**Unformatted text preview: **1 AM 1411a, Fall 2014
Lectures 12, October 1
An algorithm for …nding I ! E and E 1 ! I elem. row op. A 1 E ! I and I ! E 1 inverse row op. Theorem1.5.3. Equivalent statements. An n
(a) A is invertible.
(b) Ax = 0 has only the trivial solution.
(c) RREF of A is In (A is row equivalent to In). (d) A is expressible as a product of elementary matrices. Proof (c) ) (d)
(c) RREF of A is In ) Ek :::E2E1A = In
multiply by inverses
E1 1E2 1:::Ek 11Ek 1Ek Ek 1:::E2E1A =
E1 1E2 1:::Ek 11Ek 1In )
A = E1 1E2 1:::Ek 11Ek 1: Method of …nding the inverse
multiply (F) by A 1
Ek :::E2E1AA 1 = InA 1 ) (F) A 1 = Ek :::E2E1 Compare Ek :::E2E1A = In and Ek :::E2E1In = A 1
The same sequence of operations transforms A to In and
In to A 1
Matrix inversion Algorithm
1. Find RREF of A:
2. If RREF of A has a zero row, stop. A is not invertible.
3. If RREF of A is I , then read A 1 from
h [AjI ] ! IjA 1
2 i 1 0 6
Ex 1. A = 4 3 4 3 5 3 1
7
2 5 : Find the inverse.
2 2 1 0 6
[AjI ] = 4 3 4 2 1 0 6
4 0 4
2 0 5 1 0 6
4 0 4 1
1
1 1
3 1
4 3
4 1 0
6
4 0 4
0 0 1
1
1 1 0 1
0
1 2 6
4 0 4
2 0 0 1 0 0 6
4 0 1 0 0 0 1 1 1 0 0
7 R
2 0 1 0 5 3
R2
2 0 0 1
3 1 0 0
7
3 1 0 5 R3
3 0 1 1
1 0 0 2 3 5 3 0
1 0
0 7
5
5 1
4 1 0
3 1
3 5
1
0
3
2
0
3 3 0
4
5
5
1
5 3R1 ! R3
3R1 ! R2
! 5R ! R
3
4 1 ! 4R3 ! R3
! 3 0
7
0 5 R2
4 R1 ! R2
! 3 0
7 R + R3 ! R1
4 5 11
4 R2 ! R2
4
!
3 4
h
i
7
1
1 5 = IjA
4 2 2 6
A 1=4 0 3 5
1
5 3 4
7
1 5
4 Check AA 1 = A=1A = I
2 1 6
Ex 2. A = 4 3
2 1 6
[AjI ] = 4 3
2 1 1 0 6
4 0 0 1 2 0 0 1 1 1 0
6
4 0 0 1
0 0 0 2 2 3 1 0
7
3 1 5 : Find the inverse.
2 1
3 1 0 1 0 0
7 R + 2R1 ! R3
3 1 0 1 0 5 3
R2 3R1 ! R2
2 1 0 0 1
!
3 1 0 0
7
3 1 0 5 R3
2 0 1
1
3
5 3 R2 ! R3
! 0 0
h
i
7
1
6
IjA
1 0 5=
1 1 The inverse does not exist. ...

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- Fall '19
- Invertible matrix, Inverse function, E2E1