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Lecture 25.pdf

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Department
Mathematics
Course
MATH 136
Professor
Robert Sproule
Semester
Winter

Description
Monday, March 10 − Lecture 25 : Invertible matrices. Concepts: 1. Define the inverse of a matrix. 2. Find the inverse of a matrix. 3. Solve a system by using the inverse of a matrix. 4. Recognize that a matrix is invertible if and only if its inverse has the RREF. −1 The multiplicative matrix inverse A of A. 25.1 Definition − We say that an n × n matrix A is invertible (or non-singular) if there exists an n × n matrix B such that both conditions AB = I =nBA hold true. We call such a matrix B, an inverse of A. Note that only a pair of square matrices of equal dimension can satisfy the condition AB = BA. So if we refer to matrix A as being invertible then A must be a square matrix 25.1.1 Definition – If A is an n × m matrix for which there exists an m × n matrix B such that AB = I n then we say that B is a right inverse of A and A is a left inverse of B. If B is both a right inverse and left inverse of A then, by definition of “inverse”, both A and B are easily seen to be both (square) invertible matrices. 25.1.2 Proposition − Uniqueness of the inverse of a matrix. Let A be an n × n invertible matrix. Then there exists no more than one inverse B of A. Proof : Suppose A is invertible. Suppose both B and C are inverses of the matrix A. Then AB = I = BA and AC = I = CA. Then C = CI = C(AB) = (CA)B = IB = B. So C = B. Then if a matrix A is invertible it will have, at most, one inverse. 25.1.3 Notation − For an invertible matrix A, the unique inverse matrix of A is denoted by A -1. If a matrix has dimensions 3 × 3 or larger it is not easy to tell whether it has an inverse or not simply by looking at its entries. Even if we know a matrix is invertible finding its unique inverse can still be a considerable amount of work. However, if a matrix is 2 × 2 it is fairly straightforward as the following example shows. 25.1.4 Example − Show that if A is the matrix, -1 such that ad − bc is not zero, A exists and is equal to the matrix : Solution: We compute the product AB: Then AB = I. Computation BA = I follows similar steps. This means that if ad − bc ≠ 0 then a two by two matrices has an inverse and is of the form given to the matrix B. The converse : If ad – bc = 0 then A cannot be invertible. To see this, suppose both ad −1 – bc = 0 and A exists. Then Then Then for any matrix B, AB will produce a two by two matrix whose second row only contains zeros and so cannot be the identity matrix. −1 25.1.5 Definition − The number ad − bc obtained in the 2 × 2 matrix A above is called the determinant of the 2 × 2 matrix A. It is denoted by det A. - We can always use the derived expression as a way of finding the inverse of a 2 by 2 provided det A ≠ 0. The inverse of a matrix A, when it exists, can be quite useful when trying to solve systems of linear equations Ax = b. This is shown in the following proposition. 25.2 Theorem − If A is invertible and Ax = b is consistent then the only solution to the system Ax = b is x = A b.1 Proof : Given: A is an n × n invertible matrix and Ax = b is consistent. If the vector uis a solution to Ax = b and A is invertible then u = Inu = (A A)u = A (Au) = A b. −1 So u = A b is the only solution of Ax = b. 25.2.1 Corollary – If A is an invertible n × n matrix then A RREF = I nnd so Rank(A) = n. Proof : Given: A is an n × n invertible matrix. Since Ax = 0 is consistent and A is invertible, by the theorem, the system Ax = 0 has a unique solution u = A 0. Since the solution to Ax = 0 is unique, the matrix A RREF points to no free variables. This means A has a leading-1 in each of its n columns RREF and each of its n rows. Hence A RREF = In. By definition, Rank(A) = n, as required. 25.2.2 Corollary – If A and B are an n × n matrices and AB = I then Anis invertible. Proof : Given : A and B are an n × n matrices and AB = I n Required to show: BA = I . n If B is as defined above, we construct the homogeneous system Bx = 0. Let u be any solution o
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