Chapter 2.1, Problem 60E

In the financial pages of a newspaper, one can sometimes find a table (or matrix) listing the exchange ratesbetween currencies. In this exercise we will considera miniature version of such a table, involving onlythe Canadian dollar (C$) and the South African Rand(ZAR). Consider the matrix
$\begin{array}{l}\text{C\$}\text{ZAR}\\ A=\left[\begin{array}{cc}1& 1/8\\ 8& 1\end{array}\right]\begin{array}{c}C\$\\ \text{ZAR}\end{array}\end{array}$
representing the fact tha C$1 is worth ZAR8 (as ofSeptember 2012).
a. After a trip you have C$100 and ZAR 1,600 in yourpocket. We represent these two values in the vector $\stackrel{\to }{x}=\left[\begin{array}{l}100\\ 1,600\end{array}\right]$ .Compute $A\stackrel{\to }{x}$ . What is the practical significance of the two components of the vector $A\stackrel{\to }{x}$ ?
b. Verify that matrix A fails (o be invertible. For whichvectors $\stackrel{\to }{b}$ is the system $A\stackrel{\to }{x}=\stackrel{\to }{b}$ consistent? Whatis the practical significance of your answer? If thesystem $A\stackrel{\to }{x}=\stackrel{\to }{b}$ is consistent, how many solutions $\stackrel{\to }{x}$ are there? Again, what is the practical significanceof the answer?