3. Linear Equations3. Linear Equations

Objectives:

1. Exercise - let’s see if you can do it!

a) Overdetermined systemsb) Underdetermined systemsc) Geometric interpretation.

Refs: B&Z 4.3.

2. Different types of solutions.

Example 4:

We can perform multiple (C ) operations provided at least one row is kept constant and only multiples of it are used to perform the other operations.

€

1 2 −2 4

3 1 0 −1

2 2 −1 0

⎡

⎣

⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥

€

1 2 −2 4

0 −5 6 −13

2 2 −1 0

⎡

⎣

⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥

~

€

1 2 −2 4

0 −5 6 −13

0 −2 3 −8

⎡

⎣

⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥

~

€

R2 − 3R1 → R2

€

R3 − 2R1 → R3

Obviously, performing multiple (A) type operations causes no problem.

Exercise 1: Solve the following system of simultaneous equations:

€

2x + 4y − z = −3

x − 3y + 2z =11

4x − 2y + 5z = 21.

Example 4 (continued):

€

1 2 −2 4

3 1 0 −1

2 2 −1 0

⎡

⎣

⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥

€

R3 − 2R1 → R3

€

R2 − 3R1 → R2

€

1 2 −2 4

0 −5 6 −13

0 −2 3 −8

⎡

⎣

⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥

~No problem -We kept R1 constantAnd used it to get R2 and R3

Solution to Exercise

€

2 4 −1 −3

1 −3 2 11

⎡

⎣

⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥

€

⎡

⎣

⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥

€

2 4 −1 −3 ⎡

⎣

⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥

€

2 4 −1 −3

1 −3 2 11

4 −2 5 21

⎡

⎣

⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥€

R1 → R2

€

R2 → R1

€

R2 − 2R1 → R2

€

R3 − 4R1 → R3

€

R2 10 → R2

€

R3 2 → R3

€

R3 −10R2 → R3€

R1 + 3R2 → R1

€

€

R1 −1 2 R3 → R1

€

R2 +1 2 R3 → R2

€

1 −3 2 11

2 4 −1 −3

4 −2 5 21

⎡

⎣

⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥

~

€

1 −3 2 11

0 10 −5 −25

0 10 −3 −23

⎡

⎣

⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥

~

€

1 0 1 2 7 2

0 1 −1 2 −5 2

0 0 1 1

⎡

⎣

⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥

~€

1 0 1 2 7 2

0 1 −1 2 −5 2

0 0 2 2

⎡

⎣

⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥

~€

1 −3 2 11

0 1 −1 2 −5 2

0 10 −3 −23

⎡

⎣

⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥

~

€

1 0 0 3

0 1 0 −2

0 0 1 1

⎡

⎣

⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥

x y z

(x,y,z)=(3,-2,1)

You can now attempt

Q1 and Q2 from Exercise Sheet 1

of the Orange Book

So far we have seen examples which have the same number of equations as there are variables….but this will not always be the case!

overdetermined systems

or more variables than equations

underdetermined systems

Example (overdetermined system)

We may have more equations than variables

Solve:

€

2x − y = 3

x − y = −2

3x + 2y = −6

€

2 −1 3

1 −1 −2

3 2 −6

⎡

⎣

⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥

€

R2 → R1

€

R1 → R2

Now let’s try to read off the solutions.

x=5 y=7 0 x x + 0 x y = -35!!!!

Not possible - there is no solution here.

€

R2 − 2R1 → R2

€

R3 − 3R1 → R3

€

1 −1 −2

2 −1 3

3 2 −6

⎡

⎣

⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥

~

€

1 0 5

0 1 7

0 0 −35

⎡

⎣

⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥

~

€

R3 − 5R2 → R3

€

R1 + R2 → R1

€

1 −1 −2

0 1 7

0 5 0

⎡

⎣

⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥

~

What does this result tell us?

€

x − y = −2

2

4-2 x

y

€

3x + 2y = −6

€

2x − y = 3

4

6

-1

-3

-2

Clearly, these three lineshave NO common point ofintersection - this is reflected in the algebraic solution.

Remember we are trying to find an (x,y) that satisfies all theequations; that is, a point where the three lines intersect.

For any system of linear equations (in two variables) there are three possibilities: either the system has

A. Exactly one solution. -the lines have common point of intersection.

B. No solution. -the lines do not intersect in a common point.

C. Infinitely many solutions. -the lines are coincident.

For case (A) we say that the equations are:

consistent and independent.

Common intersection point

For case (B) we say that the equations are

inconsistent.

No common intersection point

(i)(ii)

(iii)

For case (C) we say that the equations are

consistent and dependent.

The lines are coincident

Example (overdetermined system):

Solve:

€

x + 3y = 7

2x − y = 4

3x + 2y =11.

€

1 3 7

2 −1 4

3 2 11

⎡

⎣

⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥

€

1 3 7

0 1 10 7

0 −7 −10

⎡

⎣

⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥

~ €

1 3 7

0 −7 −10

0 −7 −10

⎡

⎣

⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥

~

€

1 0 19 7

0 1 10 7

0 0 0

⎡

⎣

⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥

~

the solution is

x=19/7 y=10/7

0=0

No problem -

so

(x,y)=(19/7,10/7)

€

R3 + 7R2 → R3

€

R1 − 3R2 → R1

€

−1 7 R2 → R2

€

R3 − 3R1 → R3

€

R2 − 2R1 → R2

The difference here is the bottom row - there is a zero on the RHS

In the previous example there was a non-zero entry, indicating no solution.

Example (underdetermined system)

Solve:

€

x1 + 2x2 + 2x3 = 4

4x1 − 6x2 + x3 = 2

€

0 0 0[ ]

€

0 0 a[ ], a ≠ 0

€

1 2 2 4

4 −6 1 2

⎡

⎣ ⎢

⎤

⎦ ⎥

€

1 2 2 4

0 −14 −7 −14

⎡

⎣ ⎢

⎤

⎦ ⎥~

€

1 0 1 2

0 1 1 2 1

⎡

⎣ ⎢

⎤

⎦ ⎥~

€

R2 − 4R1 → R2

€

−1 14 R2 → R2

See slide for solution

€

1 2 2 4

0 1 1 2 1

⎡

⎣ ⎢

⎤

⎦ ⎥~

€

R1 − 2R2 → R1

€

(x1, x2 , x3) = (2 − t,1−1 2 t , t)

where t is any real number.

Example:

Solve:

€

x1 + x2 − x3 = 5

2x1 + x2 + 3x3 = 9

4x1 + 3x2 + x3 =19

€

1 1 −1 5

2 1 3 9

4 3 1 19

⎡

⎣

⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥

Solution: (

€

x1, x2, x3 )=

€

(4 − 4t,1+ 5t, t) , t is a real number.

€

1 1 −1 5

0 −1 5 −1

0 −1 5 −1

⎡

⎣

⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥

~

€

1 0 4 4

0 1 −5 1

0 0 0 0

⎡

⎣

⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥

~

€

1 1 −1 5

0 1 −5 1

0 −1 5 −1

⎡

⎣

⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥

~€

R1 − R2 → R1

€

R3 + R2 → R3

€

−R2 → R2

€

R2 − 2R1 → R2

€

R3 − 4R1 → R3

See slide for solution

Extra Note:

i. If a row of zeroes

appears somewhere in your calculation - always shift it to the bottom row.

ii. If a row

appears at any stage this tells you there are no solutions.

€

0 0 ... ... 0 0[ ]

€

0 0 ... ... 0 a[ ],a ≠ 0

You may now complete Example Sheet 1

From the Orange Book.