Assignment 1 - Solutions

Assignment 1 - Solutions - MATH 223, Linear Algebra Winter,...

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Unformatted text preview: MATH 223, Linear Algebra Winter, 2010 Solutions to Assignment 1 1. Let z = 5 + 3 i and w = 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 = 5- 3 i , w = 4 + i , z + w = (5 + 4) + (3- 1) i = 9 + 2 i , z- w = (5- 4) +(3 +1) i = 1 +4 i , z w = 5 4 +3 i 4- 5 i- 3 i 2 = 23 +7 i (since- 3 i 2 = +3) and z w = (5+3 i )(4+ i ) (4- i )(4+ i ) = 20+12 i +5 i- 3 17 = 1 + i . | z | = p 5 2 + (- 3) 2 = 34, | w | = 4 2 + 1 2 = 17, | z + w | = 9 2 + 2 2 = 85, | z- w | = 1 2 + 4 2 = 17, | z w | = 23 2 + 7 2 = 578 and | z w | = 1 2 + 1 2 = 2. 2. (a) Prove that for any complex numbers z 1 and z 2 , z 1 + z 2 = z 1 + z 2 and z 1 z 2 = z 1 z 2 You may use any properties about multiplication and addition of real numbers. Solution: The first is practically obvious, but here goes. Say z 1 = a 1 + b 1 i and z 2 = a 2 + b 2 i where each a j and b j is real. Then z 1 + z 2 = ( a 1 + a 2 )+( b 1 + b 2 ) i and z 1 + z 2 = ( a 1 + a 2 )+(- b 1- b 2 ) i . z j = a j +(- b j ) i for j = 1 , 2, so z 1 + z 2 is also ( a 1 + a 2 )+(- b 1- b 2 ) i . z 1 z 2 = ( a 1 a 2- b 1 b 2 )+( a 1 b 2 + a 2 b 1 ) i and so z 1 z 2 = ( a 1 a 2- b 1 b 2 )+ (- a 1 b 2- a 2 b 1 ) i . z 1 z 2 = ( a 1- b 1 i ) ( a 2- b 2 i ) = a 1 a 2- b 1 a 2 i- a 1 b 2 i + (- b 1 i )(- b 2 i ) = ( a 1 a 2- b 1 b 2 ) + (- a 1 b 2- a 2 b 1 ) i , too. (Since (- b 1 i )(- b 2 i ) = b 1 b 2 i 2 =- b 1 b 2 .) (b) If A is a matrix over the complex numbers, we let A be the most obvious thing it is obtained from A by replacing each entry by its conjugate. Supposing that A B is defined, show that A B = A B . Solution: I trust you see that A + B = A + B , whenever this is defined. For the given question, suppose that A has n columns, so that B has n rows, and that the ( r,s )-entry of A is a r,s and of B is b r,s . Then the ( j,k )-entry of AB is a j, 1 b 1 ,k + a j, 2 b 2 ,k + + a j,n b n,k and so the ( j,k )-entry of A B is a j, 1 b 1 ,k + a j, 2 b 2 ,k + + a j,n b n,k . By the previous part of this problem, this is a j, 1 b 1 ,k + a j, 2 b 2 ,k + + a j,n b n,k = a j, 1 b 1 ,k + a j, 2 b 2 ,k + + a j,n b n,k , which is also the ( j,k )-entry of A B. 1 3. Solve each of the following systems of equations. That is, find the unique solution if there is one, the general solution in vector parametric form if there is more than one solution, or explain why there is no solution if that is the case. Use augmented matrices. (a) This ones over the field R , the reals. x 1- 3 x 2 +5 x 3- 7 x 5 = 4- 3 x 1 +9 x 2- 15 x 3 + x 4 11 x 5 = 2 2 x 1- 6 x 2 +10 x 3- x 4- 4 x 5 =- 1 Solution: The relevant augmented matrix is 1- 3 5- 7 | 4- 3 9- 15 1 11 | 2 2- 6 10- 1- 4 | - 1 ....
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This note was uploaded on 02/10/2010 for the course MATH math 223 taught by Professor Loveys during the Winter '10 term at McGill.

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Assignment 1 - Solutions - MATH 223, Linear Algebra Winter,...

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