# TIFR2019-A-15

1k views

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
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Consider this approach.Let A^inf=X

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

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

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@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.

$\begin{bmatrix} \frac{1}{2} &\frac{1}{2} & 0\\ 0& \frac{3}{4} & \frac{1}{4}\\ 0& \frac{1}{4} & \frac{3}{4} \end{bmatrix}$.$\begin{bmatrix} \frac{1}{2} &\frac{1}{2} & 0\\ 0& \frac{3}{4} & \frac{1}{4}\\ 0& \frac{1}{4} & \frac{3}{4} \end{bmatrix}$ [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 $E)$
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mam I think A is the correct answer ??

sir can you tell me that what is the answer given ??
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All 0's not coming na?
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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 ?
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how u done?
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@srestha  mam check my answer !

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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

Option A  0

@Magma

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

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$0.5^{\infty } = 0$

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'
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it should be done with Markov chains :(

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.

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

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