91Ó°ÊÓ

In the financial pages of a newspaper, one can sometimes find a table (or matrix) listing the exchange rates between currencies. In this exercise we will consider a miniature version of such a table, involving only the Canadian dollar $\left(C\backslash (\right)$ and the South African Rand $\left(ZAR\right)$ . Consider the matrix

role="math" localid="1659786495324" $\begin{array}{rcl}& & \begin{array}{cc}\mathbf{\text{C)}}& \mathbf{\text{ZAR}}\end{array}\\ \mathit{A}& \mathbf{=}& \mathbf{\left[}\begin{array}{cc}\mathbf{1}& \mathbf{1}\mathbf{/}\mathbf{8}\\ \mathbf{8}& \mathbf{1}\end{array}\mathbf{\right]}\begin{array}{c}\mathbf{C}\mathbf{\backslash (}\\ \mathbf{\text{ZAR}}\end{array}\end{array}$

representing the fact thatrole="math" localid="1659786520551" $\mathit{C}\mathbf{\backslash )}\mathbf{1}$ is worth role="math" localid="1659786525050" $\mathit{Z}\mathit{A}\mathit{R}\mathbf{8}$ (as of September 2012).

a. After a trip you have $\mathit{C}\mathbf{\$}\mathbf{100}$ and $\mathit{Z}\mathit{A}\mathit{R}\mathbf{1}\mathbf{,}\mathbf{600}$ in your pocket. We represent these two values in the vector $\stackrel{\mathbf{→}}{\mathbf{x}}\mathbf{=}\left[\begin{array}{c}100\\ 1,600\end{array}\right]$ . Compute $\mathit{A}\stackrel{\mathbf{→}}{\mathbf{x}}$ . What is the practical significance of the two components of the vector $\mathit{A}\stackrel{\mathbf{→}}{\mathbf{x}}$ ?

b. Verify that matrix $\mathit{A}$ fails to be invertible. For which vectors$\stackrel{\mathbf{→}}{\mathbf{b}}$is the system $\mathit{A}\stackrel{\mathbf{→}}{\mathbf{x}}\mathbf{=}\stackrel{\mathbf{→}}{\mathbf{b}}$ consistent? What is the practical significance of your answer? If the system $\mathit{A}\stackrel{\mathbf{→}}{\mathbf{x}}\mathbf{=}\stackrel{\mathbf{→}}{\mathbf{b}}$ is consistent, how many solutions$\stackrel{\mathbf{→}}{\mathbf{x}}$are there? Again, what is the practical significance of the answer?

Step by step solution

01

Step by Step Explanation: Step 1: Consider the matrices

(a)

The matrices are,

$$\begin{gathered} A=\left[\begin{array}{cc}1& 1/8\\ 8& 1\end{array}\right] \\ \stackrel{→}{x}=\left[\begin{array}{c}100\\ 1,600\end{array}\right] \end{gathered}$$

02

Compute the vector

The value of components of the vector $A\stackrel{→}{x}$are,

$$\begin{gathered} A\stackrel{→}{x}=\left[\begin{array}{cc}1& 1/8\\ 8& 1\end{array}\right]×\left[\begin{array}{c}100\\ 1,600\end{array}\right] \\ ⇒A\stackrel{→}{x}=\left[\begin{array}{c}1×100+1/8×1,600\\ 8×100+1×1,600\end{array}\right] \\ ∴A\stackrel{→}{x}=\left[\begin{array}{c}300\\ 2400\end{array}\right] \end{gathered}$$

The components of the vector $A\stackrel{→}{x}$ represents the values $C\$=300,ZAR=2400$ .

03

Check for the invertible of the matrix

(b)

Consider the matrix.

$A=\left[\begin{array}{cc}1& 1/8\\ 8& 1\end{array}\right]$

The determinant of the matrix is,

$$\begin{gathered} \left|A\right|=1×1-8×\frac{1}{8} \\ \left|A\right|=0 \end{gathered}$$

As the determinant is zero, thus, the matrix is non-invertible.

04

Check for the consistency of the system

The row-echelon form of the matrix is,

$\begin{array}{rcl}{b}_2-8b_1& =& 0\\ & ⇒& b_2=8b_1\end{array}$

The condition for the consistency of the system will be,

role="math" localid="1659788422404" $\begin{array}{rcl}{b}_2-8b_1& =& 0\\ & ⇒& b_2=8b_1\end{array}$

The vector is,

$\stackrel{→}{b}=\left[\begin{array}{c}{b}_1\\ 8b_1\end{array}\right]$

Significance: the above matrix represents the relation, $1C\$=8ZAR$

If the system is consistent, then, it will have infinitely many solutions for the values of the vector $\stackrel{→}{x}$in $A\stackrel{→}{x}=\stackrel{→}{b}$

Significance: $\stackrel{→}{x}$represents the values of the total currencies.

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