DOC PREVIEW
UIUC MATH 415 - midterm3-sol

This preview shows page 1-2-3 out of 9 pages.

Save
View full document
View full document
Premium Document
Do you want full access? Go Premium and unlock all 9 pages.
Access to all documents
Download any document
Ad free experience
View full document
Premium Document
Do you want full access? Go Premium and unlock all 9 pages.
Access to all documents
Download any document
Ad free experience
View full document
Premium Document
Do you want full access? Go Premium and unlock all 9 pages.
Access to all documents
Download any document
Ad free experience
Premium Document
Do you want full access? Go Premium and unlock all 9 pages.
Access to all documents
Download any document
Ad free experience

Unformatted text preview:

Math 415 Midterm 3 Thursday November 20 2014 Circle your section Philipp Hieronymi 2pm 3pm Armin Straub 9am 11am Name NetID UIN Problem 0 1 point Write down the number of your discussion section for instance AD2 or ADH and the first name of your TA Allen Anton Mahmood Michael Nathan Pouyan Tigran Travis Section TA To be completed by the grader 0 1 2 3 4 5 MC P 1 7 8 8 8 8 20 60 Good luck 1 Instructions No notes personal aids or calculators are permitted This exam consists of 10 pages Take a moment to make sure you have all pages You have 75 minutes Answer all questions in the space provided If you require more space to write your answer you may continue on the back of the page make it clear if you do Explain your work Little or no points will be given for a correct answer with no explanation of how you got it In particular you have to write down all row operations for full credit 1 2 Problem 1 7 points Find the QR decomposition of A 1 0 Solution We start with the columns of A v1 v2 and we use Gram Schmidt to find the columns of Q q1 q2 1 1 1 v1 q1 12 kv1 k 1 2 k k 1 and 1 1 2 2 1 2 2 1 1 1 0 0 1 v2 q1 v2 q1 2 2 2 q2 1 1 1 kv2 q1 v2 q1 k 1 2 2 2 k k k 12 12 k 1 0 0 2 2 Hence Q 1 2 1 2 1 2 12 Finally T R Q A 1 2 1 2 1 1 2 1 2 1 2 2 2 0 0 2 2 0 1 Problem 2 8 points Let A 2 1 0 Find the eigenvalues of A as well as a basis for 0 0 1 the corresponding eigenspaces Solution If we expand along the last row we obtain 2 0 1 2 0 1 0 1 det det 2 1 2 2 2 1 0 0 1 Hence the eigenvalues of A are 1 with multiplicity 2 and 2 For 1 1 0 1 1 0 1 1 0 0 R1 R1 1 2R1 R2 1 2R2 R2 R2 2R1 2 0 0 0 0 2 0 0 1 0 0 0 0 0 0 0 0 0 2 Since there is only one free variable there could have been up to 2 because the eigenvalue has 0 multiplicity 2 the corresponding eigenspace is just one dimensional and has basis 1 0 For 2 0 0 1 1 21 0 R2 1 2R2 R3 R3 R1 R1 R2 2 1 0 0 0 1 0 0 1 0 0 0 1 Hence the corresponding eigenspace has the basis 2 0 1 1 1 Problem 3 Let A 1 1 and b 1 1 0 1 a 6 points Find the least squares solution of Ax b b 2 points Find the least squares line for the data points 1 1 0 1 1 1 Solution a We have to solve AT Ax AT b 1 1 3 0 1 1 1 T 1 1 A A 0 2 1 1 0 1 0 and 1 1 1 1 1 T 1 A b 0 1 1 0 1 3 0 1 R1 1 3R1 R2 1 2R2 1 0 13 0 2 0 0 1 0 1 x 3 0 Since we obtain a b Writing the least squares line as y a bx we have to find a and b so that is the b least squares solution of by changing the order of the data points 1 1 1 Ax 1 1 x 1 b 1 0 1 By part a we have 1 a 3 b 0 1 Hence the least squares line is y 3 3 Problem 4 Find the determinant of the following matrices Show all steps of your calculations a 4 points 1 0 3 0 0 1 0 1 0 0 0 2 1 1 0 0 b 4 points 2 1 3 7 0 1 0 1 2 1 3 2 0 1 1 0 Hint For this second matrix begin your calculation with a row operation to save time Solution a We 1 0 det 0 1 expand the determinant along the third column 0 3 0 0 1 1 1 1 1 0 1 3 2 6 3 det 0 0 2 3 det 0 2 0 0 2 1 1 0 1 0 0 b We use row operations to transform the matrix A into an upper triangular matrix B 2 1 3 7 2 1 3 7 0 1 0 1 R3 R3 R1 R4 R4 R2 R3 R4 0 1 0 1 A 2 1 3 2 0 0 1 1 B 0 0 0 5 0 1 1 0 Since we swap rows once we have det A det B Hence det A det B 2 1 1 5 10 1 0 1 0 Problem 5 Let W span 0 2 a 6 points Find the projection matrix corresponding to orthogonal projection onto W 1 b 2 points What is the orthogonal projection of 1 onto W 0 Solution a We have to find the orthogonal projection of elements of the standard basis 0 1 onto W Since 1 and 0 are orthogonal to each other the orthogonal projection of 0 2 the first standard basis vector onto W is 1 1 1 0 0 1 0 0 1 1 0 1 5 0 0 0 2 0 1 0 0 0 0 1 1 2 0 W 0 2 5 1 1 0 0 0 0 2 2 4 and 0 0 1 1 0 0 0 1 0 0 0 1 1 0 0 0 1 1 0 1 0 0 2 0 1 1 1 2 0 0 0 2 2 0 0 0 1 0 0 1 0 0 1 0 0 0 1 W 1 1 0 0 Hence the projection matrix is 1 0 1 0 0 2 1 5 1 2 0 0 1 1 4 2 5 0 0 2 2 0 1 0 W and 2 0 5 5 A 0 1 0 2 0 54 5 b The orthogonal projection is 1 1 1 0 25 1 5 5 1 0 1 0 1 1 2 2 0 W 0 0 45 5 5 5 MULTIPLE CHOICE 10 questions 2 points each Instructions for multiple choice questions No reason needs to be given There is always exactly one correct answer Enter your answer on the scantron sheet that is included with your exam In addition on your exam paper circle the choices you made on the scantron sheet Use a number 2 pencil to shade the bubbles completely and darkly Do NOT cross out your mistakes but rather erase them thoroughly before entering another answer Before beginning please code in your name UIN and netid in the appropriate places In the Section field on …


View Full Document

UIUC MATH 415 - midterm3-sol

Documents in this Course
disc_1

disc_1

2 pages

Load more
Download midterm3-sol
Our administrator received your request to download this document. We will send you the file to your email shortly.
Loading Unlocking...
Login

Join to view midterm3-sol and access 3M+ class-specific study document.

or
We will never post anything without your permission.
Don't have an account?
Sign Up

Join to view midterm3-sol 2 2 and access 3M+ class-specific study document.

or

By creating an account you agree to our Privacy Policy and Terms Of Use

Already a member?