Three topics for the day:Three topics for the day:

Elementary Row OperationsElementary Row Operations

Row Echelon FormRow Echelon Form

……in Sec. 7.3bin Sec. 7.3b

Reduced Row Echelon FormReduced Row Echelon Form

When using Gaussian Elimination, we work primarily on thecoefficients of the variables. Knowing this, it is often easier toperform these operations on the matrix form of a system ofequations…

Example – the system:2 7x y z

3 5 14x y z 2 2 3x y z

The augmented matrix:

1 2 1 7

3 5 1 14

2 2 1 3

The coefficient matrix:

1 2 1

3 5 1

2 2 1

Definition: Row Echelon Form (REF) of a Matrix

A matrix is in row echelon form if the following conditions aresatisfied:

1. Rows consisting entirely of 0’s (if there are any) occur at the bottom of the matrix.

1 2 1 7

0 1 2 7

0 0 1 3

2. The first entry in any row with nonzero entries is 1.

3. The column subscript of the leading 1 entries increases as the row subscript increases.

Example:

This should lookThis should looksomewhat familiar…somewhat familiar…

Triangular Form?!?!Triangular Form?!?!

Elementary Row Operations on a Matrix

A combination of the following operations will transform a matrixto row echelon form:

1. Interchange any two rows.

2. Multiply all elements of a row by a nonzero real number.

3. Add a multiple of one row to any other row.

Notation: R indicates interchanging the i-th and j-th rowsij

Notation: kR indicates multiplying the i-th row by ki

Notation: kR + R indicates adding k times the i-th row to the j-th row

i j

Example: Solve by Finding the REF

2 3x y z 2 0x y z 2 3 7x y z

AugmentedMatrix:

1 1 2 3

2 1 1 0

1 2 3 7

1 1 2 3

2 1 1 0

1 2 3 7

1 1 2 3

0 3 5 6

1 2 3 7

1 22R R

1 1 2 3

0 3 5 6

0 1 1 4

1 3R R1 1 2 3

0 1 1 4

0 3 5 6

23R

Example: Solve by Finding the REF

1 1 2 3

0 1 1 4

0 0 2 6

2 33R R 1 1 2 3

0 1 1 4

0 3 5 6

1 1 2 3

0 1 1 4

0 0 1 3

3

1

2R

z = 3

y – z = 4 y = 7

x – y + 2z = –3 x = –2

Solution: (x, y, z) = (–2, 7, 3)Solution: (x, y, z) = (–2, 7, 3)

Guided PracticeWhat elementary row operations applied to

2R + R2R + R

will yield the given matrix?

2 1 1 2

1 2 3 0

3 1 1 2

0 3 5 2

1 2 3 0

3 1 1 2

22 11

Guided PracticeWhat elementary row operations applied to

(1/4)R(1/4)R

will yield the given matrix?

2 1 1 2

1 2 3 0

3 1 1 2

2 1 1 2

1 2 3 0

0.75 0.25 0.25 0.5

33

Guided PracticeFind a row echelon form for the given matrix.

1 2 3

3 6 10

2 4 7

1 23R R

1 32R R

1 2 3

0 0 1

2 4 7

1 2 3

0 0 1

0 0 1

2 31R R 1 2 3

0 0 1

0 0 0

But first, a brief review of last few slides…

3 7 11 44x y z 2 3 12x y z

4 9 13 53x y z

AugmentedMatrix:

3 7 11 44

1 2 3 12

4 9 13 53

2 13R R 0 1 2 8

1 2 3 12

4 9 13 53

2 34R R 0 1 2 8

1 2 3 12

0 1 1 5

1 3R R 0 1 2 8

1 2 3 12

0 0 1 3

12R1 2 3 12

0 1 2 8

0 0 1 3

But first, a brief review of the last few slides…

z = –3

y – 2z = 8 y = 2

x + 2y – 3z = 12 x = –1

Solution: (x, y, z) = (–1, 2, –3)Solution: (x, y, z) = (–1, 2, –3)

1 2 3 12

0 1 2 8

0 0 1 3

It is possible to continue applying elementaryrow operations to a REF matrix until everycolumn that has a leading 1 has 0’s elsewhere.This is the Reduced Row Echelon Form(RREF) of the matrix.

Let’s see this with our previous example…

From this final RREF matrix, we canFrom this final RREF matrix, we canread the solution of (–1, 2, –3) directly…read the solution of (–1, 2, –3) directly…

We can also use thecalculator to finda RREF matrix…

Try it with our initial augmented matrix:

2 12R R 1 2 3 12

0 1 2 8

0 0 1 3

1 0 1 4

0 1 2 8

0 0 1 3

3 1R R 1 0 0 1

0 1 2 8

0 0 1 3

3 22R R1 0 0 1

0 1 0 2

0 0 1 3

3 7 11 44

1 2 3 12

4 9 13 53

More Good ExamplesSolve the system using a RREF matrix

2 3 5x y z w 2 3 16x y z w

7 3 2 2 11x y z w 5 9x y z w

Augmented Matrix:

2 1 1 3 5

1 2 3 1 16

7 3 2 2 11

1 1 1 5 9

RREF:(Calc)

1 0 0 0 2

0 1 0 0 7

0 0 1 0 1

0 0 0 1 3

Solution:Solution:(x, y, z, w) = (2, 7, –1, –3)(x, y, z, w) = (2, 7, –1, –3)

More Good ExamplesSolve the system using a RREF matrix

3x y z 2 4 8x y z

2 1x y z

RREF Matrix:

1 0 3 5

0 1 2 2

0 0 0 0

Solution: Set of all ordered triples in the formSolution: Set of all ordered triples in the form(–3z + 5, 2z – 2, z), where z is any real number(–3z + 5, 2z – 2, z), where z is any real number

3 5x z 2 2y z 0 0

3 5x z 2 2y z

For every value of z,we can use these twoequations to find adifferent x and y!!!

More Good Examples

2 3 1x y z 2 3 4 1x y z w

3 5 7 2x y z w

Augmented Matrix:

1 2 3 0 1

2 3 4 1 1

3 5 7 1 2

Solve the system using a RREF matrix

RREF Matrix:

1 0 1 2 1

0 1 2 1 1

0 0 0 0 0

2 1x z w 2 1y z w

Solution: Set of all orderedSolution: Set of all ordered4-tuples of the form4-tuples of the form

(–z – 2w + 1, 2z + w – 1, z, w), z and w any real #’s(–z – 2w + 1, 2z + w – 1, z, w), z and w any real #’s