# CH1p2.pdf - Math 4242 Applied Linear Algebra Introduction...

• Homework Help
• someoneonearth
• 10

This preview shows pages 1–4. Sign up to view the full content.

Math 4242 - Applied Linear Algebra Introduction to Linear Systems these notes follow closely Olver and Shakiban 1.5-1.9 Natalie E. Sheils 1 Matrix inverses Let A be an n × n matrix. We say B is the inverse of A if it satisfies AB = I = BA (1) where I is the n × n identity matrix. The inverse of a matrix A is commonly denoted by A - 1 . Because matrix multiplication is non-commutative we have to impose both conditions of (1). The first part AB = I implies that B is the right inverse of A and the second part BA = I implies B is the left inverse of A . Rectangular matrices might have either a left or a right inverse but they cannot have both. Thus, only square matrices have true inverses. Example 1. Let’s compute the matrix inverse for a general 2 × 2 matrix A = a 11 a 12 a 21 a 22 . Assume the inverse of A is B = b 11 b 12 b 21 b 22 . Then, AB = a 11 a 12 a 21 a 22 b 11 b 12 b 21 b 22 = a 11 b 11 + a 12 b 21 a 11 b 12 + a 12 b 22 a 21 b 11 + a 22 b 21 a 21 b 12 + a 22 b 22 = 1 0 0 1 = I. That is, a 11 b 11 + a 12 b 21 = 1 (2) a 11 b 12 + a 12 b 22 = 0 (3) a 21 b 11 + a 22 b 21 = 0 (4) a 21 b 12 + a 22 b 22 = 1 . (5) We want to solve this system for the elements b ij . To do so we set up a linear system and use Gaussian elimination: 1

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

a 11 0 a 12 0 0 a 11 0 a 12 a 21 0 a 22 0 0 a 21 0 a 22 b 11 b 12 b 21 b 22 = 1 0 0 1 . Solving for B we find B = 1 a 11 a 22 - a 12 a 21 a 22 - a 12 - a 21 a 11 . (6) It is straightforward to check that BA = I as well. Thus, B = A - 1 . The denominator appearing in (6) is called the determinant of the matrix A . It is often denoted det( A ) = a 11 a 22 - a 12 a 21 . If det( A ) = 0 then the inverse we computed does not exist. In fact, a matrix A is invertible (has an inverse) if and only if det( A ) is nonzero. Lemma 1 (Lemma 1.19 in the text) . The inverse of a square matrix (if it exists) is unique. Proof. Suppose to the contrary that both B and C are inverse of matrix A . Then, by definition, AB = I = BA and AC = I = CA . Using the associativity property of matrix multiplication, B = BI = B ( AC ) = ( BA ) C = IC = C. Hence, B = C which is a contradiction. Lemma 2 (Lemma 1.20 in the text) . If A is invertible then so is A - 1 and ( A - 1 ) - 1 = A . Proof. Since A is invertible AA - 1 = I = A - 1 A which means A - 1 is also invertible and its inverse ( A - 1 ) - 1 is A . Lemma 3 (Lemma 1.21 in the text) . If A and B are both invertible matrices of the same size then their product AB is also invertible and ( AB ) - 1 = B - 1 A - 1 . Proof. Observe by associativity ( AB )( B - 1 A - 1 ) = A ( BB - 1 ) A - 1 = AIA - 1 = AA - 1 = I. Similarly, ( B - 1 A - 1 )( AB ) = B - 1 ( A - 1 A ) B = B - 1 IB = B - 1 B = I. This result can be generalized in a straightforward way. The inverse of the product of k matrices is is the product of their inverses in reverse order. That is ( A 1 A 2 · · · A k - 1 A k ) - 1 = A - 1 k A - 1 k - 1 · · · A - 1 2 A - 1 1 . 2
1.1 Gauss-Jordan Elimination The way we found the matrix inverse previously is tedious. The principal algorithm used to find matrix inverses is Gauss-Jordan Elimination. We represent the matrix X by its column vectors ( ~ x 1 , ~ x 2 , · · · , ~ x n ) and the identity matrix by I = ( ~ e 1 , ~ e 2 , · · · , ~ e n ). The vectors ~ e j will be used in the future. They are the unit vectors with zeros in every component except for a 1 in the

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

This is the end of the preview. Sign up to access the rest of the document.
• Fall '16
• Darij Grinberg

{[ 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