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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: 1e11+3a21+2a31:7
la12+3022+2o32:1
ia.13+3(223+2033=3
2'a11+1'fL21+1'a31:1
2'(112+1‘ﬂ22+1(t32=0
2013+1ai23+1a33:3
4on+Ougl+3a31=—1
zla12+Oo.22+3(132:~3
4a13+0~a23+3ag3:7 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 , (121, 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 : Zagqbqj. Now all the terms am in this expression are 0 except for q : i, so
CU' : ailb5}. 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 205040 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 5501 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 (llj] : [ﬁg] 2 A A AA = lagg Aagjl : [ﬁllj] : 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,,,)] = [eml : 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)] : [tilj] = 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.
 Spring '09
 Ming

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