Q59E The conversion formula C=59(F-32... [FREE SOLUTION]

91影视

Linear Algebra With Applications

Otto Bretscher

$Math Studyset 91影视 Explanations$ Math

5th Edition 路 442 pages

ISBN: 9780321796974

Chapter 2: Q59E (page 57)

The conversion formula $C = \frac{5}{9} (F - 32)$ from Fahrenheit to Celsius (as measures of temperature) is nonlinear, in the sense of linear algebra (why?). Still, there is a technique that allows us to use a matrix to represent this conversion.
a. Find the $2 脳 2$ matrix $A$ that transforms the vector $[\begin{matrix} F \\ 1 \end{matrix}]$ into the vector $[\begin{matrix} C \\ 1 \end{matrix}]$ . (The second row of A will be $[\begin{matrix} 0 & 1 \end{matrix}]$ .)
b. Is the matrix $A$ in part (a) invertible? If so, find the inverse (use Exercise 13). Use the result to write a formula expressing $F$ in terms of $C$

Short Answer

Expert verified

a. The matrix is $A = [\begin{matrix} \frac{5}{9} & - \frac{160}{9} \\ 0 & 1 \end{matrix}]$ ,

b. The matrix A is invertible. The inverse is $F = \frac{9}{5} C + 32$ .

Step by step solution

Step by Step Explanation: Step 1: Compute the matrix

(a)

The vectors are,

$[\begin{matrix} F \\ 1 \end{matrix}], [\begin{matrix} C \\ 1 \end{matrix}]$

Given, the second row of the matrix A is $[\begin{matrix} 0 & 1 \end{matrix}]$ .

$A F = C 鈬� [\begin{matrix} \frac{5}{9} & - \frac{160}{9} \\ 0 & 1 \end{matrix}] [\begin{matrix} F \\ 1 \end{matrix}] = [\begin{matrix} C \\ 1 \end{matrix}] 鈭� A = [\begin{matrix} \frac{5}{9} & - \frac{160}{9} \\ 0 & 1 \end{matrix}]$

Check for the invertibility of the matrix

(b)

Consider the matrix

$A = [\begin{matrix} \frac{5}{9} & - \frac{160}{9} \\ 0 & 1 \end{matrix}]$

The determinant is,

$|A| = \frac{5}{9} 脳 1 - (- \frac{160}{9} 脳 0) 鈮� 0$

As the determinant is not zero, thus, the matrix is invertible.

Compute the inverse of the matrix.

The inverse of the matrix is,

$\begin{array}{rcl} A & = & [\begin{matrix} \frac{5}{9} & - \frac{160}{9} \\ 0 & 1 \end{matrix}] \\ 鈬� & A^{- 1} = \frac{1}{\frac{5}{9} 脳 1 - (- \frac{160}{9} 脳 0)} [\begin{matrix} 1 & \frac{160}{9} \\ 0 & \frac{5}{9} \end{matrix}] \\ 鈭� A^{- 1} & = & [\begin{matrix} \frac{9}{5} & 32 \\ 0 & 1 \end{matrix}] \end{array}$

Compute the inverse in terms of temperature scales

Consider the inverse of the matrix.

$\begin{array}{rcl} A^{- 1} & = & [\begin{matrix} \frac{9}{5} & 32 \\ 0 & 1 \end{matrix}] \end{array}$

The inverse in terms of the temperature scales is,

$F = \frac{9}{5} C + 32$

Hence, $F = \frac{9}{5} C + 32$ is the inverse of the matrix A that is invertible..

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Short Answer

Step by step solution

Step by Step Explanation: Step 1: Compute the matrix

Check for the invertibility of the matrix

Compute the inverse of the matrix.

Compute the inverse in terms of temperature scales

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