# Ch02 - 1 MP CHAPTER 2 SOLUTIONS Section 2.1 1 2 5 8 3 6 9...

This preview shows pages 1–4. Sign up to view the full content.

MP CHAPTER 2 SOLUTIONS Section 2.1 1a. -A= - - - - - - - - - 9 8 7 6 5 4 3 2 1 1b. 3A= 27 24 21 18 15 12 9 6 3 1c. A+2B is undefined. 1d. A T = 9 6 3 8 5 2 7 4 1 1e. B T = - 2 1 2 1 0 1 1f. AB= 24 16 15 10 6 4 1g. BA is undefined. 2. y y y 3 2 1 = 60 . 30 . 20 . 30 . 70 . 30 . 10 . 0 50 . x x x 3 2 1 3. Let A = (a ij ), B = (b ij ), and C = (c ij ). We must show that A(BC) = (AB)C. The i-j'th element of A(BC) is given by 1

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Σ a ix ( Σ b xy c yj ) = Σ Σ a ix b xy c yj x y x y The i-j'th element of (AB)C is given by Σ ( Σ a ix b xy )c yj = Σ Σ a ix b xy c yj = Σ Σ a ix b xy c yj y x y x x y Thus A(BC) = (AB)C 4. The i-j'th element of (AB) T = element j-i of AB = Scalar product of row j of A with column i of B. Now element i-j of B T A T = Scalar product of row i of B T with column j of A T = Scalar product of row j of A with column i of B. Thus (AB) T = B T A T . 5a. From problem 4 we know that (AA T ) T = AA T , which proves the desired result. 5b. (A + A T ) ij = a ij + a ji . Also (A + A T ) ji = a ji + a ij which proves the desired result. 6. To compute each of the n 2 elements of AB requires n multiplications. Therefore a total of n 2 multiplications are needed to compute AB. To compute each of the n 2 elements of AB requires n – 1 additions. Therefore a total of n 2 (n – 1) =n 3 –n 2 additions are needed to compute AB. 7a. Trace A+ B = + = + = + i i i ii ii ii ii B Trace A Trace b a b a . ) ( 7b. Trace(AB) = ∑∑ i k ki ik b a and Trace(BA) = ∑∑ k i ki ik a b . Note that the terms used to compute the i’th entry of BA duplicate the i’th term used to compute each entry of AB. Therefore exactly the same terms are used to compute the sum of all entries in AB and BA. 2
Section 2.2 Solutions 1. - 3 1 1 2 1 1 x x 2 1 = 8 6 4 and - 8 3 1 6 1 2 4 1 1 Section 2.3 Solutions 1. 8 4 3 1 1 2 1 0 1 1 0 1 0 1 1 5 4 3 0 1 1 0 0 1 1 0 1 0 1 1 - - 1 4 1 0 0 0 0 0 1 1 0 1 1 0 1 The last row of the last matrix indicates that the original system has no solution. 2. 6 4 0 2 1 1 1 1 - 2 4 1 1 0 1 1 1 - 2 2 1 1 0 2 0 1 This system has an infinite number of solutions of the form x 3 = k, x 1 = 2 - 2k, x 2 = 2 + k.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

## This note was uploaded on 04/02/2008 for the course IEOR 160 taught by Professor Hochbaum during the Fall '07 term at University of California, Berkeley.

### Page1 / 21

Ch02 - 1 MP CHAPTER 2 SOLUTIONS Section 2.1 1 2 5 8 3 6 9...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online