Q.Consider the matrix of transition probabilities of a product available in the market in two brands A and B $\begin{array}{c c} & \begin{array}{c c} A & B \end{array} \\ \begin{array}{c} A \\ B \end{array} & \begin{pmatrix} 0.9 & 0.1 \\ 0.3 & 0.7 \end{pmatrix} \end{array}$. Determine the market share of each brand in equilibrium position.

Solve $(a\ b)\begin{pmatrix}0.9&0.1\\0.3&0.7\end{pmatrix}=(a\ b)$ with $a+b=1$; this gives $a=0.75,\ b=0.25$.

In the TN HSC Class-12 Business Maths transition-probability topic, the equilibrium (steady-state) market shares form a state vector that is unchanged when multiplied by the transition matrix.

Step 1 — set up the equilibrium condition. Let the equilibrium shares be $a$ (brand A) and $b$ (brand B) with $a+b=1$. The transition matrix is

$T=\begin{pmatrix}0.9&0.1\\0.3&0.7\end{pmatrix}.$

Equilibrium requires $(a\ \ b)\,T=(a\ \ b)$.

Step 2 — form the equations.

$0.9a+0.3b=a,\qquad 0.1a+0.7b=b.$ …