ENGG2013 Unit 2 Linear Equations Jan, 2011.. Linear Equation in n variables a 1 x 1 + a 2 x 2 + …...

ENGG2013Unit 2 Linear Equations

Jan, 2011.

Linear Equation in n variables

• a1x1 + a2x2+ … + an xn = c– a1, a2, …, an are called coefficients (real numbers).

– x1, x2,…, xn are variables (or indeterminates).

– c is a constant term (real number).

• Example– 2x + 3y – 4z = 0.2

• Non-example– x2+y2=1

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Geometry of a linear equation

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Two variables: straight line

ax + by = c

Three variables: plane

ax + by + cz = d

System of linear equations

• A system of linear equations (or linear system) is a collection of one or more linear equations.– for example:

• A solution is a list of numbers (s1, s2, …, sn) which satisfies all equalities after substituting xi by si, for i =1,2,…,n.

• The set of all solutions is called the solution set.

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Nutrition problem• Find a combination of food A, B, C and D in order to satisfy the nutrition requirement exactly.

• Let xA, xB, xC and xD be the amount of food A, B, C and D respectively.

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Food A Food B Food C Food D Requirement

Protein 9 8 3 3 5

Carbohydrate 15 11 1 4 5

Vitamin A 0.02 0.003 0.01 0.006 0.01

Vitamin C 0.01 0.01 0.005 0.05 0.01

Formal notation

• Given a system of m linear equations in n variables

the solution set is defined as

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Doublesubscripts

Review of set notation

• Set of Greek letters = {,,,,,,,,,, ,,,,,,,,,,,,,}

• Set of prime numbers = {2,3,5,7,11,13,17,23,29,31,37,41, …}

• Sphere with radius r centered at origin= {(x,y,z): x2+y2+z2=r2}

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finite

Countably infinite

Uncountably infinite

Examples of solution sets

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{ (x,y): ax + by = c }

-1

-0.5

0

0.5

1

-1

-0.5

0

0.5

1-5

0

5

xy

z

x

y

Consistency

• A linear system is called consistent if there is at least one solution, in other words, if the solution set is non-empty.

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x

yInconsistent,no solution

x

y

Consistent

Classification

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Linear System

Inconsistent(no solution)

Consistent

Unique solution Infinitely many solutions

Tasks:Determine whether a linearsystem is consistent.If yes, find all solutions.

Short-hand notation using matrix

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(2 rows, 4 columns)

(4 rows, 3 columns)Usually called the augmented matrix

The nutrition example

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Elementary row operations

1. Interchange two rows2. Multiply a row by a non-zero constant3. Replace a row by the sum of itself and a

constant multiple of another row

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Facts: Elementary row operations do not change the solution(s). (There is no loss, and no gain, of information.)

Illustration – row interchange

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Illustration – Multiply by constant

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2 2

Illustration – Row replacement

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(1) (1) – (2) (1) (1) – (2)

How to solve?

• Idea: Apply the three kinds of information-lossless elementary row operations, and transform the linear system into one which is easier to solve.

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Linear system in upper triangular matrix form can be easily solved

by backward substitution

Carl Friedrich Gauss

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(1777~1855)

The old Deutsche 10-Mark note

Gaussian elimination

• Step 0: Write the linear system in matrix format

• Step 1: Try to transform the matrix into upper triangular form

• Step 2: Solve for the variables one by one, in backward order

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Example 1 (row operations)

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(2) (2) – (1)

(3)

(2)

(1)

(3) (3) + (2)/2

Solve

Example 1 (backward sub.)

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Upper triangular

(3) z = 7/3

(2) – 2y – (7/3) = 1 y = –5/3

(1) x+(–5/3)+(7/3) = 1 x = 1/3

Verify:x+y+z = 1/3 – 5/3 + 7/3 = 3/3 = 1x–y = (1/3) – (– 5/3) = 6/3 = 2y+2z = (– 5/3)+2(7/3) = 9/3 = 3

Solution: x=1/3, y = –5/3, z = 7/3(unique solution)

Example 2 (row operations)

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(2) (2) – (1) (3) (3) – (1)

(3) (3) – 2 (2)

Solve

Example 2 (backward sub.)

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z can be taken as a free variable.Let z to be any real number.

(1)

(2)

From (2), y = –1 – z

From (1), x +(–1 – z)+3z = 1 x = 2 – 2z

Solution: x= 2– 2z, y = –1–z, z = any real number.

Solution set = {(2 – 2z, –1–z, z): z is any real no.}(Infinitely many solutions)

Note: You can let y tobe the free variable as well,and obtain the solutions interms of y.

Example 2 (cont’d)

• Verification– x+y+3z

=(2 – 2z) + (– 1 – z) + 3z = 1

– x+2y+4z=(2 – 2z) + 2(– 1 – z) + 3z = 0

– x+3y+5z =(2 – 2z) + 3(– 1 – z) + 5z = – 1

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Solution: x= 2– 2z, y = –1–z, z = any real numberSolution set = {(2 – 2z, –1–z, z): z is any real no.}

-5

0

5

10

-50

510

-10

-8

-6

-4

-2

0

2

4

x

y

z

Example 3 (row operations)

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Solve

(2) (2) – (1) (3) (3) – (1)

(3) (3) – 2(2)

Example 3 (cont’d)

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Contradiction, cannot be true

Answer: the linear system is inconsistent

Example 3 (picture)

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Cross-section

-5-4

-3-2

-10

1

-2

-1.8

-1.6

-1.4

-1.2

-1

0

0.5

1

1.5

2

2.5

x

y

z

No common intersection

An infinitely long triangular tubeis formed by the three planes

Key concepts

• Three kinds of elementary row operations– The solution set is invariant under any elementary

row operation• Gaussian elimination

– Transform a linear system to upper triangular form– Backward substitution

• Three types of solutions– No solution– Unique solution– Infinitely many solutions

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