A matrix  is involutory if , or, equivalently, if . Multiply  by itself by multiplying rows by columns - add the products of corresponding elements - as follows:

If we set this equal to , the equation is

The only condition required to set these matrices equal is

,

or

,

making this condition necessary and sufficient for  to be involutory.

A stochastic matrix for the game can be formed from the probability that the player will land on each space given the space he is on currently.

If he is on pink, he will roll a six-sided die; his probabilities are:

Pink:  (1 only)

Blue:  (2, 3, or 5)

Yellow:  (4 or 6)

If he is on blue, he will roll an eight-sided die; his probabilities are:

Pink:  (1, 4, 6, 8)

Orange:  (2, 3, 5, 7)

If he is on orange, he will roll a ten-sided die; his probabilities are:

Pink:  (1 only)

Blue:  (4, 6, 8, 9, 10)

Yellow:  (2, 3, 5, 7)

If he is on yellow, he will roll a twelve-sided die; his probabilities are:

Pink:  (1, 2, 3, 5, 7, 11)

Orange:  (4, 6, 8, 9, 10, 12 )

With the rows/columns representing pink, blue, orange, and yellow states, in that order, the stochastic matrix is 

Any stochastic matrix has 1 as its dominant eigenvalue; the eigenvector corresponding to that eigenvalue is its steady-state vector. We can find this vector directly, but we can also calculate it numerically.

The matrix that shows the probabilities after  turns is . If we raise  to a sufficiently large power - say,  - we see that the result approaches the matrix , where 

This is the steady state vector of the system. The greatest value appears in the first row, which represents the pink circle. It follows that a player will tend to spend most of his time on the pink circle as the game progresses.

This is not as daunting a task as it seems. 

First, find .

Multiply rows of  by columns of  by adding the products of corresponding elements, as follows:

Note that 

It easily follows that

 can be calculated as follows:

A stochastic matrix for the game can be formed from the probability that the player will land on each space given the space he is on currently.

If he is on pink, he will roll a six-sided die; his probabilities are:

Blue:  (2, 3, or 5)

Orange:  (1 only)

Yellow:  (4 or 6)

If he is on blue, he will roll an eight-sided die; his probabilities are:

Pink:  (4, 6, 8)

Orange:  (2, 3, 5, 7)

Yellow:  (1 only)

If he is on orange, he will roll a ten-sided die; his probabilities are:

Pink:  (1 only)

Blue:  (4, 6, 8, 9, 10)

Yellow:  (2, 3, 5, 7)

If he is on yellow, he will roll a twelve-sided die; his probabilities are:

Pink:  (2, 3, 5, 7, 11)

Blue:  (1 only)

Orange:  (4, 6, 8, 9, 10, 12)

With the rows/columns representing pink, blue, orange, and yellow states, in that order, the stochastic matrix is 

Any stochastic matrix has 1 as its dominant eigenvalue; the eigenvector corresponding to that eigenvalue is its steady-state vector. We can find this vector directly, but we can also calculate it numerically.

The matrix that shows the probabilities after  turns is . If we raise  to a sufficiently large power - say,  - we see that the result approaches the matrix , where 

This is the steady state vector of the system. We see that the greatest value appears in the third row, which represents the orange circle. It follows that a player will tend to spend most of his time on the orange circle as the game progresses.

 is involutory if ; it is idempotent if . Therefore, to determine which, if either, describes , square . This can be done by multiplying rows of  by columns of  - adding the products of corresponding entries, as follows:

 is therefore involutory. 

 is an involutory matrix if , so square  to answer this question. 

This matrix can be squared more easily by noting that this matrix can be blocked as:

,

Where , , and  is the  zero matrix. 

Now, find the squares of the two blocks:

, making  involutory.

 is an idempotent matrix if . 

For any  for which  is defined, . Setting , if  is idempotent, this becomes

It follows that  is idempotent.

 is an involutory matrix if . 

For any  for which  is defined, . Setting , if  is involutory, this becomes

Thus,  is involutory, making the statement true. 

A matrix  is nilpotent if there exists  so that , the zero matrix. 

Suppose  is nilpotent.

Case 1: If , which is trivially nilpotent, then , which is trivially nilpotent.

Case 2: Suppose . For any  for which  is defined, . Setting , if  is nilpotent, this becomes

It follows that  is nilpotent. This reasoning can easily be extended to the case , since .

Form a stochastic (probability) matrix for this system, where the columns represent current political affiliation (Republicans, Democrats, independents, and Libertarians, in that order), and the rows represent affiliation after one year (same order). This matrix

Raise this matrix to the power of 40 to obtain the stochastic matrix with the probabilities that political affiliations will change over 40 years. Observe the entries in the main diagonal, which represent voters who have the same party affiliation at the beginning and the end of the forty-year period:

The greatest number appears in the fourth entry - the voters who were Libertarians at the beginning and end of the forty-year period. The correct choice is Libertarians.