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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$ ?

  1. $\begin{bmatrix} \ 0 & 0 & 0\\ 0& 0 & 0\\ 0 & 0 & 0 \end{bmatrix}$
  2. $\begin{bmatrix} \frac{1}{4} &\frac{1}{2} & \frac{1}{2}\\ \frac{1}{4}& \frac{1}{2} & \frac{1}{2}\\ \frac{1}{4}& \frac{1}{2} & \frac{1}{2}\end{bmatrix}$
  3. $\begin{bmatrix} \frac{1}{2} &\frac{1}{4} & \frac{1}{4}\\ \frac{1}{2}& \frac{1}{4} & \frac{1}{4}\\ \frac{1}{2}& \frac{1}{4} & \frac{1}{4}\end{bmatrix}$
  4. $\begin{bmatrix} 0 &\frac{1}{2} & \frac{1}{2}\\ 0 & \frac{1}{2} & \frac{1}{2}\\ 0 & \frac{1}{2} & \frac{1}{2}\end{bmatrix}$
  5. $\text{The limit exists, but it is none of the above}$
in Calculus
edited by
Consider this approach.Let A^inf=X





Substitute for X from options

Now B and C are not possible

oly A and D are possible.

Still cant eliminate A.


5 Answers

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:

@severustux , i don't able to understand your answer, it seems complex to understand. Can you please explain in simple way.
1 vote

It's  answer should be A i.e ZERO MATRIX,  By simple observation,  values are in fraction will lead to 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 $E)$
mam I think A is the correct answer ??

sir can you tell me that what is the answer given ??
All 0's not coming na?
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 ?
how u done?

@srestha  mam check my answer !


This is wrong answer as D is the answer. Here is a wolfram alpha calculation



0 votes

Option A




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

$0.5^{\infty } = 0$

yeah A or E

but we have generalized the matrix in terms of  'n' but in your case you ain't generalized wrt to 'n'
it should be done with Markov chains :(
0 votes

edited by
Okay, but how will you solve it in the exam?

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:

Okay. This is a TIFR question. calculators are not allowed in the TIFR exam.

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.


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?
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.

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