Classify the following system, whether (a) intersecting, (b) parallel, or (c) coinciding lines 1. 3 x 4 y 1 0 3 x 4 y 2 0 3 x 4 y 1 0 6 x 8 y 2 0
Solve the following systems in three variables: 1. 3 x 4 y z 1 2. x y 2 x 4 y 3z 3 3 x 2 y 2 z 0
________
3 y z 1 x 2 z 7
2.
________
3.
2 x 5 y 1 0 5 x 2 y 2 0 2 x y 1 4 x 2 y 3 x 2 y 1 0 2 x y 1
________
4.
________
5.
________
1 x Solve 1 x
2 3 y 3 2 y
Problem solving Form a system of equations from the problems given below. A) (MIXTURE PROBLEM 1) How many pounds of a 35% salt solution and a 14% salt solution should be combined so that a 50 pounds of a 20% solution is obtained? B) (UNIFORM MOTION) Two motorists start at the same time from two places 128 km apart and drive toward each other. One drives 10kph than the other. If they met after 48 minutes (that is, 4/5 hr), find the average speed of each. C) A dietician is preparing a meal consisting of foods A, B, and C as shown in the table below. Fat Protein Carbohydrate If the meal must provide exactly 24 units of fat, 25 Food A 3 2 4 units of protein, and 21 units of carbohydrate, how Food B 2 3 1 many ounces of each food should be used? Food C 3 3 2...
...The purpose of this project is to solve the game of Light’s Out! by using basic knowledge of Linear algebra including matrix addition, vector spaces, linear combinations, and row reducing to reduced echelon form. 
Lights Out! is an electronic game that was released by Tiger Toys in 1995. It is also now a flash game online. The game consists of a 5x5 grid of lights. When the game stats a set of lights are switched to on randomly or in a pattern. Pressing one light will toggle it and the lights adjacent to it on and off. The goal of the game is to switch all the lights off in as few button presses as possible. In the folling examples, 1 will represent a “on” light and 0 will represent an “off” light. Yellow represents a button pressed and changed and green represents a button that was not pressed but was changed as a result of the pressed button.
Example 1
1  1  0  1  1 
1  0  1  0  1 
0  1  1  1  0 
1  0  1  0  1 
1  1  0  1  1 
0  0  0  1  1 
0  0  1  0  1 
0  1  1  1  0 
1  0  1  0  1 
1  1  0  1  1 
0  0  0  1  1 
0  0  0  0  1 
0  0  0  0  0 
1  0  0  0  1 
1  1  0  1  1 
Starting Grid Pressing button 1 Pressing button 18
Example 2
1  1  0  1  1 
1  0  1  0  1 
0  1  1  1  0 
1  0  1  0  1 
1  1  0  1  1 
0  0  0  1  1 
0  0  1  0  1 ...
...Solving systems of linearequations
7.1 Introduction
Let a system of linearequations of the following form:
a11 x1
a21 x1
a12 x2
a22 x2
ai1x1 ai 2 x2
am1 x1 am2 x2
a1n xn
a2 n x n
ain xn
amn xn
b1
b2
bi
bm
(7.1)
be considered, where x1 , x2 , ... , xn are the unknowns, elements aik (i = 1, 2, ..., m;
k = 1, 2, ..., n) are the coefficients, bi (i = 1, 2, ..., m) are the free terms of the system. In
matrix notation, this system has the form:
Ax b ,
(7.2)
where A is the matrix of coefficients of the system (the main matrix), A = [aik]mn, b is the
column vector of the free terms, bT [b1 , b2 , ... , bm ] , x is the column vector of the
unknowns, xT [ x1 , x2 , ... , xn ] ; the symbol () T denotes transposition.
It is assumed that aik and bi are known numbers. An ordered set, {x1, x2, ..., xn}, of real
numbers satisfying (7.1) is referred to as the solution of the system, and the individual
numbers, x1, x2, ..., xn, are roots of the system.
A system of linearequations is:
consistent  if it has at least one solution. At the same time it can be

determined  if it has exactly one, unique solution,
undetermined  if it has...
...TR 3923 Programming Design in Solving Biology Problems Semester 1, 2011/2012
Elankovan Sundararajan School of Information Technology Faculty of Information Science and Technology
TR 3923 Elankovan Sundararajan 1
Lecture 3
System of LinearEquations
TR 3923
Elankovan Sundararajan
2
Introduction
• Solving sets of linearequations is the most frequently used numerical procedure when realworld situations are modeled. modeled Linearequations are the basis for mathematical models of
1. 2. 2 3. 4. 5. Economics, Computational Biology Comp tational Biolog and Bioinformatics Bioinformatics, Weather prediction, Heat and mass transfer, Statistical analysis, and a myriad of other application.
•
•
TR 3923
The methods for solving ODEs and PDEs also depend on them.
Elankovan Sundararajan
3
System of LinearEquations
• Consider the following general set of n equations in n unknowns: a11 x1 a12 x2 a1n xn c1 : R1
a21 x1 a22 x2 a2 n xn c2 , an 1,1 x1 an 1, 2 x2 an 1,n xn cn 1 , an ,1 x1 an , 2 x2 an ,n xn cn .
Which can be written in matrix form as:
: R2
: R n1 : Rn
A x b.
~ ~
TR 3923 Elankovan Sundararajan 4
where, A is the nxn matrix, ,
a11 a12 a1n a21 a22 a2 n A . a an 2 ann n1 ...
...MODULE  1
LinearEquations
Algebra
5
Notes
LINEAREQUATIONS
You have learnt about basic concept of a variable and a constant. You have also learnt
about algebraic exprssions, polynomials and their zeroes. We come across many situations
such as six added to twice a number is 20. To find the number, we have to assume the
number as x and formulate a relationship through which we can find the number. We shall
see that the formulation of such expression leads to an equation involving variables and
constants. In this lesson, you will study about linearequations in one and two variables.
You will learn how to formulate linearequations in one variable and solve them algebraically.
You will also learn to solve linearequations in two variables using graphical as well as
algebraic methods.
OBJECTIVES
After studying this lesson, you will be able to
•
identify linearequations from a given collection of equations;
•
cite examples of linearequations;
•
write a linearequation in one variable and also give its solution;
•
cite examples and write linearequations in two variables;
•
draw graph of a linearequation in two...
...✎ ✍
5.2
✌
Cramer’s Rule
Introduction
Cramer’s rule is a method for solving linear simultaneous equations. It makes use of determinants and so a knowledge of these is necessary before proceeding.
1. Cramer’s Rule  two equations
If we are given a pair of simultaneous equations a1 x + b1 y = d1 a2 x + b2 y = d2 then x, and y can be found from d1 b1 d2 b2 a1 b1 a2 b2 a1 d1 a2 d2 a1 b 1 a2 b 2
x=
y=
Example Solve the equations 3x + 4y = −14 −2x − 3y = 11
Solution Using Cramer’s rule we can write the solution as the ratio of two determinants. −14 4 11 −3 3 4 −2 −3 −2 = 2, −1 3 −14 −2 11 3 4 −2 −3
x=
=
y=
=
5 = −5 −1
The solution of the simultaneous equations is then x = 2, y = −5.
5.2.1
copyright c Pearson Education Limited, 2000
2. Cramer’s rule  three equations
For the case of three equations in three unknowns: If a1 x + b1 y + c1 z = d1 a2 x + b2 y + c2 z = d2 a3 x + b3 y + c3 z = d3
then x, y and z can be found from d1 d2 d3 a1 a2 a3 b1 b2 b3 b1 b2 b3 c1 c2 c3 c1 c2 c3 a1 a2 a3 a1 a2 a3 d1 d2 d3 b1 b2 b3 c1 c2 c3 c1 c2 c3 a1 a2 a3 a1 a2 a3 b1 b2 b3 b1 b2 b3 d1 d2 d3 c1 c2 c3
x=
y=
z=
Exercises Use Cramer’s rule to solve the following sets of simultaneous equations. a)
7x + 3y = 15 −2x + 5y = −16
b)
x + 2y + 3z = 17 3x + 2y + z = 11 x − 5y + z = −5
Answers a) x = 3,...
...Universitatea Politehnica Bucuresti  FILS
Systems of Differential Equations and Models in Physics, Engineering and Economics
Coordinating professor: Valeriu Prepelita
Bucharest,
July, 2010
Table of Contents
1. Importance and uses of differential equations 4
1.1. Creating useful models using differential equations 4
1.2. Reallife uses of differential equations 5
2. Introduction to differentialequations 6
2.1. First order equations 6
2.1.1. Homogeneous equations 6
2.1.2. Exact equations 8
2.2. Second order linearequations 10
3. Systems of differential equations 14
3.1. Systems of linear differential equations 16
3.1.1. Systems of linear differential equations with constant coefficients 22
3.2. Systems of first order equations 27
3.2.1. General remarks on systems 27
3.2.2. Linearsystems. Case n=2 31
3.2.3. Nonlinear systems. Volterra’s Preypredator equations 38
3.3. Critical points and stability for linearsystems 44
3.3.1. Bounded input bounded output stability 44
3.3.2. Critical points 44
3.3.3. Methods of...
...LINEAR ALGEBRA
Paul Dawkins
Linear Algebra
Table of Contents
Preface............................................................................................................................................. ii Outline............................................................................................................................................ iii Systems of Equations and Matrices.............................................................................................. 1
Introduction ................................................................................................................................................ 1 Systems of Equations ................................................................................................................................. 3 Solving Systems of Equations .................................................................................................................. 15 Matrices .................................................................................................................................................... 27 Matrix Arithmetic & Operations .............................................................................................................. 33 Properties of Matrix Arithmetic and the Transpose...
...for row 1 (or equation 1), row 2 (or equation 2), and so on. Additional notes are at the end of the section.
1.
x1 + 5 x2 = 7 −2 x1 − 7 x2 = −5
1 −2
5 −7
7 −5
x1 + 5 x2 = 7
Replace R2 by R2 + (2)R1 and obtain:
3x2 = 9
x1 + 5 x2 = 7 x2 = 3 x1
1 0 1 0 1 0
5 3 5 1 0 1
7 9 7 3 −8 3
Scale R2 by 1/3: Replace R1 by R1 + (–5)R2: The solution is (x1, x2) = (–8, 3), or simply (–8, 3).
2.
= −8
x2 = 3
2 x1 + 4 x2 = −4 5 x1 + 7 x2 = 11
2 5
4 7
−4 11
x1 + 2 x2 = −2
Scale R1 by 1/2 and obtain: Replace R2 by R2 + (–5)R1: Scale R2 by –1/3: Replace R1 by R1 + (–2)R2: The solution is (x1, x2) = (12, –7), or simply (12, –7).
5 x1 + 7 x2 = 11
x1 + 2 x2 = −2
1 5 1 0 1 0 1 0
2 7 2 −3 2 1 0 1
−2 11 −2 21 −2 −7 12 −7
−3x2 = 21
x1 + 2 x2 = −2 x2 = −7 x1
= 12
x2 = −7
1
2
CHAPTER 1
•
LinearEquations in Linear Algebra
3. The point of intersection satisfies the system of two linearequations: x1 + 5 x2 = 7 x1 − 2 x2 = −2
1 1
5 −2
7 −2
x1 + 5 x2 = 7
Replace R2 by R2 + (–1)R1 and obtain: Scale R2 by –1/7: Replace R1 by R1 + (–5)R2: The point of intersection is (x1, x2) = (4/7, 9/7).
−7 x2 = −9
x1 + 5 x2 = 7 x2 = 9/7 x1
1 0 1 0 1 0
5 −7 5 1 0 1
7 −9 7 9/7 4/7 ...
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