ass1 - MATH 223, Linear Algebra Fall, 2007 Solutions to...

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Unformatted text preview: MATH 223, Linear Algebra Fall, 2007 Solutions to Assignment 1 1. Let z = 2- 7 i and w = 3 + 4 i . Find z , w , z + w , z- w , z w and z w (all in the form a + bi with a and b real numbers). Find the absolute value of each of these 6 numbers. Solution: z = 2 + 7 i , w = 3- 4 i , z + w = 5- 3 i , z- w =- 1- 11 i , z w = (2- 7 i ) (3+4 i ) = 2 3+2 4 i- 7 i 3- 7 i 4 i = 6+8 i- 21 i +28 = 34- 13 i and z w = 2- 7 i 3 + 4 i = (2- 7 i )(3- 4 i ) (3 + 4 i )(3- 4 i ) = 6- 8 i- 21 i- 28 3 2 + 4 2 =- 22 25 +- 29 25 i. The absolute vales are, in order, 2 2 + 7 2 = 53, p 3 2 + (- 4) 2 = 25 = 5, p 5 2 + (- 3) 2 = 34, p (34) 2 + (- 13) 2 = 1325 = 5 53 and r (- 22 25 ) 2 + (- 29 25 ) 2 = 1 25 p 22 2 + 29 2 = 1325 25 = 53 5 . (Full marks for the second-last answer.) 2. Show that if z and w are any two complex numbers, then z w = z w . Use this to show that if A and B are any complex matrices, then A B = A B . [N.B. The conjugate A of a matrix A is the most obvious thing you just replace each entry of A by its conjugate. Also, we of course assume here that A B is defined.] Solution: Suppose that z = a + bi and w = c + di , where a,b,c,d are real. Then z w = ( ac- bd ) + ( ad + bc ) i , so z w = ( ac- bd )- ( ad + bc ) i . z = a- bi and w = c- di , so z w = ac- (- b )(- d ) + ( a (- d ) + (- b ) c ) i = ac- bd- ( ad + bc ) i ; that does the first part. Obviously, all of the matrices A B , A B and A B have the same number of rows and columns. We must check that the ( j,k )-entry of A B is the same as the ( j,k )-entry of A B for every j and k . Suppose that A has n columns and therefore B has n rows. The ( j,k )-entry of A B is then n m =1 a j,m b m,k where of course a j,m is the ( j,m )-entry of A and b m,k is the ( m,k )-entry of B . The ( j,k )-entry of A B is the conjugate of n j =1 a j,m b m,k . We havent shown that the conjugate of a sum is the sum of the conjugates, but this is very easy. (Try it yourself, or ask. Dont just take my word for it.) So the entry were interested in is n m =1 a j,m b m,k and by the first part of the problem, this is n m =1 a j,m b m,k . This is indeed the (....
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This note was uploaded on 04/29/2008 for the course MATH 223 taught by Professor Loveys during the Fall '07 term at McGill.

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ass1 - MATH 223, Linear Algebra Fall, 2007 Solutions to...

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