and  are both elementary matrices, in that each can be formed from the (four-by-four) identity matrix   by a single row operation. It also holds that each can be formed from  by a single column operation.

Since  differs from  in that the entry  is in Row 2, Column 2, the column operation is . Since  differs from  in that entry  is in Row 3, Column 2, the column operation is .

Postmultiplying a matrix by an elementary matrix has the effect of performing that column operation on the matrix. Looking at  as :

 Postmultiply  by  by performing the operation :

Postmultiply  by  by performing the operation :

.

This is the correct product.

Since the number of columns in the first matrix equals the number of rows in the second matrix, we know that these matrices can be multiplied together. To determine the dimensions of the product matrix, we take the number of rows in the first matrix and the number of columns in the second matrix. For this example, our product matrix will have dimensions of (2x2). Next, we notice that matrix A is the identity matrix. Any matrix multiplied by the identity matrix remains unchanged.  

Each column of a stochastic matrix has entries whose sum is 1. For each column in , set the sum of the entries equal to 1; it is easiest to work with Columns 1, 2, and 3 in that order, for reasons that become apparent:

Column 1: 

Column 2: 

It is known that , so substitute:

Column 3: 

It is known that , so substitute:

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:

Find  similarly:

This suggests a pattern:

 for any .

This can be confirmed by mathematical induction as follows:

Suppose .

Then 

Setting :

.

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.