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Unformatted text preview: MATH 223, Linear Algebra Winter, 2008 Solutions to Assignment 1 1. Let z = 4 2 i and w = 8 + i . Find Â¯ z , Â¯ w , z + w , z w , z Â· 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: Â¯ z = 4 + 2 i ; it has absolute value âˆš 4 2 + 2 2 = âˆš 20 = 2 âˆš 5. Â¯ w = 8 i and it has absolute value p 8 2 + ( 1) 2 = âˆš 65. z + w = 12 i with absolute value âˆš 145. z w = 4 3 i with absolute value âˆš 25 = 5. z Â· w = 32 16 i + 4 i + 2 = 34 12 i ; its absolute value is âˆš 1300 = 10 âˆš 13. Finally, z w = 4 2 i 8+ i = (4 2 i )(8 i ) 65 = 30 20 i 65 = 6 13 4 13 i . Its absolute value is âˆš 52 13 = 2 âˆš 13 13 . 2. Show that if z and w are any two complex numbers, then z Â· w = Â¯ z Â· Â¯ w . Use this to show that if A and B are any complex matrices, then A Â· B = Â¯ A Â· Â¯ B . [N.B. The conjugate Â¯ A of a matrix A is the most obvious thing â€” you just replace each entry of A by its conjugate. Also, we of course assume here that A Â· B is defined.] Solution: Suppose that z = a + bi and w = c + di , where a,b,c,d are real. Then z Â· w = ( ac bd ) + ( ad + bc ) i , so z Â· w = ( ac bd ) ( ad + bc ) i . Â¯ z = a bi and Â¯ w = c di , so Â¯ z Â· Â¯ 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 Â· B , A Â· B and Â¯ A Â· Â¯ B have the same number of rows and columns. We must check that the ( j,k )entry of A Â· B is the same as the ( j,k )entry of Â¯ A Â· Â¯ B for every j and k . Suppose that A has n columns and therefore B has n rows. The ( j,k )entry of A Â· B is then âˆ‘ 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 Â· B is the conjugate of âˆ‘ n j =1 a j,m b m,k . We havenâ€™t shown that the conjugate of a sum is the sum of the conjugates, but this is very easy. (Try it yourself, or ask. Donâ€™t just take my word for it.) So the entry weâ€™re interested in is âˆ‘ n m =1 a j,m b m,k and by the first part of the problem, this is âˆ‘ n m =1 Â¯ a j,m Â¯ b m,k . This is indeed the ( j,k )entry of Â¯ A Â· Â¯ B ....
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 Fall '07
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 Linear Algebra, Algebra, Real Numbers, Complex number, Jaguar Racing, augmented matrix

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