Log In Start studying!

Select your language

Suggested languages for you:
Deutsch (DE)
Deutsch (UK)
Deutsch (US)

Americas

Europe

Answers without the blur. Sign up and see all textbooks for free! → Illustration

Q7E

Expert-verified
Linear Algebra and its Applications
Found in: Page 331
Linear Algebra and its Applications

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

Answers without the blur.

Just sign up for free and you're in.

Register for Free I’ll do it later
Illustration

Short Answer

A certain experiment produces the data \(\left( {1,1.8} \right),\left( {2,2.7} \right),\left( {3,3.4} \right),\left( {4,3.8} \right),\left( {5,3.9} \right)\). Describe the model that produces a least-squares fit of these points by a function of the form

\(y = {\beta _1}x + {\beta _2}{x^2}\)

Such a function might arise, for example, as the revenue from the sale of \(x\) units of a product, when the amount offered for sale affects the price to be set for the product.

a. Give the design matrix, the observation vector, and the unknown parameter vector.

b. Find the associated least-squares curve for the data.

a. The required matrix and vectors are,

Design Matrix: \(X = \left( {\begin{aligned}1&1\\2&4\\3&9\\4&{16}\\5&{25}\end{aligned}} \right)\)

Observation vector: \({\bf{y}} = \left( {\begin{aligned}{1.8}\\{2.7}\\{3.4}\\{3.8}\\{3.9}\end{aligned}} \right)\)

Parameter vector: \(\beta = \left( {\begin{aligned}{{\beta _1}}\\{{\beta _2}}\end{aligned}} \right)\)

b. The least-squares equation is \({\bf{y}} = 1.76x - 0.20{x^2}\).

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

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 parameter vector, and \( \in = \left( {\begin{aligned}{{ \in _1}}\\{{ \in _2}}\\ \vdots \\{{ \in _n}}\end{aligned}} \right)\) is a residual vector

Find design matrix, observation vector, parameter vector for the given data set 

(a)

The given data points are: \(\left( {1,1.8} \right),\left( {2,2.7} \right),\left( {3,3.4} \right),\left( {4,3.8} \right),\left( {5,3.9} \right)\).

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

Design matrix:

\(\begin{aligned}X = \left( {\begin{aligned}1&{{1^2}}\\2&{{2^2}}\\3&{{3^2}}\\4&{{4^2}}\\5&{{5^2}}\end{aligned}} \right)\\ = \left( {\begin{aligned}1&1\\2&4\\3&9\\4&{16}\\5&{25}\end{aligned}} \right)\end{aligned}\)

Observational vector:

\({\bf{y}} = \left( {\begin{aligned}{1.8}\\{2.7}\\{3.4}\\{3.8}\\{3.9}\end{aligned}} \right)\)

And, the parameter vector for the given equation is shown below:

\(\beta = \left( {\begin{aligned}{{\beta _1}}\\{{\beta _2}}\end{aligned}} \right)\)

The normal equation

(b)

The general least-squares equation is given by \({\bf{y}} = {\beta _1}x + {\beta _2}{x^2}\), and to find the associated least-squares curve, the values of \({\beta _1},{\rm{ }}{\beta _2}\) are required, so find the values of \({\beta _1},{\rm{ }}{\beta _2}\) by using the normal equation.

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

\(\left( {{{\left( {\begin{aligned}1&1\\2&4\\3&9\\4&{16}\\5&{25}\end{aligned}} \right)}^T}\left( {\begin{aligned}1&1\\2&4\\3&9\\4&{16}\\5&{25}\end{aligned}} \right)} \right)\left( {\begin{aligned}{{\beta _1}}\\{{\beta _2}}\end{aligned}} \right) = {\left( {\begin{aligned}1&1\\2&4\\3&9\\4&{16}\\5&{25}\end{aligned}} \right)^T}\left( {\begin{aligned}{1.8}\\{2.7}\\{3.4}\\{3.8}\\{3.9}\end{aligned}} \right)\)

That implies;

\(\left( {\begin{aligned}{{\beta _1}}\\{{\beta _2}}\end{aligned}} \right) = {\left( {{{\left( {\begin{aligned}1&1\\2&4\\3&9\\4&{16}\\5&{25}\end{aligned}} \right)}^T}\left( {\begin{aligned}1&1\\2&4\\3&9\\4&{16}\\5&{25}\end{aligned}} \right)} \right)^{ - 1}}{\left( {\begin{aligned}1&1\\2&4\\3&9\\4&{16}\\5&{25}\end{aligned}} \right)^T}\left( {\begin{aligned}{1.8}\\{2.7}\\{3.4}\\{3.8}\\{3.9}\end{aligned}} \right)\)

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

  1. Enter the data \(\left( {\begin{aligned}{{\beta _1}}\\{{\beta _2}}\end{aligned}} \right) = {\left( {{{\left( {\begin{aligned}1&1\\2&4\\3&9\\4&{16}\\5&{25}\end{aligned}} \right)}^T}\left( {\begin{aligned}1&1\\2&4\\3&9\\4&{16}\\5&{25}\end{aligned}} \right)} \right)^{ - 1}}{\left( {\begin{aligned}1&1\\2&4\\3&9\\4&{16}\\5&{25}\end{aligned}} \right)^T}\left( {\begin{aligned}{1.8}\\{2.7}\\{3.4}\\{3.8}\\{3.9}\end{aligned}} \right)\) in the tab in the form of \({\left( {{{\left( {\begin{aligned}1&1\\2&4\\3&9\\4&{16}\\5&{25}\end{aligned}} \right)}^T}\left( {\begin{aligned}1&1\\2&4\\3&9\\4&{16}\\5&{25}\end{aligned}} \right)} \right)^{ - 1}}{\left( {\begin{aligned}1&1\\2&4\\3&9\\4&{16}\\5&{25}\end{aligned}} \right)^T}\left( {\begin{aligned}{1.8}\\{2.7}\\{3.4}\\{3.8}\\{3.9}\end{aligned}} \right)\).
  2. Use colons after that and press ENTER.

So, the value of \(\left( {\begin{aligned}{{\beta _1}}\\{{\beta _2}}\end{aligned}} \right)\) is \(\left( {\begin{aligned}{1.76}\\{ - 0.20}\end{aligned}} \right)\).

Now, substitute the obtained values into \({\bf{y}} = {\beta _1}x + {\beta _2}{x^2}\).

\({\bf{y}} = 1.76x - 0.20{x^2}\)

So, the required equation is \({\bf{y}} = 1.76x - 0.20{x^2}\).

Most popular questions for Math Textbooks

For a matrix program, the Gram–Schmidt process works better with orthonormal vectors. Starting with \({x_1},......,{x_p}\) as in Theorem 11, let \(A = \left\{ {{x_1},......,{x_p}} \right\}\) . Suppose \(Q\) is an \(n \times k\) matrix whose columns form an orthonormal basis for

the subspace \({W_k}\) spanned by the first \(k\) columns of A. Then for \(x\) in \({\mathbb{R}^n}\), \(Q{Q^T}x\) is the orthogonal projection of x onto \({W_k}\) (Theorem 10 in Section 6.3). If \({x_{k + 1}}\) is the next column of \(A\), then equation (2) in the proof of Theorem 11 becomes

\({v_{k + 1}} = {x_{k + 1}} - Q\left( {{Q^T}T {x_{k + 1}}} \right)\)

(The parentheses above reduce the number of arithmetic operations.) Let \({u_{k + 1}} = \frac{{{v_{k + 1}}}}{{\left\| {{v_{k + 1}}} \right\|}}\). The new \(Q\) for the next step is \(\left( {\begin{aligned}{{}{}}Q&{{u_{k + 1}}}\end{aligned}} \right)\). Use this procedure to compute the \(QR\) factorization of the matrix in Exercise 24. Write the keystrokes or commands you use.

In Exercises 1-4, find the equation \(y = {\beta _0} + {\beta _1}x\) of the least-square line that best fits the given data points.

  1. \(\left( {1,0} \right),\left( {2,1} \right),\left( {4,2} \right),\left( {5,3} \right)\)

In Exercises 7–10, let \[W\] be the subspace spanned by the \[{\bf{u}}\]’s, and write y as the sum of a vector in \[W\] and a vector orthogonal to \[W\].

7. \[y = \left[ {\begin{aligned}1\\3\\5\end{aligned}} \right]\], \[{{\bf{u}}_1} = \left[ {\begin{aligned}1\\3\\{ - 2}\end{aligned}} \right]\], \[{{\bf{u}}_2} = \left[ {\begin{aligned}5\\1\\4\end{aligned}} \right]\]

Question: In Exercises 9-12, find (a) the orthogonal projection of b onto \({\bf{Col}}A\) and (b) a least-squares solution of \(A{\bf{x}} = {\bf{b}}\).

11. \(A = \left( {\begin{aligned}{{}{}}{\bf{4}}&{\bf{0}}&{\bf{1}}\\{\bf{1}}&{ - {\bf{5}}}&{\bf{1}}\\{\bf{6}}&{\bf{1}}&{\bf{0}}\\{\bf{1}}&{ - {\bf{1}}}&{ - {\bf{5}}}\end{aligned}} \right)\), \({\bf{b}} = \left( {\begin{aligned}{{}{}}{\bf{9}}\\{\bf{0}}\\{\bf{0}}\\{\bf{0}}\end{aligned}} \right)\)

In Exercises 7–10, let \[W\] be the subspace spanned by the \[{\bf{u}}\]’s, and write y as the sum of a vector in \[W\] and a vector orthogonal to \[W\] .

8. \[y = \left[ {\begin{aligned}{ - 1}\\4\\3\end{aligned}} \right]\], \[{{\bf{u}}_1} = \left[ {\begin{aligned}1\\1\\{\bf{1}}\end{aligned}} \right]\], \[{{\bf{u}}_2} = \left[ {\begin{aligned}{ - 1}\\3\\{ - 2}\end{aligned}} \right]\]
Icon

Want to see more solutions like these?

Sign up for free to discover our expert answers
Get Started - It’s free

Recommended explanations on Math Textbooks

Decision Maths

Read explanation

Calculus

Read explanation

Probability and Statistics

Read explanation

Statistics

Read explanation

Mechanics Maths

Read explanation

Geometry

Read explanation

Pure Maths

Read explanation

94% of StudySmarter users get better grades.

Sign up for free
94% of StudySmarter users get better grades.