# 2250-4 12:25pm Lecture Record Week 5 F2010

Last Modified: September 24, 2010, 20:17 MDT.    Today: July 17, 2018, 17:23 MDT.

## 20 Sep: Michal

Exam 1 day
Sep 20 at 12:50pm in JWB 335. Exam problems 1,2,3 only.

Sample Exam: Exam 1 key from S2010. See also F2009, exam 1.
Answer Keys: Exam 1, F2009 and S2010 (17.7 K, html, 18 Dec 2010)

## 21 Sep: Augmented Matrix for System Ax=b. RREF. Last Frame Algorithm. Sections 3.3, 3.4.

Review
The three possibilities
Frame sequence analysis and the general solution.
Last frame test.
Last frame algorithm.
Scalar form of the solution.
Lecture: 3.3 and 3.4
Translation of equation models
Equality of vectors
Scalar equations translate to augmented matrix
Augmented matrix translate to scalar equations
Matrix toolkit: Combo, swap and multiply
Frame sequences for matrix models.
Special matrices
Zero matrix
identity matrix
diagonal matrix
upper and lower triangular matrices
square matrix
THEOREM. Homogeneous system with a unique solution.
THEOREM. Homogeneous system with more variables than equations.
Equation ideas can be used on a matrix A.
View matrix A as the set of coefficients of a homogeneous
linear system Ax=0. The augmented matrix B for this homogeneous
system would be the given matrix with a column of zeros appended:
B=aug(A,0).
matlab, maple and mathematica.
Pitfalls.
Answer checks should also use the online FAQ.

html: Problem notes F2010 (4.6 K, html, 26 Nov 2010)

## Last Frame Algorithm

How to use maple to compute a frame sequence. Example is Exercise 3.2-14 from Edwards-Penney. Frame sequences with symbol k.
Matrices
Vector.
Matrix multiply
The college algebra definition
Examples.
Matrix rules
Vector space rules.
Matrix multiply rules.
Examples: how to multiply matrices on paper.

Slides: Matrix add, scalar multiply and matrix multiply (122.5 K, pdf, 02 Oct 2009)
Manuscript: Vectors and Matrices (266.8 K, pdf, 09 Aug 2009)
Manuscript: Matrix Equations (162.6 K, pdf, 09 Aug 2009)
General structure of linear systems.
Superposition.
General solution
X=X0+t1 X1 + t2 X2 + ... + tn Xn.
Matrix formulation Ax=b of a linear system
Properties of matrices: addition, scalar multiply.
Matrix multiply rules. Matrix multiply Ax for x a vector.
Linear systems as the matrix equation Ax=b.

## 21 Sep: Elementary matrices. Section 3.5

Elementary matrices.
How to write a frame sequence as a matrix product
Fundamental theorem on frame sequences
THEOREM. If A1 and A2 are the first two frames of a sequence,
then A2=E A1, where E is the elementary matrix built
from the identity matrix I by applying one toolkit
operation combo(s,t,c), swap(s,t) or mult(t,m).
THEOREM. If a frame sequence starts with A and ends with B, then
B = (product of elementary matrices) A.
The meaning: If A is the first frame and B a later frame in a
sequence, then there are elementary swap, combo
and mult matrices E1 to En such
that the frame sequence A ==> B can be written as
the matrix multiply equation
B=En En-1 ... E1 A.

## 22 Sep: Michal

Exam 1 day
Sep 22 at 12:50pm in JWB 335. Exam problems 1,2,3 only.

Sample Exam: Exam 1 key from S2010. See also F2009, exam 1.
Answer Keys: Exam 1, F2009 and S2010 (17.7 K, html, 18 Dec 2010)

## 23 Sep: Elementary Matrices. Inverses. Rank and nullity. Sections 3.4, 3.5.

Discussion of 3.4 problems.
Elementary matrices
Inverses of elementary matrices.
Solving B=E3 E2 E1 A for matrix A = (E3 E2 E1)^(-1) B.
This problem is the basis for the fundamental theorem on
elementary matrices (see above). While 3.5-44 is a difficult
technical proof, the extra credit problems on this subject
replace the proofs by a calculation. See Xc3.5-44a and Xc3.5-44b.
Ideas of rank, nullity, dimension in examples.

Slides: Rank, nullity and elimination (111.6 K, pdf, 29 Sep 2009) More on Rank, Nullity dimension 3 possibilities elimination algorithm Question answered: What did I just do, by finding rref(A)? Problems 3.4-17 to 3.4-22 are homogeneous systems Ax=0 with A in reduced echelon form. Apply the last frame algorithm then write the general solution in vector form.

## 23 Sep: Inverses. Sections 3.4, 3.5.

How to compute the inverse matrix
Def: AB=BA=I means B is the inverse of A.
Frame sequences method.
Inverse rules
Web References: Construction of inverses. Theorems on inverses.
THEOREM. A square matrix A has a inverse if and only if
one of the following holds:
1. rref(A) = I
2. Ax=0 has unique solution x=0.
3. det(A) is not zero.
4. rank(A) = n =row dimension of A.
5. There are no free variables in the last frame.
6. All variables in the last frame are lead variables.
7. nullity(A)=0.
THEOREM. The inverse matrix is unique and written A^(-1).
THEOREM. If A, B are square and AB = I, then BA = I.
THEOREM. The inverse of inverse(A) is A itself.
THEOREM. If C and D have inverses, then so does CD and
inverse(CD) = inverse(D) inverse(C).
THEOREM.  The inverse of a 2x2 matrix is given by the formula
[a  b]           1     [ d  -b]
inverse [     ]  =  -------  [        ]
[c  d]      ad - bc  [-c   a]
THEOREM.  The inverse B of any square matrix A can be
found from the frame sequence method
augment(A,I)
toolkit steps combo, swap, mult
.
.
augment(I,B)
in which the inverse B of A is read-off from the right panel of
the last frame.

Slides: Inverse matrix, frame sequence method (71.6 K, pdf, 02 Oct 2009)
Slides: Matrix add, scalar multiply and matrix multiply (122.5 K, pdf, 02 Oct 2009)
How to do 3.5-16 in maple.
with(linalg):#3.5-16
A:=matrix([[1,-3,-3],[-1,1,2],[2,-3,-3]]);
A1:=augment(A,diag(1,1,1));
rref(A1); # Expected answer in right panel
evalm(A&*B);
See problem notes chapter 3

html: Problem notes F2010 (4.6 K, html, 26 Nov 2010)

## 28 Sep: Determinants. Sections 3.6.

slides for 3.6 determinant theory and Cramer's Rule
Slides: Determinants 2008 (167.7 K, pdf, 23 Sep 2010)
Manuscript: Determinants, Cramer's rule, Cayley-Hamilton (186.5 K, pdf, 09 Aug 2009)
Methods for computing a determinant
Sarrus' rule, 2x2 and 3x3 cases.
Four rules for determinants
Triangular Rule (one-arrow Sarrus' Rule): The determinant of
a triangular matrix is the product of the diagonal elements.
Multiply rule: B=answer after mult(t,m), then |A| = (1/m) |B|
Swap rule: B=answer after swap(s,t), then |A| = (-1) |B|
Combo rule: B=answer after combo(s,t,c), then |A| = |B|