lecture 13, August 14, 2018.pdf

# lecture 13, August 14, 2018.pdf - Finding matrix inverses...

• Notes
• 19

This preview shows page 1 - 7 out of 19 pages.

Finding matrix inverses, and more on the invertible matrix theorem Jacob Shapiro MATH1115 Lecture 13, August 14, 2018
Matrix inversion algorithm I A I n E N · · · E 1 A E N · · · E 1 = R E N · · · E 1 . Example We use the algorithm discussed in the previous lecture to find the inverse of the matrix A = 1 2 3 2 5 3 1 0 8 . 1 2 3 1 0 0 2 5 3 0 1 0 1 0 8 0 0 1 1 2 3 1 0 0 0 1 - 3 - 2 1 0 0 - 2 5 - 1 0 1 1 2 3 1 0 0 0 1 - 3 - 2 1 0 0 0 - 1 - 5 2 1 1 2 3 1 0 0 0 1 - 3 - 2 1 0 0 0 1 5 - 2 - 1
Matrix inversion algorithm Example (continued) 1 2 3 1 0 0 0 1 - 3 - 2 1 0 0 0 1 5 - 2 - 1 1 2 0 - 14 6 3 0 1 0 13 - 5 - 3 0 0 1 5 - 2 - 1 1 0 0 - 40 16 9 0 1 0 13 - 5 - 3 0 0 1 5 - 2 - 1 Therefore A - 1 = - 40 16 9 13 - 5 - 3 5 - 2 - 1 .
Matrix inversion algorithm Example We determine whether the matrix A = 1 6 4 2 4 - 1 - 1 2 5 is invertible. 1 6 4 1 0 0 2 4 - 1 0 1 0 - 1 2 5 0 0 1 1 6 4 1 0 0 0 - 8 - 9 - 2 1 0 0 8 9 1 0 1 1 6 4 1 0 0 0 - 8 - 9 - 2 1 0 0 0 0 - 1 1 1 Continued row reductions will not affect the last row, so we conclude that the reduced row echelon form of A is not the identity. Hence A is not invertible.
Theorem (invertible matrix theorem) Let A be an n × n matrix. Then the following statements are equivalent (TFAE). a. A is invertible. b. The homogeneous equation A x = 0 has only the trivial solution x = 0 . c. The reduced row echelon form of A is I n . d. A is expressible as a product of elementary matrices. We will add even more equivalent statements as we go.
Theorem A system of linear equations has zero, one, or infinitely many solutions.

Want to read all 19 pages?

Want to read all 19 pages?

#### You've reached the end of your free preview.

Want to read all 19 pages?

• Three '08
• ME
• ax, Invertible matrix

{[ snackBarMessage ]}

### What students are saying

• 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.

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

• 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.

Dana University of Pennsylvania ‘17, Course Hero Intern

• 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.

Jill Tulane University ‘16, Course Hero Intern