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Math 307 Spring, 2003 Hentzel Time: 1:10-2:00 MWF Room: 1324 Howe Hall Instructor: Irvin Roy Hentzel Office 432 Carver Phone 515-294-8141 E-mail: hentzel@iastate.edu. http: //www.math.iastate.edu/hentzel/class.307.ICN Text: Linear Algebra With Applications,
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Math 307 Spring, 2003 Hentzel Time: 1:10-2:00 MWF Room: 1324 Howe Hall Instructor: Irvin Roy Hentzel Office 432 Carver Phone 515-294-8141 E-mail: hentzel@iastate.edu
http: //www.math.iastate.edu/hentzel/class.307.ICN • Text: Linear Algebra With Applications, • Second Edition, Otto Bretscher
Wednesday, Feb 5 Chapter 2 Page 94 Problems 1 through 50 • Main Idea: Apply the row operation to the identity. • Key Words: Differentiation, Tidbits • Goal: Learn some new tricks.
Previous Assignment • Page 85 Problem 34. • Consider two nxn matrices A and B such that • the product AB in invertible. Show that the • matrices A and B are both invertible.
(AB)-1A B = I • Since B has a left inverse, B is invertible. • A B(AB)-1 = I • Since A has a right inverse, A is invertible.
Page 85 Problem 50. • | 1 0 0 | • Consider the matrix E = |-3 1 0 | • | 0 0 1 | • and an arbitrary 3x3 matrix • | a b c | • A = | d e f | • | g h k |
(a) EA has row two replaced by • -3 Row 1 + Row 2.
|1 0 0 | Consider the matrix E = |0 1/4 0 | |0 0 1 |
(c) Can you think of a 3x3 matrix E such that EA is obtained from A by swapping the last two rows (for any 3x3 matrix A)?
| 1 0 0 | • | 0 0 1 | • | 0 1 0 |
(d) The three types of elementary matrices are • i j • | 1 0 0 ... 0 | • | 0 1 0 ... 0 | • | 0 0 1 ... 0 | • | . . | • i | . 0 1 . | Switch row i and row j. • | . . | • j | 1 0 | • | 1 0| • | 0 0 ... 0 1|
Multiply row i by c. • i • | 1 0 0 ... 0| • | 0 1 0 ... 0| • | 0 0 1 ... 0| • | . .| • i| . c 0 .| • | . .| • | 0 1 | • | 1 0| • | 0 0 0 1|
Add c times row i to row j. • | 1 0 0 ... 0 | • | 0 1 0 ... 0 | • | 0 0 1 ... 0 | • | . . | • i | . 1 0 . | • | . . | • j | c 1 | • | 1 0 | • | 0 0 ... 0 1 | • i j
Page 85 Problem 52. • Justify the following: • If A is an mxn matrix, then • there are elementary mxm matrices • E1 E2, ... Ep • such that RREF(A) = E1 E2 ... EpA.
Solution: We can reduce A to row canonical form by using the elementary row operations. Each of these elementary row operations can be viewed as an action performed by mxm matrices multiplying A on the left. • I would have preferred that the actions start by counting from right to left as • Ep ... E3 E2 E1 A = RCF(A).
But the numbering in the book goes the other way. How they know to start with p and work downwards is beyond me.
Find such elementary matrices E1 E2 ... Ep for • A = | 0 2 | • | 1 3 |
E1 = | 0 1 | E1 A = | 1 3 | | 1 0 | | 0 2 | E2 = | 1 0 | E2E1A = | 1 3 | | 0 1/2 | | 0 1 | • E3 = | 1 -3 | E3E2E1A = | 1 0 | • | 0 1 | | 0 1 |
Write the matrix of the linear transformation of differentiation with respect to the basis e 2x , xe2x , x2 e2x , x3 e 2x . Find some way to use the matrix from part to compute the sixth derivative of {2 - x + x 2 - x 3} e2x
e2x x e2x x2 e2x x3 e2x • e2x 2 0 0 0 • x e2x 1 2 0 0 • x2 e2x 0 2 2 0 • x3 e2x 0 0 3 2
Take the TRANSPOSE. • | 2 1 0 0 | • D = | 0 2 2 0 | • | 0 0 2 3 | • | 0 0 0 2 |
0 1 2 3 4 5 6 2 1 0 0 2 3 6 10 0 -96 -544 0 2 2 0 -1 0 -2 -20 -96 -352 -1120 0 0 2 3 1 -1 -8 -28 -80 -208 -512 0 0 0 2 -1 -2 -4 -8 -16 -32 -64
The sixth derivative is • e2x (-544 - 1120 x -512 x2 - 64 x3)
(a) The rows of AB are linear combinations of what? (b) What is the rank of a matrix? (c) What is the relation between the rank, the nullity, and the number of columns of a matrix?
(d) What is the relationship between the Row Canonical Form of a matrix and the existence of the inverse of the matrix? • (e) What is the 2x2 matrix which rotates the plane through 60 degrees?
(f) Give a non zero 2x2 matrix which is not invertible. • (g) What are the three elementary row operations. • (h) When is a function a linear transformation?
Multiply these two matrices. | 1 0 0 0 1 0 | | 3 8 2 4 | | 0 1 0 0 0-1 | | 1 0 1 2 | | 0 0 2 0 0 0 | | 3 3 1 2 | | 1 1 1 0 0 0 | | 4 3 1 0 | | 0 0 0 0 1 1 | | 5 4 3 2 | | 1 3 9 2 |
Solve AX = B and write the answer as • X = X0 + a1X1 + a2X2 + ... arXr and check • your answer using • A[X0 X1 X2 ... Xr] = [B 0 0 0 ... 0]. • | 1 2 0 1 2 3 | | x | | 2 | | 1 3 0 0 1 2 | | y | | 1 | | 2 5 0 1 3 5 | | z | = | 3 | | 4 10 0 2 6 11 | | w | | 7 | | u | | v |
Find the inverse of this matrix. • | 1 1 2 0 | • | 1 2 3 0 | • | 2 0 5 1 | • | 2 3 4 0 |