Science:Math Exam Resources/Courses/MATH152/April 2015/Question A 10

MATH152 April 2015

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Question A 10

Consider the following lines of MATLAB code: A = zeros(10, 10); for i=1:9 A(i, i) = 1/2; A(i+1, i) = 1/2; end A(10, 10) = 1; Circle the answer below that best describes the resulting matrix $A$: (a) an error occurs in the last line above. (b) $A$ is the transition matrix for a random walk. (c) $A$ is an upper triangular matrix. (d) $A$ is a diagonal matrix with diagonal entries 1/2. (e) $A$ contains the solution of a differential equation system.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint

Recall the following definitions: The transition matrix for a random walk (in the convention used in MATH 152) is a left-stochastic matrix, i.e., a nonnegative real matrix wherein each column sums to 1. An upper triangular matrix $A$ has $a_{ij}=0$ for all $i>j$, i.e., all entries below its diagonal are zero. A diagonal matrix $A$ has $a_{ij}=0$ for all $i\neq j$, i.e., all its off-diagonal entries are zero.

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. We see from the first line A = zeros(10, 10); that $A$ is initialized as a 10-by-10 matrix of zeroes. In the first iteration of the loop (when i = 1), we set $a_{1,1}={\tfrac {1}{2}}$ and the entry below it, $a_{2,1}$, to ${\tfrac {1}{2}}$ also. Continuing in this manner for $i=1,2,\dots ,9$, at the end of the loop we obtain ${\begin{bmatrix}{\frac {1}{2}}&0&\cdots &\cdots &\cdots &0\\{\frac {1}{2}}&{\frac {1}{2}}&\ddots &&&\vdots \\0&{\frac {1}{2}}&\ddots &\ddots &&\vdots \\\vdots &\ddots &\ddots &{\frac {1}{2}}&\ddots &\vdots \\\vdots &&\ddots &{\frac {1}{2}}&{\frac {1}{2}}&0\\0&\cdots &\cdots &0&{\frac {1}{2}}&0\end{bmatrix}}$ where the last two entries modified by the loop are $a_{9,9}$ and $a_{10,9}$. We finally set $a_{10,10}=1$, forming the matrix $A={\begin{bmatrix}{\frac {1}{2}}&0&\cdots &\cdots &\cdots &0\\{\frac {1}{2}}&{\frac {1}{2}}&\ddots &&&\vdots \\0&{\frac {1}{2}}&\ddots &\ddots &&\vdots \\\vdots &\ddots &\ddots &{\frac {1}{2}}&\ddots &\vdots \\\vdots &&\ddots &{\frac {1}{2}}&{\frac {1}{2}}&0\\0&\cdots &\cdots &0&{\frac {1}{2}}&1\end{bmatrix}}$ which is seen to be left-stochastic, and hence (b) the transition matrix for a random walk.