Elimination using matrices - Boise State University · Elimination using matrices Recall :...
Transcript of Elimination using matrices - Boise State University · Elimination using matrices Recall :...
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”
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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