# Science:Math Exam Resources/Courses/MATH152/April 2022/Question B2 (c)/Solution 1

Suppose that, in the long run, ${\displaystyle v_{1},v_{2},v_{3}}$ are the proportions of buyers who buy Ambrosia, Braeburn and Cameo apples, respectively. Then the transition matrix ${\displaystyle A}$ should leave the vector ${\displaystyle \left[{\begin{array}{c}v_{1}\\v_{2}\\v_{3}\end{array}}\right]}$ fixed. In other words, we are looking for an eigenvector with eigenvalue 1, or an element of the null-space of the matrix

${\displaystyle A-\mathrm {Id} =\left[{\begin{array}{ccc}-0.3&0.2&0.2\\0.2&-0.4&0.4\\0.1&0.2&-0.6\end{array}}\right].}$

We can find the elements of the null-space by row-reducing. Again, we indicate the row operations as in B1 (a), but we also use ${\displaystyle T_{i,j}}$ for the operation "swap rows ${\displaystyle i}$ and ${\displaystyle j}$".

{\displaystyle {\begin{aligned}\left[{\begin{array}{ccc}-0.3&0.2&0.2\\0.2&-0.4&0.4\\0.1&0.2&-0.6\end{array}}\right]\xrightarrow {D_{\text{all}}(10),\ T_{1,3}} \left[{\begin{array}{ccc}1&2&-6\\2&-4&4\\-3&2&2\end{array}}\right]\xrightarrow {L_{2,1}(-2),\ L_{3,1}(3)} \left[{\begin{array}{ccc}1&2&-6\\0&-8&16\\0&8&-16\end{array}}\right]\xrightarrow {L_{3,2}(1),\ D_{2}(1/8),L_{1,2}(2)} \left[{\begin{array}{ccc}1&0&-2\\0&-1&2\\0&0&0\end{array}}\right].\end{aligned}}}

We can easily see from the row-reduced matrix above that eigenvectors of ${\displaystyle A}$ with eigenvalue 1 are of the form ${\displaystyle \left[{\begin{array}{c}v\\v\\{\frac {v}{2}}\end{array}}\right].}$ If we want the entries of this vector to be proportions of people buying the three different kinds of apples, then the entries of the eigenvector must sum up to 1. The equation ${\displaystyle v+v+{\frac {v}{2}}=1}$ implies that ${\displaystyle v=0.4}$, so we have that the long-run shares of each apple variety are

${\displaystyle v_{1}=0.4,\qquad v_{2}=0.4,\qquad v_{3}=0.2.}$