ch09 - CHAPTER 9 1. With A + = 0 0 0 0 0 1 0 0 0 0 0 2 0 0...

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Unformatted text preview: CHAPTER 9 1. With A + = 0 0 0 0 0 1 0 0 0 0 0 2 0 0 0 0 0 3 0 0 0 0 0 4 we have ( A + ) 2 = 0 0 0 0 0 1 0 0 0 0 0 2 0 0 0 0 0 3 0 0 0 0 0 4 0 0 0 0 0 1 0 0 0 0 0 2 0 0 0 0 0 3 0 0 0 0 0 4 = 0 0 0 0 0 0 0 0 0 0 2 0 0 0 0 0 6 0 0 0 0 0 12 0 0 It follows that ( A + ) 3 = 0 0 0 0 0 1 0 0 0 0 0 2 0 0 0 0 0 3 0 0 0 0 0 4 0 0 0 0 0 0 0 0 0 0 2 0 0 0 0 0 6 0 0 0 0 0 12 0 0 = 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 3.2.1 0 0 0 0 0 4.3.2 0 0 0 The next step is obvious: In the 5 x 5 format, there is only one entry in the bottom left- most corner, and it is 4.3.2.1 . 2. [The reference should be to Eq. (6-36) instead of Eq. (6-4) x = h 2 m ( A + A + ) = h 2 m 0 1 0 0 0 1 0 2 0 0 0 2 0 3 0 0 3 0 4 0 0 0 4 from which it follows that x 2 = h 2 m 1 0 2.1 0 0 0 3 0 3.2 2.1 0 5 0 4.3 0 3,2 0 7 0 0 0 4.3 0 9 3. The procedure here is exactly the same. We have p = i m h 2 ( A +- A ) = i m h 2- 1 0 0 0 1- 2 0 0 0 2- 3 0 0 3- 4 0 0 0 4 from which it follows that p 2 = m h 2 1 0- 2.1 0 0 0 3 0- 3.2- 2.1 0 5 0- 4.3- 3.2 0 7 0 0 0- 4.3 0 9 4. We have u 1 = A + u = 0 0 0 0 0 1 0 0 0 0 0 2 0 0 0 0 0 3 0 0 0 0 0 4 1 = 1 5. We write u 2 = 1 2!...
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ch09 - CHAPTER 9 1. With A + = 0 0 0 0 0 1 0 0 0 0 0 2 0 0...

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