The final question for this project is below. Screenshot the question and answer all parts, showing the matrices thatare being multiplied together, and the basic components of what you are computing. If you do it in Sage, ascreenshot of the computations there will suffice.Consider the matrix:2-1 1A = 1 2 -11 -1 21.Find Q and D such that A.2. Find the maximum and minimum values of q(u)= x¹Ax among all unit vectors u.3. Describe the direction in which the minimum value occurs. What can you say about the direction in which themaximum occurs?

Given Information:To find:1) The matrix .2) The maximum and minimum values of 3) The Direction of…

Answered: Consider the matrix: 2-1 1 A 1 2 -1 1…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Expert Solution

Step 1: Analysis and Introduction

Given Information:

$A equals open square brackets table row 2 cell negative 1 end cell 1 row cell negative 1 end cell 2 cell negative 1 end cell row 1 cell negative 1 end cell 2 end table close square brackets$

To find:

1) The matrix $Q comma D$ .

2) The maximum and minimum values of $q open parentheses bold u close parentheses equals bold x to the power of T A bold x$

3) The Direction of minimum and maximum occurs.

Concept used:

Any symmetric matrix is orthogonally diagonalizable.

i.e., $A equals Q D Q to the power of T$ , where $Q$ is a matrix formed by the orthonormal eigenvectors as columns and $D$ is a matrix which is formed by the eigenvalues as the diagonal entry.

In orthogonal diagonalization, the maximum value is equal to the maximum eigenvalue and it is in the direction of the unit eigenvector corresponds to it. The minimum value is equal to the minimum eigenvalue and it is in the direction of the unit eigenvector corresponds to it.

Gram Schmidt Orthogonalization Process:

If the vectors are given as $open curly brackets w subscript 1 comma w subscript 2 comma w subscript 3 close curly brackets$ , then it can be orthogonalized as,

$table row cell v subscript 1 end cell equals cell w subscript 1 end cell row cell v subscript 2 end cell equals cell w subscript 2 minus fraction numerator v subscript 1 times w subscript 2 over denominator open vertical bar v subscript 1 close vertical bar squared end fraction v subscript 1 end cell row cell v subscript 3 end cell equals cell w subscript 3 minus fraction numerator v subscript 1 times w subscript 3 over denominator open vertical bar v subscript 1 close vertical bar squared end fraction v subscript 1 minus fraction numerator v subscript 2 times w subscript 3 over denominator open vertical bar v subscript 2 close vertical bar squared end fraction v subscript 2 end cell end table$

Now, the vectors $open curly brackets v subscript 1 comma v subscript 2 comma v subscript 3 close curly brackets$ are orthogonal to each other and $open curly brackets fraction numerator v subscript 1 over denominator open vertical bar v subscript 1 close vertical bar end fraction comma fraction numerator v subscript 2 over denominator open vertical bar v subscript 2 close vertical bar end fraction comma fraction numerator v subscript 3 over denominator open vertical bar v subscript 3 close vertical bar end fraction close curly brackets$ are orthonormal vectors.