Archive for the ‘Matrix’ Category

Problem 1187

A study of students taking a 20-question exam ranked their progress from one testing period to the next. Students scoring 0 to 5 from group 1, those scoring 6 to 10 for group 2, those scoring 11 to 15 from group 3 and those scoring 15 to 20 form group 4. The transition matrix to the right shows the result.

\begin{bmatrix}  0.188 & 0.4 & 0.26 & 0.156\\  0.376 & 0.11 & 0.17 &0.344 \\  0.056& 0.002 &0.351  & 0.591\\  0&  0.395& 0.222 & 0.383  \end{bmatrix}

 

Solution:-

 

Find the long-range prediction for the proportion of the students in each group.

Let V = [w x y z ] be the vector for the long-range predication. If P is the given matrix, solve the equation VP=V together with the equation w + x + y + z=1 to get w ,x ,y, and z.

w = 0.1247

x = 0.2325

y = 0.2464

z = 0.3964

The proportion of students will be 12.47% in group 1, 23.25% in group 2, 24.64 in group 3, and 39.64% in group 4.

 

Problem 1186

Identify the absorbing states in the transition matrix.

P = \begin{bmatrix}  0.5 &  0.2& 0.1 &0.2 \\  0& 0 & 1 & 0\\  0& 0 & 1 & 0\\  0& 0 &0  & 1  \end{bmatrix}

 

Solution:-

 

Absorbing States and Transition Matrices

A state in a Markov chain is absorbing if and only if the row of the transition matrix corresponding to the state has a 1 on the main diagonal and 0’s elsewhere.

The row that corresponds to A does not have a 1 on the main  diagonal, so therefore, A is not an absorbing state.

The row that corresponds to B does not have a 1 on the main diagonal, so therefore, B is not an absorbing state.

The row that corresponds to C has a 1 on the main diagonal, so therefore, C is an absorbing state.

The row that corresponds to D has a 1 on the main diagonal, so therefore, D is an absorbing state.

 

 

Problem 1185

Find the absorbing state(s) for the transition matrix shown.

\begin{bmatrix}  0.0 & 0.00 &1.00 \\  0.0&1.0  & 0.0\\  0.0& 0.0 & 1.0  \end{bmatrix}

 

Solution:-

 

A state is an absorbing state of a Markov chain if Pii=1. Thus, check the entries P11,  P22, and P33 to see if any of them are equal to 1.

P11 = 0.0

P22 = 1.0

P33 = 1.0

Since P22 and P33 equals 1, state 2 and 3 are absorbing state.

 

 

Problem 1184

Find the absorbing state(s) for the transition matrix shown.

\begin{bmatrix}  1.0 & 0.00 &0.00 \\  1.0&0.0  & 0.0\\  0.3& 0.7 & 0.0  \end{bmatrix}

 

Solution:-

 

A state is an absorbing state of a Markov chain if Pii=1. Thus, check the entries P11,  P22, and P33 to see if any of them are equal to 1.

P11 = 1.1

P22 = 0.0

P33 = 0.0

Since P11 equals 1, state 1 is the absorbing state.

 

 

Problem 1183

Identify the absorbing states in the transition matrix.

P = \begin{bmatrix}  0.5&0.2  &0.1  &0.2 \\  0& 0 &1  &0 \\  0& 0 & 1 &0 \\  0& 0 &0  &1  \end{bmatrix}

 

Solution:-

 

Absorbing  States and Transition Matrices

A state in a Markov chain to A does not have a 1 on the main diagonal, so therefore, A is not an absorbing state.

The row that corresponds to B does not have a 1 on the main diagonal, so therefore, B is not an absorbing state.

The row that corresponds to C has a 1 on the main diagonal, so therefore, C is an absorbing state.

The row that corresponds to D has a 1 on the main diagonal, so therefore, D is an absorbing state.

 

Problem 1182

The expression simplifies to a constant, a single circular function, or a power of a circular function.

Use fundamental identities to simplify the expression.

cot x tan x

 

Solution:-

 

First, write all of the function in terms of sines and cosines.

tan x = \frac{sinx}{cosx} and cotx = \frac{cosx}{sinx}.

This allows us to simplify the expression.

cot x tan x  = \frac{cos x}{sin x} \frac{sin x}{cos x} = 1

Thus, cot x tan x = 1.

 

Problem 1181

Find the area of a sector of a circle having radius r and central angle θ.

r =  70.0 mi, θ = 100°

 

Solution:-

 

First , multiply the degree measure of the angle by \frac{\pi}{180} and simplify to convert to radians.

100°(\frac{\pi}{180}) = \frac{5\pi}{9} radian

The area A of a sector of a circle of radius r and central angle θ (in radians) is given by the following formula.

A = \frac{1}{2}r2θ,θ in radians

Substitute the values r  = 70.0 and θ = \frac{5\pi}{9} radian into the formula shown above and simplify.

A = \frac{1}{2}(70.0)2 (\frac{\5pi}{9}) \approx 4276.1 mi2

Therefore, the area of the sector is approximately 4276.1 mi2.

 

Problem 1180

Find the equilibrium vector for the given matrix.

P = \begin{bmatrix}  0.3 & 0.7  \\  0.4 & 0.6  \end{bmatrix}

 

Solution:-

 

First, we should determine if the transition matrix P is regular.

Since all of the entries are positive, P is regular.

Since it is regular, we can find a probability vector V where VP = V.

V will be the equilibrium vector. Let V = [x, y]. Then

VP = [x y] \begin{bmatrix}  0.3 & 0.7  \\  0.4 & 0.6  \end{bmatrix}

= [0.3x + 0.4y   0.7x+0.6y]

Set the entries equal to each other

0.3x + 0.4y = x and 0.7x +0.6y = y

Simplify both equations.

-0.7x + 0.4y = 0  and 0.7x – 0.4y = 0

Therefore are really the same equation, so we need a second equation.

Since V is a probability vector, x + y = 1.

Now solve the system.

-0.7x + 0.4y = 0

x + y = 1

[x y ] = [\frac{4}{11} \frac{7}{11}]

Problem 1179

Is the following transition matrix regular?

A = \begin{bmatrix}  1 & 0  \\  0.75& 0.25  \end{bmatrix}

 

Solution:-

 

A transition matrix is regular if same power of the matrix contains all positive entries.

The given matrix is A, or A1. Since 0 is not a positive number, A1 does not contain all positive entries.

Thus, other powers of a must be examined to determine whether or not A is regular.

Next, look at A2, which if found by multiplying A by itself using matrix multiplication.

A2 = \begin{bmatrix}  1 & 0  \\  0.9375& 0.0625  \end{bmatrix}

 

Notice that both A and A2 contain only one zero and in both matrices this zero is in row 1 and column 2. For any transition matrix P, if all zeros occur in the identical places in both Pn and Pn+1 for any n, they will appear in those places for all higher powers of P. Therefore there would be no power of P that contain all positive entries, so P is  not regular.

Thus, the given transition matrix A is nor regular because the only 0 occurs in the identical place in both A1 and A2, which means there is no power of A that contains all positive entries.