Elimination using matrices

Recall : Multiplication on the left by a row vector results in a row vector.

Multiplication on the left can be thought of as a linear combination of rows.

Result is a row vector

⇥a b c

⇤ 2

42 4 51 9 �3

�2 1 7

3

5 =

a⇥2 4 5

⇤+ b

⇥1 9 �3

⇤+ c

⇥�2 1 7

⇤

1Monday, October 7, 13

Elimination using matrices

(a, b, c) = ? (0, 1, 0)

(a, b, c) = (1, 0, 1)We managed to eliminate the first coefficient

⇥a b c

⇤ 2

42 4 51 9 �3

�2 1 7

3

5 =⇥1 9 �3

⇤

What about ⇥a b c

⇤ 2

42 4 51 9 �3

�2 1 7

3

5 =⇥0 5 12

⇤

2Monday, October 7, 13

Elimination matrices

2x+ 4y � 2z = 2

4x+ 9y � 3z = 8

�2x� 3y + 7z = 10

2x+ 4y � 2z = 2

y + z = 4

4z = 8

matrix

multiplication?

2

42 4 �24 9 �3

�2 �3 7

3

5

2

42 4 �20 1 10 0 4

3

5

elimination

Find a matrix E so that EA = U

A UE

3Monday, October 7, 13

Elimination matrices : First step

How do we express ”(equation 2)�(2)(equation 1)”

2x+ 4y � 2z = 2

4x+ 9y � 3z = 8

�2x� 3y + 7z = 10

elimination2x+ 4y � 2z = 2

y + z = 4

y + 5z = 12

2

41 0 0? ? ?? ? ?

3

5

2

42 4 �24 9 �3

�2 �3 7

3

5 =

2

42 4 �20 1 10 1 5

3

5eliminate

How do we express ”(equation 3)�(�1)(equation 1)”

4Monday, October 7, 13

Elimination matrices : First step

�`21

How do we express ”(equation 2)�(2)(equation 1)”

�`31

How do we express ”(equation 3)�(�1)(equation 1)”2

41 0 0

�2 1 01 0 1

3

5

2

42 4 �24 9 �3

�2 �3 7

3

5 =

2

42 4 �20 1 10 1 5

3

5

2

41 0 0

�2 1 0? ? ?

3

5

2

42 4 �24 9 �3

�2 �3 7

3

5 =

2

42 4 �20 1 10 1 5

3

5

E31E21 “Elimination matrices”

5Monday, October 7, 13

Elimination matrix : Second step

2x+ 4y � 2z = 2

y + z = 4

y + 5z = 12

2x+ 4y � 2z = 2

y + z = 4

4z = 8

elimination

How do we express ”(equation 3)�(1)(equation 2)”

eliminate

2

41 0 00 1 0? ? ?

3

5

2

42 4 �20 1 10 1 5

3

5 =

2

42 4 �20 1 10 0 4

3

5

6Monday, October 7, 13

Elimination matrix : Second step

How do we express ”(equation 3)�(1)(equation 2)”

2

41 0 00 1 00 �1 1

3

5

2

42 4 �20 1 10 1 5

3

5 =

2

42 4 �20 1 10 0 4

3

5

�`32 E32 E31E21A U

We found a sequence of matrices that will take A to U

7Monday, October 7, 13

Elimination matrices

2

41 0 00 1 00 �1 1

3

5

2

41 0 0

�2 1 01 0 1

3

5

2

42 4 �24 9 �3

�2 �3 7

3

5 =

2

42 4 �20 1 10 0 4

3

5

We have constructed elimination matrices

E31E21E32 A U

E31E21 =

2

41 0 00 1 01 0 1

3

5

2

41 0 0

�2 1 00 0 1

3

5 =

2

41 0 0

�2 1 01 0 1

3

5

E32E31E21A = U

U is an upper triangular matrix

Check

Product of elimination matrices is lower triangular

8Monday, October 7, 13