Relating two vectors:

Suppose that Rt be the number of rural residents at the beginning of the year t, and Ct be the number of city residents at the beginning of the year t.

Since 10% of the rural residents moved to city, remaining 90% of the rural decided to continue to stay at the rural residents at the beginning of the year t+1.

Thus, the number of rural residents in the beginning of the year t+1 is given below:

Rt+1=0.9Rt+0.2Ct

Also, 20% of city residents moved to the rural residents in the beginning of the year t+1.

Therefore, the number of city residents in the beginning of the year t+1 is as follows:

Ct+1=0

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Type in Latex **Problem**. Let $$A = \begin{bmatrix} .5 & .2 & .3 \\ .3 & .8 & .3 \\ .2 & 0 & .4 \end{bmatrix}.$$ This matrix is an example of a **stochastic matrix**: its column sums are all equal to 1. The vectors $$\mathbf{v}_1 = \begin{bmatrix} .3 \\ .6 \\ .1 \end{bmatrix}, \mathbf{v}_2 = \begin{bmatrix} 1 \\ -3 \\ 2 \end{bmatrix}, \mathbf{v}_3 = \begin{bmatrix} -1 \\ 0 \\ 1\end{bmatrix}$$ are all eigenvectors of $A$. * Compute $\left[\begin{array}{rrr} 1 & 1 & 1 \end{array}\right]\cdot\mathbf{x}_0$ and deduce that $c_1 = 1$.* Finally, let $\mathbf{x}_k = A^k \mathbf{x}_0$. Show that $\mathbf{x}_k \longrightarrow \mathbf{v}_1$ as $k$ goes to infinity. (The vector $\mathbf{v}_1$ is called a **steady-state vector** for $A.$) **Solution**. To prove that $c_1 = 1$, we first left-multiply both sides of the above equation by $[1 \, 1\, 1]$ and then simplify both sides:$$\begin{aligned}[1 \, 1\, 1]\mathbf{x}_0 &= [1 \, 1\, 1](c_1\mathbf{v}_1 +…

Problem 1) Change the diagonal elements of a 3X3 matrix to 1. Diagonal elements of the example matrix shown here are 10, 8, 2 10 -4 0 7 8 3 0 0 2 Read the given values from input (cin) using for loops and then write your code to solve the diagonal.

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Chapter 2 Solutions

Introduction to mathematical programming

Ch. 2.1 - Prob. 1P Ch. 2.1 - Prob. 2P Ch. 2.1 - Prob. 3P Ch. 2.1 - Prob. 4P Ch. 2.1 - Prob. 5P Ch. 2.1 - Prob. 6P Ch. 2.1 - Prob. 7P Ch. 2.2 - Prob. 1P Ch. 2.3 - Prob. 1P Ch. 2.3 - Prob. 2P

Ch. 2.3 - Prob. 3P Ch. 2.3 - Prob. 4P Ch. 2.3 - Prob. 5P Ch. 2.3 - Prob. 6P Ch. 2.3 - Prob. 7P Ch. 2.3 - Prob. 8P Ch. 2.3 - Prob. 9P Ch. 2.4 - Prob. 1P Ch. 2.4 - Prob. 2P Ch. 2.4 - Prob. 3P Ch. 2.4 - Prob. 4P Ch. 2.4 - Prob. 5P Ch. 2.4 - Prob. 6P Ch. 2.4 - Prob. 7P Ch. 2.4 - Prob. 8P Ch. 2.4 - Prob. 9P Ch. 2.5 - Prob. 1P Ch. 2.5 - Prob. 2P Ch. 2.5 - Prob. 3P Ch. 2.5 - Prob. 4P Ch. 2.5 - Prob. 5P Ch. 2.5 - Prob. 6P Ch. 2.5 - Prob. 7P Ch. 2.5 - Prob. 8P Ch. 2.5 - Prob. 9P Ch. 2.5 - Prob. 10P Ch. 2.5 - Prob. 11P Ch. 2.6 - Prob. 1P Ch. 2.6 - Prob. 2P Ch. 2.6 - Prob. 3P Ch. 2.6 - Prob. 4P Ch. 2 - Prob. 1RP Ch. 2 - Prob. 2RP Ch. 2 - Prob. 3RP Ch. 2 - Prob. 4RP Ch. 2 - Prob. 5RP Ch. 2 - Prob. 6RP Ch. 2 - Prob. 7RP Ch. 2 - Prob. 8RP Ch. 2 - Prob. 9RP Ch. 2 - Prob. 10RP Ch. 2 - Prob. 11RP Ch. 2 - Prob. 12RP Ch. 2 - Prob. 13RP Ch. 2 - Prob. 14RP Ch. 2 - Prob. 15RP Ch. 2 - Prob. 16RP Ch. 2 - Prob. 17RP Ch. 2 - Prob. 18RP Ch. 2 - Prob. 19RP Ch. 2 - Prob. 20RP Ch. 2 - Prob. 21RP Ch. 2 - Prob. 22RP

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Solution Summary: The author explains how the relation between the matrices is given below.

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Chapter 2 Solutions