3.8 (evens) - SECTION 3.8 Matrices 2. We just add entr)r by...

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Unformatted text preview: SECTION 3.8 Matrices 2. We just add entr)r by entry. 749210 74—540 4. To multiply matrices A and B, we compute the (i,j)”‘ entry of the product AB by adding all the products of elements from the in‘ row of A with the corresponding element. in the j Lh column of B, that is 213:1 aikbkj. This can only be done, of course, when the number of columns of A equals the number of rows of B (called n in the formula showu here). a) 71 1 0 b) 4 —1 —7 b c) 2 0 —3 —4. -1 0 1 —1 —7 75 8 5 24 —7 20 29 2 1 72 1 4 0 7 '1 —10 4 717 —24 —3 Section 3.8 ' 6. 10. 12. 14. 16. 18. Allatrices 85 First note that A must be a 3 x 3 matrix in order for the sizes to work out as shown. If we name the elements of A in the usual way as [013-], then the given equation is really nine equations in the nine unknowns a”, obtained simply by writing down what the matrix multiplication on the left means: 1-e11+3-a21+2-a31:7 l-a12+3-022+2-o32:1 i-a.13+3-(223+2-033=3 2'a11+1'fL21+1'a-31:1 2'(112+1‘fl22+1-(t32=0 2-013+1-ai23+1-a33:3 4-on+O-ugl+3-a31=—1 zl-a12+O-o.22+3-(132:~3 4-a13+0~a23+3-ag3:7 This is really not as had as it looks, since each variable only appears in three equatioas. For example, the first, fourtlL and seventh equations are a system of three equations in the three variables {in , (1-21, and (131. We can solve them using standard algebraic techniques to obtain a.“ = -1, (121 2 2 and (3.31 : 1. By similar reasoning we also obtain (ng : 0, (1'22 : 1 and 032 : el; and 013 = 1, (123 = 0 and (133 = 1. Thus our answer is As a check we can carry out the matrix multiplication and verify that we obtain the given right—hand side. . Since the entries of A + B are on + big; and the entries of B + A are bU- + Gij, that A + B : B + A follows from the commutativity of addition of real numbers. a) This product is a 3 x 5 matrix. b) This is not defined since the number of columns of B does not equal the number of rows of A. c) This product is a 3 x 4 matrix. (1) This is not defined since the number of columns of C does not equal the number of rows of A. e) This is not defined since the number of columns of B does not equal the number of rows of C. f) This product is a 4 X 5 matrix. “re use the definition of matrix addition and multiplication. All summations here are from 1 to k. a) (A + B)C : [Elem + mac“) : aichj + Z bgchj] : AC + BC b) C(A + B) = [ECMGQJ' + bqjll : [2621:an + Zciqbqjl : CA + CB Let A and B be two diagonal n. x11 matrices. Let C : [CU] be the product AB. From the definition of matrix multiplication, CU- : Za-gqbqj. Now all the terms am in this expression are 0 except for q : i, so CU' : ail-b5}. But by] = 0 unless 1' : j, so the only nonzero entries of C are the diagonal entries ch = nab“. The (2'..j)"l‘ entry of (Air is the (j,i)t‘l‘ entry of A“, which is the (i,j)”' entry of A. We need to multiply these two matrices together in both directions and check that both products are 13. Indeed, they are. 86 20. 22. 24. 26. 28. 30. 32. 34. 36. Chapter 3 The Fundamentals: Algorithms, the Integers, and Alatrices a) Using Exercise 19, noting that ad. 7 be 2 —5, we write down the inverse immediately: —3/5 2/5 1/5 1/5 ' _. r . 2 _ 3 4 3 _ 1 18 b) We multiply to obtain A i [2 11] and then A f 9 37 . I _ . . ,1 2 i 11/25 —4/25 ,1 3 _ 437/125 18/125 c) We multiply to obtain (A ) — [_2/25 3/25 and then (A ) _ 9/125 71/125 . (‘1) Applying the method of Exercise 19 for obtaining inverses to the answer in part (b), we Obtain the answer in Part- Therefore (A3)_1 = (A4)? A matrix is symmetric if and only if it equals its transpose. So let us compute the transpose of AA: and see if we get this matrix back. Using Exercise 17b and then Exercise 16, we have (AA‘)’ : ({A')‘)A* = AA‘, as desired. a) If we compute the product as A1(A2A3), then by the result of Exercise ‘23 it will take 50 - 10 - 40 multiplications for the first product and then 2050-40 for the second. This is a total of 60,000 multiplications. If we compute the product as (AIA2)A3, then it will take 20 - 50 - 10 multiplications for the first product and then 20 - 10 - 40 for the second. This is a total of 18,000 multiplications. Therefore the second method .is more efficient. b) If we compute the product as A1(A2 A3), then by the result of Exercise ‘23 it will take 5-50-1 multiplications for the first product and then 10 - 5 - 1 for the second. This is a total of 300 multiplications. If we compute the product as (A1_A2)A3, then it will take 10 - 5 -50 multiplications for the first product and then 10- 50 -1 for the second. This is a total of 1000 multiplications. Therefore the first method is more efficient. ' a) “is simply note that under the given definitions of A, X, and B, the definition of matrix multiplication is exactly the system of equations shown. b) The given system is the matrix equation AX = B. If A is invertible with inverse A4, then we can multiply both sides of this equation by A‘1 to obtain A‘IAX = A‘IB. The left—hand side simplifies to IX, however, by the definition of inverse, and this is simplyr X. Thus the given system is equivalent to the system X : A‘-IB, which obviously tells us exactly what X is (and therefore what all the values 3:,- are). “We follow the definitions. a) 1 1 b) 0 1 c) 1 1 1 1 0 0 l 0 1 0 We follow the definition and obtain 1 1 1 1 a) AVA =[r1gj V (ll-j] : [fig] 2 A A AA = lagg- Aagjl : [fill-j] :- A a) V B) V C : [(0,3) V bu) V Cg] = [flij V (0,-3- V GUN : A V (B V b) This is identical to part (a), with /\ replacing V. Since the 1‘“ row of I consists of all 0’s except for a 1 in the (i, 1')"" position, we have I G) A = [(0 An”) V V (1 A a”) V - - - V (0 /\ a,,,)] = [em-l : A. Similarly, since the 3"" column of I consists of all 0’s except for a 1 in the (j,j)l‘h position, we have A (D I :[(cr,-1/\ 0) V - - - V (aij /\ 1) V - -- V (a.,-,, /\ 0)] : [til-j] = A. ...
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This note was uploaded on 02/18/2012 for the course MATH 640:244 taught by Professor Ming during the Spring '09 term at Rutgers.

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3.8 (evens) - SECTION 3.8 Matrices 2. We just add entr)r by...

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