How to get to Reduced Row Echelon Form

Use rows with leading 1’s Get the leading 1’s starting at the pivot Use the pivot to get zeros below

and/or above Use rows with zeros on top of each

other as you move right

Interchange two rows

Multiply a row throughby a non-zero constant

Add a multiple of onerow to another.

Allowable Row Operations Notation

Ri ↔Rj

kRi → Ri

kRi + Rj → Rj

PIVOT

1 1 2 7

3 2 1 10

1 3 1 2

-3R1 + R2 → R2

R1 + R3 → R3

Use this one, called the pivot, to get zeros below.

1 1 2 7

3 2 1 10

1 3 1 2

-3R1 + R2 → R2

1 1 2 7

0 5 7 31

1 3 1 2

1 1 2 7

0 5 7 31

0 2 3 5

R1 + R3 → R3

1 1 2 7

0 5 7 31

1 3 1 2

New pivot location

1 1 2 7

0 5 7 31

0 2 3 5

-2R3 + R2 → R2

We want a 1 here but we also want to avoid fractions

New pivot

1 1 2 7

0 1 13 41

0 2 3 5

Use this new pivot to get zeros above and below

R2 + R1 → R1

-2R2 + R3 → R3

New pivot location

1 0 11 34

0 1 13 41

0 0 29 87 Now get a 1 here.

R3 → R31

29

New pivot

1 0 11 34

0 1 13 41

0 0 1 3

Use this new pivot to make zeros above.

11R3 + R1→ R1

13R3 + R2→ R2

1 0 0 1

0 1 0 2

0 0 1 3

1

2

3

1

2

3

x

x

x

1 3 4

1 2 4

7 5

2 3 18 12

x x x

x x x

Solve the system:

Augmented matrix:

1 0 1 7 5

2 3 0 18 12

1 2 3 4x x x x

There are two equations and

four unknowns so we need a 2 x 5

matrix.

1 0 1 7 5

2 3 0 18 12-2R1 + R2 → R2

1 0 1 7 5

0 3 2 4 2R2 → R2

13

42 23 3 3

1 0 1 7 5

0 1←RREF

42 23 3 3

1 0 1 7 5

0 1

Expressing the solution:

If a system is consistent, any

column that does not contain a leading one will have a parameter.

1 2 3 4x x x x

Parameters required

for x3 and x4

42 23 3 3

1 0 1 7 5

0 1

Expressing the solution:

1 2 3 4x x x x

s t

3

4

x s

x t

LET

42 23 3 3

1 0 1 7 5

0 1

Converting back to equations in terms of s and t: s t

1

42 22 3 3 3

7 5x s t

x s t

142 2

2 3 3 3

3

4

5 7x s t

x s t

x s

x t

Solving for x1 and x2 leads to the final solution: