lecture7 - Lecture 7 Intersection of Hyperplanes and Matrix...

Info icon This preview shows pages 1–18. Sign up to view the full content.

View Full Document Right Arrow Icon
Lecture 7 Intersection of Hyperplanes and Matrix Inverse Shang-Hua Teng
Image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Elimination Methods for 2 by 2 Linear Systems 2 by 2 linear system can be solved by eliminating the first variable from the second equation by subtracting a proper multiple of the first equation and then by backward substitution Sometime, we need to switch the order of the first and the second equation Sometime we may not be able to complete the elimination
Image of page 2
Singular Systems versus Non-Singular Systems A singular system has no solution or infinitely many solution Row Picture: two line are parallel or the same Column Picture: Two column vectors are co- linear A non-singular system has a unique solution Row Picture: two non-parallel lines Column Picture: two non-colinear column vectors
Image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Gaussian Elimination in 3D Using the first pivot to eliminate x from the next two equations 10 7 3 2 8 3 9 4 2 2 4 2 = + - - = - + = - + z y x z y x z y x
Image of page 4
Gaussian Elimination in 3D Using the second pivot to eliminate y from the third equation 12 5 4 2 2 4 2 = + = + = - + z y z y z y x
Image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Gaussian Elimination in 3D Using the second pivot to eliminate y from the third equation 8 4 4 2 2 4 2 = = + = - + z z y z y x
Image of page 6
Now We Have a Triangular System From the last equation, we have 8 4 4 2 2 4 2 = = + = - + z z y z y x
Image of page 7

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Backward Substitution And substitute z to the first two equations 2 4 2 2 4 2 = = + = - + z z y z y x
Image of page 8
Backward Substitution We can solve y 2 4 2 2 4 4 2 = = + = - + z y y x
Image of page 9

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Backward Substitution Substitute to the first equation 2 2 2 4 4 2 = = = - + z y y x
Image of page 10
Backward Substitution We can solve the first equation 2 2 2 4 8 2 = = = - + z y x
Image of page 11

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Backward Substitution We can solve the first equation 2 2 1 = = - = z y x
Image of page 12
Generalization How to generalize to higher dimensions? What is the complexity of the algorithm? Answer: Express Elimination with Matrices
Image of page 13

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Step 1 Build Augmented Matrix 10 7 3 2 8 3 9 4 2 2 4 2 = + - - = - + = - + z y x z y x z y x Ax = b [ ] - - - - = 10 7 3 2 8 3 9 4 2 2 4 2 b A [ A b ]
Image of page 14
Pivot 1: The elimination of column 1 - 1 2 - - - - 10 7 3 2 8 3 9 4 2 2 4 2 - - - 10 7 3 2 4 1 1 0 2 2 4 2 - 12 5 1 0 4 1 1 0 2 2 4 2
Image of page 15

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Pivot 2: The elimination of column 2 - 1 - 12 5 1 0 4 1 1 0 2 2 4 2 - 8 4 0 0 4 1 1 0 2 2 4 2 Upper triangular matrix
Image of page 16
Backward Substitution 1: from the last column to the first - 8 4 0 0 4 1 1 0 2 2 4 2 Upper triangular matrix - 2 1 0 0 4 1 1 0 2 2 4 2 - 2 1 0 0 2 0 1 0 2 2 4 2 2 1 0 0 2 0 1 0 6 0 4 2 - 2 1 0 0 2 0 1 0 2 0 0 2
Image of page 17

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 18
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern