Lecture #9:1. Linear Equations: y=mx +b

2. Solution System: N.S., U.S., I.S.

3. Augmented Matrix

4. Solving a System of Linear Equations

Today:1. Echelon Form, Reduced Echelon Form

2. Gauss-Jordan Elimination Method

3. Homogeneous Linear Equations

4. Matrix Operations

Lecture #10: System of Linear Equations & Matrices

Announcements:

• Review Class on Tuesday 28:– Room: Ricketts 203– Time: 6:30-8:00pm

• Exam #1 Next Wednesday: 9/29

Unit Vectors, Cartesian Vector Form.

System of linear equations

• The general form:

A11x1+A12x2+A13x3+…..A1nxn=B1 Rx=0

A21x1+A22x2+A23x3+…..A2nxn=B2 Ry=0

A31x1+A32x2+A33x3+…..A3nxn=B3 Rz=0

. . . . .

. . . . .

Am1x1+Am2x2+Am3x3+…Amnxn=Bm

Matrix Form:

• Coefficient Matrix

mnmmm

n

n

n

aaaa

aaaa

aaaa

aaaa

.......

.......

.......

.......

321

3333231

2232221

1131211

ROW #

Column #

Augmented Matrix:

• System of linear eqns.

1x + y + 2z = 9

2x + 4y – 3z = 1

3x + 6y –5z = 0

Remember:

Rx=0 Ry=0 Rz=0

• Augmented Matrix:(array of numbers of the system of

eqns)

0563

1342

9211

Solving a System of Linear Eqns.

• GOAL– FIND the solution for x, y,z (TA, TB, TC, TD, TE)

• The idea is to replace a given system by a system which has the same solution set, but it is easier to solve.

Basic Operations to find Unknown

• Multiply a row by a nonzero constant.(the row you multiply by a number after adding the two rows will not

change)

• Interchange two rows.

• Add a multiple of one row to another row.

Gauss-Jordan Elimination

• Goal: to reduce the augmented matrix into a form

simple enough such that system of equation are solved by inspection.

Reduced row-echelom form

1. If row does not consist entirely of zeros, then the first non-zero number in row is 1.

2. If a row consist of zeros, then they are moved to the bottom of matrix.

3. In any two successive rows that do not consist entirely of zeros, the leading 1 in the lower row occur farther to the right of above row.

4. Each column that contains a leading 1 has zero everywhere.

IMPORTANT

• Reduced Row echelom– Must have zeros above

and below each leading 1.

• Row-echelom form– Must have zeros below

each leading 1.

0100

1010

9001

0000

7100

1010

9001

0100

1010

9041

1000

7100

1010

912101

Gauss-Jordan Elimination Method

• Step1: Locate the leftmost column that does not consist entirely of zero.

• Step 2: Interchange the top row with another row, if necessary, to bring a nonzero entry to the

top from step 1.• Step 3: If the entry that is now at the top is a

constant, divide entire row by it.• Step 4: Add multiples to top row to the rows below

such that all entries have 1 as leading term.• Step 5: Cover top row and begin with step 1 applied

to submatrix.

• For problem 3.22 find FAB, FAC, FAD using Gauss-Jordan method.

Example #1

125044

2

7

2

45

3

0

044

6

7

6

45

6

0

044

2

7

3

45

0

0x

ABACAD

z

ABACAD

y

ABACAD

FFF

F

FFF

F

FFF

F

Activity:#1

• For Problem in example #1 solve using Gauss-Jordan Method.

-TA (0.766) + TB (0.866) = 1699

TA (0.643) + TB (0.500) = 2943

Homogeneous System of Linear Equations

• Non-homogeneous

A11x1+A12x2+A13x3+…..A1nxn=B1

A21x1+A22x2+A23x3+…..A2nxn=B2

A31x1+A32x2+A33x3+…..A3nxn=B3

. . . .

. . . .

Am1x1+Am2x2+Am3x3+…Amnxn=Bm

The Constants B not equal to 0

• Homogeneous

A11x1+A12x2+A13x3+…..A1nxn= 0

A21x1+A22x2+A23x3+…..A2nxn= 0

A31x1+A32x2+A33x3+…..A3nxn= 0

. . . .

. . . .

Am1x1+Am2x2+Am3x3+…Amnxn= 0

The Constants B’s Equal to 0

Solutions in Homogeneous System

• Trivial Solution

X1 = 0

X2 = 0

X3 = 0

….

….

Xn = 0• For same # equations and

same # unknowns

• Non trivial solution:X1 = C1

X2 = C2

X3 = C3

….

….

Xn = C4

• When there is more unknowns

than equations .