conditions:
A^T = BCD
B^T = CDA
C^T = DAB
D^T = ABC

We are asked to find M^2016. We can start by trying to simplify M:

M = ABCD
M^2 = ABCDABCD
M^3 = ABCDABCDABCD
.
.
.

We can notice that M^n has the form ABCDABCD...ABCD (n times). To simplify this expression, we can use the given conditions. Let's take M^2 as an example:

M^2 = ABCDABCD
M^2 = ABCD(BCDA)(DABC)
M^2 = ABCDBCDA^2B^2C^2D^2

Using the given conditions, we can simplify this expression further:

M^2 = ABCDBCDA^2B^2C^2D^2
M^2 = ABCDDAB^2C^2A^2D^2
M^2 = ABCDD^2A^2B^2C^2

We can notice that this is the same as M, except with D^2 instead of D. We can use a similar method to simplify M^3, M^4, and so on. In general, we can say that:

M^n = ABCD(D^n-1A)(B^n-1C)(A^n-1D)

Using this formula, we can easily calculate M^2016:

M^2016 = ABCD(D^2015A)(B^2015C)(A^2015D)

We can notice that D^2015A is the same as D^3A, since D^4 = I (the identity matrix). Similarly, B^2015C is the same as B^3C, A^2015D is the same as A^3D. Using this, we can simplify the expression:

M^2016 = ABCD(D^3A)(B^3C)(A^3D)
M^2016 = (ABCD)^3

Therefore, M^2016 is equal to (ABCD)^3.