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### Linear Algebra With Applications

Book edition 5th
Author(s) Otto Bretscher
Pages 442 pages
ISBN 9780321796974

# Three holy men (let’s call them Anselm, Benjamin, and Caspar) put little stock in material things; their only earthly possession is a small purse with a bit of gold dust. Each day they get together for the following bizarre bonding ritual: Each of them takes his purse and gives his gold away to the two others, in equal parts. For example, if Anselm has 4 ounces one day, he will give 2 ounces each to Benjamin and Caspar.(a) If Anselm starts out with 6 ounces, Benjamin with 1 ounce, and Caspar with 2 ounces, find formulas for the amounts a(t), b(t), and c(t) each will have after t distributions.Hint: The vector $\left[\begin{array}{c}1\\ 1\\ 1\end{array}\right]{\mathbf{,}}\left[\begin{array}{c}1\\ -1\\ 0\end{array}\right]{\mathbf{}}{\mathbf{and}}{\mathbf{}}\left[\begin{array}{c}1\\ 0\\ -1\end{array}\right]$, and will be useful. (b) Who will have the most gold after one year, that is, after 365 distributions?

The solutions are,

$a\left(t\right)=3.{1}^{t}+2.{\left(-\frac{1}{2}\right)}^{t}+{\left(-\frac{1}{2}\right)}^{t}=3+3\left(-\frac{1}{2}\right)\phantom{\rule{0ex}{0ex}}b\left(t\right)=3-2.\left(-\frac{1}{2}\right)\phantom{\rule{0ex}{0ex}}\left(a\right)\phantom{\rule{0ex}{0ex}}c\left(t\right)=3-\left(-\frac{1}{2}\right)$

(b) Benjamin

See the step by step solution

## Step 1: Solving for (a)

This is dynamical system, with,

$A=\left[\begin{array}{ccc}0& \frac{1}{2}& \frac{1}{2}\\ \frac{1}{2}& 0& \frac{1}{2}\\ \frac{1}{2}& \frac{1}{2}& 0\end{array}\right]$

This matrix’s eigenvectors are,

${v}_{1}=\left[\begin{array}{c}1\\ 1\\ 1\end{array}\right],{v}_{2}=\left[\begin{array}{c}1\\ -1\\ 0\end{array}\right],{v}_{3}\left[\begin{array}{c}1\\ 0\\ -1\end{array}\right]\phantom{\rule{0ex}{0ex}}{\lambda }_{1}=1,{\lambda }_{2,3}=-\frac{1}{2}.$

With the corresponding eigenvalues being .

We also have

${x}_{0}=\left[\begin{array}{c}6\\ 1\\ 2\end{array}\right]=3\left[\begin{array}{c}1\\ 1\\ 1\end{array}\right]+2\left[\begin{array}{c}1\\ -1\\ 0\end{array}\right]+\left[\begin{array}{c}1\\ 0\\ -1\end{array}\right]$

Therefore,

role="math" localid="1659583142610" $a\left(t\right)=3.{1}^{t}+2.{\left(-\frac{1}{2}\right)}^{t}+{\left(-\frac{1}{2}\right)}^{t}=3+3{\left(-\frac{1}{2}\right)}^{t}\phantom{\rule{0ex}{0ex}}b\left(t\right)=3-2.{\left(-\frac{1}{2}\right)}^{t}\phantom{\rule{0ex}{0ex}}\left(a\right)\phantom{\rule{0ex}{0ex}}c\left(t\right)=3-{\left(-\frac{1}{2}\right)}^{t}$

## Step 2: Solving for (b)

We have,

$a\left(365\right)=3-3{\left(\frac{1}{2}\right)}^{365},b\left(365\right)=3+2{\left(\frac{1}{2}\right)}^{365},c\left(365\right)=3+{\left(\frac{1}{2}\right)}^{365}$

So Benjamin will have the most ounces of gold.

Hence final solution is,

(a)

(b) Benjamin

$a\left(t\right)=3.{1}^{t}+2.{\left(-\frac{1}{2}\right)}^{t}+{\left(-\frac{1}{2}\right)}^{t}=3+3{\left(-\frac{1}{2}\right)}^{t}\phantom{\rule{0ex}{0ex}}b\left(t\right)=3-2.{\left(-\frac{1}{2}\right)}^{t}\phantom{\rule{0ex}{0ex}}\left(a\right)\phantom{\rule{0ex}{0ex}}c\left(t\right)=3-{\left(-\frac{1}{2}\right)}^{t}$