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Q10E

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Found in: Page 331

### Linear Algebra and its Applications

Book edition 5th
Author(s) David C. Lay, Steven R. Lay and Judi J. McDonald
Pages 483 pages
ISBN 978-03219822384

# Suppose radioactive substance A and B have decay constants of $$.02$$ and $$.07$$, respectively. If a mixture of these two substances at a time $$t = 0$$ contains $${M_A}$$ grams of $$A$$ and $${M_B}$$ grams of $$B$$, then a model for the total amount of mixture present at time $$t$$ is$$y = {M_A}{e^{ - .02t}} + {M_B}{e^{ - .07t}}$$ (6)Suppose the initial amounts $${M_A}$$ and are unknown, but a scientist is able to measure the total amounts present at several times and records the following points $$\left( {{t_i},{y_i}} \right):\left( {10,21.34} \right),\left( {11,20.68} \right),\left( {12,20.05} \right),\left( {14,18.87} \right)$$ and $$\left( {15,18.30} \right)$$. a. Describe a linear model that can be used to estimate $${M_A}$$ and $${M_B}$$.b. Find the least-squares curved based on (6).

(a) The required matrix and vectors are shown below:

Design Matrix: X = \left( {\begin{aligned}{{e^{ - 0.02\left( {10} \right)}}}&{{e^{ - 0.07\left( {10} \right)}}}\\{{e^{ - 0.02\left( {11} \right)}}}&{{e^{ - 0.07\left( {11} \right)}}}\\{{e^{ - 0.02\left( {12} \right)}}}&{{e^{ - 0.07\left( {12} \right)}}}\\{{e^{ - 0.02\left( {14} \right)}}}&{{e^{ - 0.07\left( {14} \right)}}}\\{{e^{ - 0.02\left( {15} \right)}}}&{{e^{ - 0.07\left( {15} \right)}}}\end{aligned}} \right)

Observation vector: {\bf{y}} = \left( {\begin{aligned}{21.34}\\{20.68}\\{20.05}\\{18.87}\\{18.30}\end{aligned}} \right)

Parameter vector: \beta = \left( {\begin{aligned}{{M_A}}\\{{M_B}}\end{aligned}} \right)

(b) The least-squares equation is $$y = 19.94{e^{ - 0.02t}} + 10.10{e^{ - 0.07t}}$$.

See the step by step solution

## The General Linear Model

The equation of the general linear model is defined as:

$${\bf{y}} = X\beta + \in$$

Here, {\bf{y}} = \left( {\begin{aligned}{{y_1}}\\{{y_2}}\\ \vdots \\{{y_n}}\end{aligned}} \right) is an observational vector, X = \left( {\begin{aligned}1&{{x_1}}& \cdots &{x_1^n}\\1&{{x_2}}& \cdots &{x_2^n}\\ \vdots & \vdots & \ddots & \vdots \\1&{{x_n}}& \cdots &{x_n^n}\end{aligned}} \right) is the design matrix, \beta = \left( {\begin{aligned}{{\beta _1}}\\{{\beta _2}}\\ \vdots \\{{\beta _n}}\end{aligned}} \right) is the parameter vector, and \in = \left( {\begin{aligned}{{ \in _1}}\\{{ \in _2}}\\ \vdots \\{{ \in _n}}\end{aligned}} \right) is the residual vector.

## Find design matrix, observation vector, parameter vector for given data

(a)

The given equation is $$y = {M_A}{e^{ - 0.02t}} + {M_B}{e^{ - 0.07t}}$$ , and the given data sets are $$\left( {10,21.34} \right)$$, $$\left( {11,20.68} \right)$$, $$\left( {12,20.05} \right)$$, $$\left( {14,18.87} \right)$$ and $$\left( {15,18.30} \right)$$.

Write the Design matrix, observational vector, and the parameter vector for the given equation and data set by using the information given in step 1.

Design matrix:

X = \left( {\begin{aligned}{{e^{ - 0.02\left( {10} \right)}}}&{{e^{ - 0.07\left( {10} \right)}}}\\{{e^{ - 0.02\left( {11} \right)}}}&{{e^{ - 0.07\left( {11} \right)}}}\\{{e^{ - 0.02\left( {12} \right)}}}&{{e^{ - 0.07\left( {12} \right)}}}\\{{e^{ - 0.02\left( {14} \right)}}}&{{e^{ - 0.07\left( {14} \right)}}}\\{{e^{ - 0.02\left( {15} \right)}}}&{{e^{ - 0.07\left( {15} \right)}}}\end{aligned}} \right)

Observational vector:

{\bf{y}} = \left( {\begin{aligned}{21.34}\\{20.68}\\{20.05}\\{18.87}\\{18.30}\end{aligned}} \right)

And the parameter vector for the given equation is,

\beta = \left( {\begin{aligned}{{M_A}}\\{{M_B}}\end{aligned}} \right)

Thus, the above values are the best fit for the given data set and equation.

## Normal equation

The normal equation is given by,

$${X^T}X\beta = {X^T}{\bf{y}}$$

## Find the least-squares curve

The general least-squares equation is given by $$y = {M_A}{e^{ - 0.02t}} + {M_B}{e^{ - 0.07t}}$$, and to find the associated least-squares curve, the values of $${M_A},{M_B}$$ are required, so find the values of $${M_A},{M_B}$$ by using normal equation.

By using the obtained information from step 2, the normal equation will be,

$$\beta = {\left( {{X^T}X} \right)^{ - 1}}{X^T}{\bf{y}}$$

That implies;

\left( {\begin{aligned}{{M_A}}\\{{M_B}}\end{aligned}} \right) = {\left( {{{\left( {\begin{aligned}{{e^{ - 0.02\left( {10} \right)}}}&{{e^{ - 0.07\left( {10} \right)}}}\\{{e^{ - 0.02\left( {11} \right)}}}&{{e^{ - 0.07\left( {11} \right)}}}\\{{e^{ - 0.02\left( {12} \right)}}}&{{e^{ - 0.07\left( {12} \right)}}}\\{{e^{ - 0.02\left( {14} \right)}}}&{{e^{ - 0.07\left( {14} \right)}}}\\{{e^{ - 0.02\left( {15} \right)}}}&{{e^{ - 0.07\left( {15} \right)}}}\end{aligned}} \right)}^T}\left( {\begin{aligned}{{e^{ - 0.02\left( {10} \right)}}}&{{e^{ - 0.07\left( {10} \right)}}}\\{{e^{ - 0.02\left( {11} \right)}}}&{{e^{ - 0.07\left( {11} \right)}}}\\{{e^{ - 0.02\left( {12} \right)}}}&{{e^{ - 0.07\left( {12} \right)}}}\\{{e^{ - 0.02\left( {14} \right)}}}&{{e^{ - 0.07\left( {14} \right)}}}\\{{e^{ - 0.02\left( {15} \right)}}}&{{e^{ - 0.07\left( {15} \right)}}}\end{aligned}} \right)} \right)^{ - 1}}{\left( {\begin{aligned}{{e^{ - 0.02\left( {10} \right)}}}&{{e^{ - 0.07\left( {10} \right)}}}\\{{e^{ - 0.02\left( {11} \right)}}}&{{e^{ - 0.07\left( {11} \right)}}}\\{{e^{ - 0.02\left( {12} \right)}}}&{{e^{ - 0.07\left( {12} \right)}}}\\{{e^{ - 0.02\left( {14} \right)}}}&{{e^{ - 0.07\left( {14} \right)}}}\\{{e^{ - 0.02\left( {15} \right)}}}&{{e^{ - 0.07\left( {15} \right)}}}\end{aligned}} \right)^T}\left( {\begin{aligned}{21.34}\\{20.68}\\{20.05}\\{18.87}\\{18.30}\end{aligned}} \right)

Use the following steps to find the associated values for the obtained data in MATLAB.

1. Enter the data \left( {\begin{aligned}{{M_A}}\\{{M_B}}\end{aligned}} \right) = {\left( {{{\left( {\begin{aligned}{{e^{ - 0.02\left( {10} \right)}}}&{{e^{ - 0.07\left( {10} \right)}}}\\{{e^{ - 0.02\left( {11} \right)}}}&{{e^{ - 0.07\left( {11} \right)}}}\\{{e^{ - 0.02\left( {12} \right)}}}&{{e^{ - 0.07\left( {12} \right)}}}\\{{e^{ - 0.02\left( {14} \right)}}}&{{e^{ - 0.07\left( {14} \right)}}}\\{{e^{ - 0.02\left( {15} \right)}}}&{{e^{ - 0.07\left( {15} \right)}}}\end{aligned}} \right)}^T}\left( {\begin{aligned}{{e^{ - 0.02\left( {10} \right)}}}&{{e^{ - 0.07\left( {10} \right)}}}\\{{e^{ - 0.02\left( {11} \right)}}}&{{e^{ - 0.07\left( {11} \right)}}}\\{{e^{ - 0.02\left( {12} \right)}}}&{{e^{ - 0.07\left( {12} \right)}}}\\{{e^{ - 0.02\left( {14} \right)}}}&{{e^{ - 0.07\left( {14} \right)}}}\\{{e^{ - 0.02\left( {15} \right)}}}&{{e^{ - 0.07\left( {15} \right)}}}\end{aligned}} \right)} \right)^{ - 1}}{\left( {\begin{aligned}{{e^{ - 0.02\left( {10} \right)}}}&{{e^{ - 0.07\left( {10} \right)}}}\\{{e^{ - 0.02\left( {11} \right)}}}&{{e^{ - 0.07\left( {11} \right)}}}\\{{e^{ - 0.02\left( {12} \right)}}}&{{e^{ - 0.07\left( {12} \right)}}}\\{{e^{ - 0.02\left( {14} \right)}}}&{{e^{ - 0.07\left( {14} \right)}}}\\{{e^{ - 0.02\left( {15} \right)}}}&{{e^{ - 0.07\left( {15} \right)}}}\end{aligned}} \right)^T}\left( {\begin{aligned}{21.34}\\{20.68}\\{20.05}\\{18.87}\\{18.30}\end{aligned}} \right) in the tab in the form of {\left( {{{\left( {\begin{aligned}{{e^{ - 0.02\left( {10} \right)}}}&{{e^{ - 0.07\left( {10} \right)}}}\\{{e^{ - 0.02\left( {11} \right)}}}&{{e^{ - 0.07\left( {11} \right)}}}\\{{e^{ - 0.02\left( {12} \right)}}}&{{e^{ - 0.07\left( {12} \right)}}}\\{{e^{ - 0.02\left( {14} \right)}}}&{{e^{ - 0.07\left( {14} \right)}}}\\{{e^{ - 0.02\left( {15} \right)}}}&{{e^{ - 0.07\left( {15} \right)}}}\end{aligned}} \right)}^T}\left( {\begin{aligned}{{e^{ - 0.02\left( {10} \right)}}}&{{e^{ - 0.07\left( {10} \right)}}}\\{{e^{ - 0.02\left( {11} \right)}}}&{{e^{ - 0.07\left( {11} \right)}}}\\{{e^{ - 0.02\left( {12} \right)}}}&{{e^{ - 0.07\left( {12} \right)}}}\\{{e^{ - 0.02\left( {14} \right)}}}&{{e^{ - 0.07\left( {14} \right)}}}\\{{e^{ - 0.02\left( {15} \right)}}}&{{e^{ - 0.07\left( {15} \right)}}}\end{aligned}} \right)} \right)^{ - 1}}{\left( {\begin{aligned}{{e^{ - 0.02\left( {10} \right)}}}&{{e^{ - 0.07\left( {10} \right)}}}\\{{e^{ - 0.02\left( {11} \right)}}}&{{e^{ - 0.07\left( {11} \right)}}}\\{{e^{ - 0.02\left( {12} \right)}}}&{{e^{ - 0.07\left( {12} \right)}}}\\{{e^{ - 0.02\left( {14} \right)}}}&{{e^{ - 0.07\left( {14} \right)}}}\\{{e^{ - 0.02\left( {15} \right)}}}&{{e^{ - 0.07\left( {15} \right)}}}\end{aligned}} \right)^T}\left( {\begin{aligned}{21.34}\\{20.68}\\{20.05}\\{18.87}\\{18.30}\end{aligned}} \right).
2. Use colons after that and press ENTER.

So, the value of \left( {\begin{aligned}{{M_A}}\\{{M_B}}\end{aligned}} \right) is \left( {\begin{aligned}{19.94}\\{10.10}\end{aligned}} \right).

Now, substitute the obtained values into $$y = {M_A}{e^{ - 0.02t}} + {M_B}{e^{ - 0.07t}}$$.

$$y = 19.94{e^{ - 0.02t}} + 10.10{e^{ - 0.07t}}$$

So, the required equation is $$y = 19.94{e^{ - 0.02t}} + 10.10{e^{ - 0.07t}}$$.