Math 307 Spring, 2005 Hentzel Time: 10:00 to 10:40 MWF Room: 205 Carver Hall Instructor: Irvin Roy Hentzel Office 432 Carver Phone 515-294-8141 E-mail: hentzel@iastate.edu http://orion.math.iastate.edu/hentzel/class.307.05 Text: Linear Algebra and its Applications, Third Edition David C. Lay Friday, January 26, Chapter 1.9 ans 1.10 Work the practice test for Friday. The actual test is on Monday. Main Idea: Find out how to get back to where you started. Key Words: Linear Transformation, Matrix Inverse, Inverse. Goal: Be able to set up the matrix of a linear transformation and be able to compute a matrix inverse. ----------------------------------------------------------- Previous Assignment: Page 80 Problem 6 Find a vector x whose image under T is b and determine whether x is unique | 1 -2 1 | | 1 | | 3 -4 5 | | 9 | | 0 1 1 | | 3 | | -3 5 -4 | | -6 | A = { { 1, -2, 1, 1 }, { 3, -4, 5, 9 }, { 0, 1, 1, 3 }, { -3, 5, -4, -6 }}; x y z=a 1 0 3 7 | 7 | | -3 | 0 1 1 3 | 3 | + a | -1 | 0 0 0 0 | 0 | | 1 | 0 0 0 0 Check | 1 -2 1 | | 7 -2 | | 1 0 | | 3 -4 5 | | 3 -1 | = | 9 0 | | 0 1 1 | | 0 1 | | 3 0 | | -3 5 -4 | | -6 0 | It checks. Page 80 Problem 17 Let T be a linear transformation that maps u = 5 ---> 2 2 1 1 ---> -1 v = 3 3 3u --------> 6 3 2v --------> -2 6 3u+2v -------> 4 9 Page 80 Problem 20 v1 = -2 5 v2 = 7 -3 x1 --------> x1 v1 + x2 v2 x2 Find the matrix which does this | v1 v2 | = | -2 7 | | 7 -3 | Page 80 Problem 37 Find the kernel 4 -2 5 -5 -9 7 -8 0 -6 4 5 3 5 -3 8 -4 A = { { 4, -2, 5, -5}, {-9, 7, -8, 0}, {-6, 4, 5, 3}, { 5, -3, 8, -4}}; x y z w=a 1 0 0 -7/2 0 1 0 -9/2 0 0 1 0 0 0 0 0 7/2 a 9/2 0 1 4 -2 5 -5 7/2 -9 7 -8 0 9/2 -6 4 5 3 0 5 -3 8 -4 1 -------------------------------------------------- Definition: A linear transformation does just what any reasonable person would expect. These are (a) T(V+W) = T(V)+T(W) for all vectors V, W. (b) T(cV) = cT(V) for all numbers c and all vectors V. ------------------------------------------------------------ Linear transformations can be represented by a matrix. | | | | | | The matrix | c1 c2 c3 ... cn | sends the first basis | | | | | | vector to c1, the second basis vector to c2, etc. ------------------------------------------------------- A matrix A has an inverse if there exists a matrix B such that A B = B A = I. B is called the inverse of A. A matrix has an inverse if and only if the row canonical form of A is the identity matrix. ------------------------------------------------------ Finding matrix inverses ======================= The technique for computing the inverse of a matrix A is the following. Write [A I] and reduce it to row canonical form. If the result is [I B] for some B, then the inverse of A is B. If the Row Canonical Form does not start with I, the matrix has no inverse. This presupposes that A is square. A non square matrix will never have an inverse. Find the inverse of the matrix of the above function f. | 1 2 1 | | 2 5 -1 | | 5 4 24 | | 1 2 1 1 0 0 | | 2 5 -1 0 1 0 | | 5 4 24 0 0 1 | | 1 2 1 1 0 0 | | 0 1 -3 -2 1 0 | | 0 -6 19 -5 0 1 | | 1 0 7 5 -2 0 | | 0 1 -3 -2 1 0 | | 0 0 1 -17 6 1 | | 1 0 0 124 -44 -7 | | 0 1 0 -53 19 3 | | 0 0 1 -17 6 1 | | 124 -44 -7 | The inverse of A is | -53 19 3 | | -17 6 1 | check. | 1 2 1 | | 124 -44 -7 | | 1 0 0 | | 2 5 -1 | | -53 19 3 | = | 0 1 0 | | 5 4 24 | | -17 6 1 | | 0 0 1 | It checks. The theory why this process works. Suppose that En ... E3 E2 E1 A = RCF(A). If RCF(A) is the identity I then (En...E3 E2 E1) A = I and so (En...E3 E2 E1) is the inverse of A. Notice that if we start with [A I], then we end up with [I | En...E3 E2 E1] and the inverse is captured in the back half of the matrix. If RCF(A) =/= I, then RCF(A) has a row of zeros. Since A is square, then RCF(A) has an unknown which is not above a stairstep one. Therefore, there is some V=/= 0 such that AV = 0. The existence of such a V implies that A cannot be invertible. --------------------------------------------------------------- Find the inverse of | 1 2 1 | | 1 3 2 | | 1 0 1 | Solution: | 1 2 1 | 1 0 0 | | 1 3 2 | 0 1 0 | | 1 0 1 | 0 0 1 | | 1 2 1 | 1 0 0 | | 0 1 1 | -1 1 0 | | 0 -2 0 | -1 0 1 | | 1 0 -1 | 3 -2 0 | | 0 1 1 | -1 1 0 | | 0 0 2 | -3 2 1 | | 1 0 -1 | 3 -2 0 | | 0 1 1 | -1 1 0 | | 0 0 1 | -3/2 1 1/2 | | 1 0 0 | 3/2 -1 1/2 | | 0 1 0 | 1/2 0 -1/2 | | 0 0 1 | -3/2 1 1/2 | -1 | 1 2 1 | | 3 -2 1 | | 1 3 2 | = (1/2) | 1 0 -1 | | 1 0 1 | |-3 2 1 | | 2 0 0 | check 1/2 | 0 2 0 | = I. It checks. | 0 0 2 |