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

Q5E

Expert-verified
Linear Algebra and its Applications
Found in: Page 395
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

Find the matrix of the quadratic form. Assume x is in \({\mathbb{R}^2}\).

a. \(3x_1^2 + 2x_2^2 - 5x_3^2 - 6{x_1}{x_2} + 8{x_1}{x_3} - 4{x_2}{x_3}\)

b. \(6{x_1}{x_2} + 4{x_1}{x_3} - 10{x_2}{x_3}\)

  1. The matrix for the quadratic form \(3x_1^2 + 2x_2^2 - 5x_3^2 - 6{x_1}{x_2} + 8{x_1}{x_3} - 4{x_2}{x_3}\) is \(\left( {\begin{aligned}{{}}3&{ - 3}&4\\{ - 3}&2&{ - 2}\\4&{ - 2}&{ - 5}\end{aligned}} \right)\).

  1. The matrix for the quadratic form \(6{x_1}{x_2} + 4{x_1}{x_3} - 10{x_2}{x_3}\) is \(\left( {\begin{aligned}{{}}0&3&2\\{ - 3}&0&{ - 5}\\2&{ - 5}&0\end{aligned}} \right)\).
See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1: Matrix of the quadratic form

The coefficients of the square of variables, that is \(x_i^2\), are to be divided on the main diagonal as per the order, and the coefficient of term \({x_i}{x_j}{\rm{ }}\left( {i \ne j} \right)\), divided by 2 to split properly between the entries \(\left( {i,j} \right)\) and \(\left( {j,i} \right)\).

Step 2: Find the corresponding matrix of \(5x_1^2 + 16{x_1}{x_2} - 5x_2^2\)

(a)

The given quadratic expression is \(3x_1^2 + 2x_2^2 - 5x_3^2 - 6{x_1}{x_2} + 8{x_1}{x_3} - 4{x_2}{x_3}\).

For this expression, the order of the square matrix is 3.

So, let \(A = \left( {\begin{aligned}{{}}{{a_{11}}}&{{a_{12}}}&{{a_{13}}}\\{{a_{21}}}&{{a_{22}}}&{{a_{23}}}\\{{a_{31}}}&{{a_{32}}}&{{a_{33}}}\end{aligned}} \right)\)

According to the rule in steps (1), 3, 2, and \( - 5\), will be on the main diagonal of the matrix, like \({a_{11}} = 3\), \({a_{22}} = 2\) and \({a_{33}} = - 5\).

And,\({a_{12}} = \frac{{ - 6}}{2}\), \({a_{21}} = \frac{{ - 6}}{2}\), \({a_{13}} = \frac{8}{2}\), \({a_{31}} = \frac{8}{2}\) and \({a_{23}} = \frac{{ - 4}}{2}\), \({a_{32}} = \frac{{ - 4}}{2}\).

Substitute the required value into \(A = \left( {\begin{aligned}{{}}{{a_{11}}}&{{a_{12}}}&{{a_{13}}}\\{{a_{21}}}&{{a_{22}}}&{{a_{23}}}\\{{a_{31}}}&{{a_{32}}}&{{a_{33}}}\end{aligned}} \right)\).

\(A = \left( {\begin{aligned}{{}}3&{ - 3}&4\\{ - 3}&2&{ - 2}\\4&{ - 2}&{ - 5}\end{aligned}} \right)\)

So, the required matrix is \(\left( {\begin{aligned}{{}}3&{ - 3}&4\\{ - 3}&2&{ - 2}\\4&{ - 2}&{ - 5}\end{aligned}} \right)\).

Step 3: Find the corresponding matrix of \(2{x_1}{x_2}\)

(b)

The given quadratic expression is \(6{x_1}{x_2} + 4{x_1}{x_3} - 10{x_2}{x_3}\), which can be written as \(0x_1^2 + 0x_2^2 + 0x_3^2 + 6{x_1}{x_2} + 4{x_1}{x_3} - 10{x_2}{x_3}\).

For this expression, the order of the square matrix is 3.

So, let \(A = \left( {\begin{aligned}{{}}{{a_{11}}}&{{a_{12}}}&{{a_{13}}}\\{{a_{21}}}&{{a_{22}}}&{{a_{23}}}\\{{a_{31}}}&{{a_{32}}}&{{a_{33}}}\end{aligned}} \right)\)

According to the rule in step (1), 0, 0 and 0, will be on the main diagonal of the matrix, like \({a_{11}} = 0\), \({a_{22}} = 0\) and \({a_{33}} = 0\).

And \({a_{12}} = \frac{6}{2}\), \({a_{21}} = \frac{6}{2}\), \({a_{13}} = \frac{2}{2}\), \({a_{31}} = \frac{2}{2}\) and \({a_{23}} = \frac{{ - 10}}{2}\), \({a_{32}} = \frac{{ - 10}}{2}\) .

Substitute the required value into \(A = \left( {\begin{aligned}{{}}{{a_{11}}}&{{a_{12}}}&{{a_{13}}}\\{{a_{21}}}&{{a_{22}}}&{{a_{23}}}\\{{a_{31}}}&{{a_{32}}}&{{a_{33}}}\end{aligned}} \right)\).

\(A = \left( {\begin{aligned}{{}}0&3&2\\{ - 3}&0&{ - 5}\\2&{ - 5}&0\end{aligned}} \right)\)

So, the required matrix is \(\left( {\begin{aligned}{{}}0&3&2\\{ - 3}&0&{ - 5}\\2&{ - 5}&0\end{aligned}} \right)\).

Most popular questions for Math Textbooks

Show that if A is an \(n \times n\) symmetric matrix, then \(\left( {A{\bf{x}}} \right) \cdot {\bf{y}} = {\bf{x}} \cdot \left( {A{\bf{y}}} \right)\) for x, y in \({\mathbb{R}^n}\).

Question 11: Prove that any \(n \times n\) matrix A admits a polar decomposition of the form \(A = PQ\), where P is a \(n \times n\) positive semidefinite matrix with the same rank as A and where Q is an \(n \times n\) orthogonal matrix. (Hint: Use a singular value decomposition, \(A = U\sum {V^T}\), and observe that \(A = \left( {U\sum {U^T}} \right)\left( {U{V^T}} \right)\).) This decomposition is used, for instance, in mechanical engineering to model the deformation of a material. The matrix P describe the stretching or compression of the material (in the directions of the eigenvectors of P), and Q describes the rotation of the material in space.

In Exercises 17–24, \(A\) is an \(m \times n\) matrix with a singular value decomposition \(A = U\Sigma {V^T}\) ,

\(\)

where \(U\) is an \(m \times m\) orthogonal matrix, \({\bf{\Sigma }}\) is an \(m \times n\) “diagonal” matrix with \(r\) positive entries and no negative entries, and \(V\) is an \(n \times n\) orthogonal matrix. Justify each answer.

18. Suppose \(A\) is square and invertible. Find a singular value decomposition of \({A^{ - 1}}\)

Let \(A = \left( {\begin{aligned}{{}}{\,\,\,2}&{ - 1}&{ - 1}\\{ - 1}&{\,\,\,2}&{ - 1}\\{ - 1}&{ - 1} &{\,\,\,2}\end{aligned}} \right)\),\({{\rm{v}}_1} = \left( {\begin{aligned}{{}}{ - 1}\\{\,\,\,0}\\{\,\,1}\end{aligned}} \right)\) and and\({{\rm{v}}_2} = \left( {\begin{aligned}{{}}{\,\,\,1}\\{\, - 1}\\{\,\,\,\,1}\end{aligned}} \right)\). Verify that\({{\rm{v}}_1}\), \({{\rm{v}}_2}\) an eigenvector of \(A\). Then orthogonally diagonalize \(A\).

Question: In Exercises 1 and 2, convert the matrix of observations to mean deviation form, and construct the sample covariance matrix.

\(2.\,\,\left( {\begin{array}{*{20}{c}}1&5&2&6&7&3\\3&{11}&6&8&{15}&{11}\end{array}} \right)\)
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.