math 307 spring, 2003 hentzel time: 1:10-2:00 mwf room: 1324 howe hall
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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: [email protected]. http: //www.math.iastate.edu/hentzel/class.307.ICN Text: Linear Algebra With Applications, - PowerPoint PPT PresentationTRANSCRIPT
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: [email protected]
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 |
• EA multiplies row 2 by 1/4.
• (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 |