Now

A^inf=A*A^inf=A^inf*A

So

X=A*X=X*A

Substitute for X from options

Now B and C are not possible

oly A and D are possible.

Still cant eliminate A.

@srestha

5 votes

Consider the matrix

$$A = \begin{bmatrix} \frac{1}{2} &\frac{1}{2} & 0\\ 0& \frac{3}{4} & \frac{1}{4}\\ 0& \frac{1}{4} & \frac{3}{4} \end{bmatrix}$$

What is $\lim_{n→\infty}$$A^n$ ?

3 votes

We observe that given matrix A is a right stochastic matrix. That is, for every row the row sum is 1. What does that mean? It means the given matrix corresponds to some Markov system. which is: A 3 state system as follows : (with each edge being the probability of the system going from a state to other.)

Now we find the steady-state probability vector as follows :

[$x_1 , x_2, x_3$] $\begin{bmatrix} \frac{1}{2} & \frac{1}{2} & 0 \\\\ 0 & \frac{3}{4} & \frac{1}{4} \\ \\ 0 & \frac{1}{4} & \frac{3}{4} \end{bmatrix}$ = [$x_1, x_2, x_3$]

As well as, we have $x_1 + x_2 + x_3 = 1$

Solving the above equation, we get $x_1 = 0, x_2 =\frac{1}{2}, x_3 = \frac{1}{2}$

We know that, in a Markov chain, $A^\infty$ is nothing but all rows steady-state vector.

Therefore, $A^\infty = \begin{bmatrix} 0 & \frac{1}{2} & \frac{1}{2} \\\\ 0 & \frac{1}{2} & \frac{1}{2} \\ \\ 0 & \frac{1}{2} & \frac{1}{2} \end{bmatrix} $

Ref: If you want to know more about Markov chains and stochastic matrices: https://en.wikipedia.org/wiki/Stochastic_matrix#Definition_and_properties

1 vote

**0.00 something after multiplication, Hence keep multiplying them to infinity(some large value) will lead to ZERO MATRIX. **

0 votes

$\begin{bmatrix} \frac{1}{2} &\frac{1}{2} & 0\\ 0& \frac{3}{4} & \frac{1}{4}\\ 0& \frac{1}{4} & \frac{3}{4} \end{bmatrix}$.$\begin{bmatrix} \frac{1}{2} &\frac{1}{2} & 0\\ 0& \frac{3}{4} & \frac{1}{4}\\ 0& \frac{1}{4} & \frac{3}{4} \end{bmatrix}$ [calculating $A^{2}$]

=$\begin{bmatrix} \frac{1}{4} &\frac{5}{8} & \frac{1}{8}\\ 0& \frac{5}{8} & \frac{3}{8}\\ 0& \frac{3}{8} & \frac{5}{8} \end{bmatrix}$

=$(\frac{1}{8})^{3}\begin{bmatrix} 2&5 &1 \\ 0&5 &3 \\ 0& 3 & 5 \end{bmatrix}$.$\begin{bmatrix} \frac{1}{2} &\frac{1}{2} & 0\\ 0& \frac{3}{4} & \frac{1}{4}\\ 0& \frac{1}{4} & \frac{3}{4} \end{bmatrix}$ [calculating $A^{3}$]

=$(\frac{1}{8})^{3}\begin{bmatrix} 1&5 &2 \\ 0&\frac{9}{2} &\frac{7}{2} \\ 0& \frac{7}{2} & \frac{9}{2} \end{bmatrix}$

=$\left ( \frac{1}{8} \right )^{3}\left ( \frac{1}{2} \right )^{2}\begin{bmatrix} 1&5 &2 \\ 0&9 &7 \\ 0& 7 &9 \end{bmatrix}$.$\begin{bmatrix} \frac{1}{2} &\frac{1}{2} & 0\\ 0& \frac{3}{4} & \frac{1}{4}\\ 0& \frac{1}{4} & \frac{3}{4} \end{bmatrix}$ [calculating $A^{4}$]

=$\left ( \frac{1}{8} \right )^{3}\left ( \frac{1}{2} \right )^{2}\begin{bmatrix} \frac{1}{2}&\frac{17}{4} &\frac{11}{4} \\ 0&\frac{34}{4} &\frac{30}{4} \\ 0& \frac{30}{4} &\frac{34}{4} \end{bmatrix}$

So, we can see here limit exists for matrix ,as value is not tends to infinity

but values of matrix are not all 0s or all 1s

So, ans will be $E)$

=$\begin{bmatrix} \frac{1}{4} &\frac{5}{8} & \frac{1}{8}\\ 0& \frac{5}{8} & \frac{3}{8}\\ 0& \frac{3}{8} & \frac{5}{8} \end{bmatrix}$

=$(\frac{1}{8})^{3}\begin{bmatrix} 2&5 &1 \\ 0&5 &3 \\ 0& 3 & 5 \end{bmatrix}$.$\begin{bmatrix} \frac{1}{2} &\frac{1}{2} & 0\\ 0& \frac{3}{4} & \frac{1}{4}\\ 0& \frac{1}{4} & \frac{3}{4} \end{bmatrix}$ [calculating $A^{3}$]

=$(\frac{1}{8})^{3}\begin{bmatrix} 1&5 &2 \\ 0&\frac{9}{2} &\frac{7}{2} \\ 0& \frac{7}{2} & \frac{9}{2} \end{bmatrix}$

=$\left ( \frac{1}{8} \right )^{3}\left ( \frac{1}{2} \right )^{2}\begin{bmatrix} 1&5 &2 \\ 0&9 &7 \\ 0& 7 &9 \end{bmatrix}$.$\begin{bmatrix} \frac{1}{2} &\frac{1}{2} & 0\\ 0& \frac{3}{4} & \frac{1}{4}\\ 0& \frac{1}{4} & \frac{3}{4} \end{bmatrix}$ [calculating $A^{4}$]

=$\left ( \frac{1}{8} \right )^{3}\left ( \frac{1}{2} \right )^{2}\begin{bmatrix} \frac{1}{2}&\frac{17}{4} &\frac{11}{4} \\ 0&\frac{34}{4} &\frac{30}{4} \\ 0& \frac{30}{4} &\frac{34}{4} \end{bmatrix}$

So, we can see here limit exists for matrix ,as value is not tends to infinity

but values of matrix are not all 0s or all 1s

So, ans will be $E)$

0

In my case all 0's is coming means answer A

But I have to clarify that whether the Option "A" is right or wrong ?

But I have to clarify that whether the Option "A" is right or wrong ?

0

This is wrong answer as D is the answer. Here is a wolfram alpha calculation https://www.wolframalpha.com/input/?i=%7B%7B0.5%2C+0.5%2C0%7D%2C+%7B0%2C+0.75%2C0.25%7D+%2C%7B0%2C+.25%2C+.75%7D%7D%5E100

0 votes

0

ok

B),C),D) sort out for sure

now, u have some assumption $\lim_{n\rightarrow \infty }\left ( \frac{1}{2} \right )^{n}$ tends to $0$ then A) is true

otherwise E) right?

I truely confused what will they consider

0 votes

D is the correct answer.

Here is a wolfram alpha calculation https://www.wolframalpha.com/input/?i=%7B%7B0.5%2C+0.5%2C0%7D%2C+%7B0%2C+0.75%2C0.25%7D+%2C%7B0%2C+.25%2C+.75%7D%7D%5E100

In exam one may calculate till 8th power and see the trend.

For a proper way to solve it: https://math.stackexchange.com/questions/3239468/computing-the-matrix-powers-of-a-non-diagonalizable-matrix/3239469#3239469

0

Like I said, "In exam one may calculate till 8th power and see the trend. "

I did the same to get to the answer while practicing. But there is a proper mathematical way to solve it but its above my pay grade. This is the detailed solution: https://math.stackexchange.com/questions/3239468/computing-the-matrix-powers-of-a-non-diagonalizable-matrix/3239469#3239469

0

Okay, but you do not really need calculator to calculate till 8th power. (Three matrix multiplication is enough to see the trend)

See below what I did while practicing (bottom half of the page):

In fact the trend is clear in just 4th power.

1

The point is, the question is not meant to be solved by simply multiplying the matrices. Well, the question could be asked in multiple ways. Let's say it was a fill in the blanks question. Where determinant of $A^\infty$ was asked. If you solve it the right way, you will get right answer. Or you can stick with multiplying the matrices and brute force the answers. What if a 4x4 matrix is given? or if the numbers were messy? will you keep multiplying?

0

I wish you had given this comment as the first time, instead of the one you did. Would have saved us a lot of time. I agree with you. The question didn't have any correct answer when I answered it, all the rest were wrong answers. I just saw your answer which is probably the intended solution. Not really sure if markov matrix is part of the syllabus though (as in one hand they ask 5 red balls 4 red balls and here markov matrix, quite a difference in difficulty level) but who knows what the prof is smoking.