is an idempotent matrix, by definition, if . Since the determinant of the product of two matrices is equal to the product of their determinants, it follows that

and, since , 

 .

By transitivity, 

.

The only two numbers equal to their own squares are 0 and 1, so 

 or .

This makes the statement true.

, the conjugate transpose of , can be found by first taking the transpose of :

, 

so

then changing each element to its complex conjugate:

Find the product  by multiplying the rows of  by the columns of ; that is, add the product of the terms in corresponding positions:

, the transpose of , is the result of switching the rows of  with the columns. 

, 

so

Find the product  by multiplying the rows of  by the columns of ; that is, add the product of the terms in corresponding positions:

The statement cannot be true for nonsquare matrices. For  to be defined, the number of columns in  must be equal to the number of rows in ; for  to be defined, the reverse must hold. It follows that  and  must be  and  matrices, respectively.

The product of two matrices has the same number of rows as the former matrix and the same number of columns as the latter. Therefore,  is an  matrix and  is an  matrix. If , then  and  do not even have the same dimensions. Therefore,  is always false for nonsquare matrices.

We now show that  for some, but not all square matrices. If, it easily follows that , since . Now let 

 and .

The products are

and

. Since at least one case exists in which  and at least one case exists in which , the statement  is sometimes true for square matrices.

Since the coin is fair, each outcome will come up with probability 0.5. Therefore, a player on orange end up on orange or pink with 0.5 probability each; a player on pink will end up on orange or blue with probability 0.5 each;  a player on blue will end up on pink or green with probability 0.5 each; and a player on green will end up on orange or pink with probability 0.5 each. The stochastic matrix that models this is

.

The matrix that models the probabilities that the player will end up on a given square after eight moves, given that he starts on a particular square, is , which through calculation is 

Since he wants to land on green after eight turns, examine the fourth row. The greatest probability of ending up on green, albeit by very little, appears in the second row. Therefore, to maximize his chances of ending up on green, the player should start on pink.

, the transpose of , is the result of switching the rows of  with the columns. 

, 

so

Find the product  by multiplying the rows of  by the columns of ; that is, add the product of the terms in corresponding positions:

, the conjugate transpose of , can be found by first taking the transpose of :

, 

so

then changing each element to its complex conjugate:

Find the product  by multiplying the rows of  by the columns of ; that is, add the product of the terms in corresponding positions:

 can be eliminated as a choice;  is not a square matrix, so its inverse, , is undefined.

 can be eliminated as a choice, since it is given that  is singular - that is,  does not exist.

For the product of three matrices to be defined, they must be, in order, an  matrix, an  matrix, and a  matrix. We examine the three remaining choices.

:

 is a  matrix, so  is as well;  is a  matrix, so its transpose  is a  matrix;  is a   matrix. These are incompatible, so  is undefined.

: 

 is a  matrix;  is a  matrix;  is a  matrix; These are also incompatible, so  is undefined.

:

 is a  matrix;  is a  matrix, as stated before;  is a  matrix. These are compatible, so  is defined. This is the correct choice.

If a player is on the red space, he will end up on each of the other spaces with the indicated rolls:

Red: 12; 1 roll out of 36  - probability .

Blue: 7; 6 rolls out of 36 possible - probability .

Green: 2 (1 roll) or 8 (5 rolls); 6 rolls out of 36 - probability .

Orange: 3 (2 rolls) or 9 (4 rolls); 6 rolls out of 36 - probability .

Lavender: 4 (3 rolls) or 10 (3 rolls); 6 rolls out of 36 - probability .

Brown: 5 (4 rolls) or 11 (2 rolls); 6 rolls out of 36 - probability .

Gray ("Time out"): 6; 5 rolls out of 36  - probability .

The  stochastic matrix with the first column only filled in is

The rules of the rolls are analogous for each of the other spaces, keeping the probabilities, relatively speaking, the same - the probability of staying on the same space is , that of landing on a given other space on the hexagon is , and the probability of ending up on the gray square is . The next five columns can be filled in as follows:

A player on the gray square can only end up any given space on the hexagon by rolling one specific pair of doubles, making each of these probabilities equal to . The other 30 rolls will keep him on the gray square, so there is a  probability he will stay on the gray square on that turn. Therefore, the final column can be filled in as follows:

This matrix is the correct choice.

If a player is on one of the spaces on the hexagon, he will end up on each of the other spaces with probability ; he cannot stay on his current space, since it would require a roll of 6 - this, by the rules, takes him to the gray square. Therefore, the first six columns of the stochastic matrix for this game can be filled in as follows:

A player on the gray square can only end up any given space on the hexagon by rolling one specific pair of doubles, making each of these probabilities equal to . The other 30 rolls will keep him on the gray square, so there is a  probability he will stay on the gray square on that turn. Therefore, the final column can be filled in as follows:

This is the correct choice.