# lecture 12 finding inverse.pdf - 1 AM 1411a Fall 2014...

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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

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