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The limiting matrix is [tex]\bar{P}[/tex]=[tex]\left[\begin{array}{ccc}0.75&0.25\\0.75&0.25\\\end{array}\right][/tex].

What is meant by transition matrix?

Mathematicians frequently refer to "transition matrices" in a variety of situations. The term "change of coordinates matrix" is occasionally used in the context of linear algebra. As an alternative term for a stochastic matrix, or a matrix that describes transitions, it is employed in the theory of Markov chains. Use row operations to try to create the identity matrix to the left while augmenting the matrix of the second basis with the matrix of the first basis. Afterward, the transition matrix will be on the right. Row 1 should be multiplied by 1 as follows: R1=R1 R 1=-R 1 R1=R1.

The given transition matrix is:

P=[tex]\left[\begin{array}{ccc}0.9&0.1\\0.3&0.7\\\end{array}\right][/tex]

Let S=[tex]\left[\begin{array}{ccc}a\\b\\\end{array}\right][/tex]

[tex]\left[\begin{array}{ccc}0.9&0.1\\0.3&0.7\\\end{array}\right][/tex][tex]\left[\begin{array}{ccc}a\\b\\\end{array}\right][/tex] =[tex]\left[\begin{array}{ccc}a\\b\\\end{array}\right][/tex]

[tex]\left[\begin{array}{ccc}0.9a+0.1b\\0.3a+0.7b\\\end{array}\right][/tex]=[tex]\left[\begin{array}{ccc}a\\b\\\end{array}\right][/tex]

Here, 0.9a+0.1b=0, 0.3a-0.3b=0

So, a=b

a+b=1

a=b=1/2

S=[tex]\left[\begin{array}{ccc}0.5\\0.5\\\end{array}\right][/tex]

This S is called the stationary matrix

P=[tex]\left[\begin{array}{ccc}0.9&0.1\\0.3&0.7\\\end{array}\right][/tex]

P=det[tex]\left[\begin{array}{ccc}0.9-\lambda&0.1\\0.3&0.7-\lambda\\\end{array}\right][/tex]=0

λ²-1.6λ+0.6=(λ-1)(λ-0.6)=0

λ₁=1, λ₂=0.6

x=[tex]\left[\begin{array}{ccc}a\\b\\\end{array}\right][/tex]

Px=λx

[tex]\left[\begin{array}{ccc}0.9&0.1\\0.3&0.7\\\end{array}\right][/tex][tex]\left[\begin{array}{ccc}a\\b\\\end{array}\right][/tex] =1[tex]\left[\begin{array}{ccc}a\\b\\\end{array}\right][/tex]

0.9a+0.1b=a

a=b

v₁=[tex]\left[\begin{array}{ccc}1\\1\\\end{array}\right][/tex]

[tex]\left[\begin{array}{ccc}0.9&0.1\\0.3&0.7\\\end{array}\right][/tex][tex]\left[\begin{array}{ccc}a\\b\\\end{array}\right][/tex] =0.6[tex]\left[\begin{array}{ccc}a\\b\\\end{array}\right][/tex]

0.9a+0.1b=0.6a

b=-3a

v₂=[tex]\left[\begin{array}{ccc}1\\-3\\\end{array}\right][/tex]

V=[tex]\left[\begin{array}{ccc}1&1\\1&-3\\\end{array}\right][/tex]

V⁻¹=(1/4)[tex]\left[\begin{array}{ccc}3&1\\1&-1\\\end{array}\right][/tex]

D=[tex]\left[\begin{array}{ccc}\lambda_{1} &0\\0&\lambda_{2}\\\end{array}\right][/tex]

D=[tex]\left[\begin{array}{ccc}1&0\\0&0.6\\\end{array}\right][/tex]

P=VDV⁻¹

Pⁿ=VDⁿV⁻¹

[tex]\bar{P}[/tex]=[tex]\left[\begin{array}{ccc}0.75&0.25\\0.75&0.25\\\end{array}\right][/tex]

This is the required limit matrix.

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