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Solution: a) For “hand computation,” the best choice is to interchange equations 3 and 4. Another possibility is to multiply equation 3 ...
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FOR THE ELECTRONICS AND TELECOMMUNICATION STUDENTS
2
Preface
This is the complementary text to my Linear Algebra Lecture Notes for the telecommunication students at Technical University in Pozna´n. It is designed to help you succeed in your linear algebra course, and shows you how to study mathematics, to learn new material, and to prepare effec- tive review sheets for tests. This text guide you through each section, with summaries of important ideas and tables that connect related ideas. Detailed solutions to many of exercises allow you to check your work or help you get started on a difficult problem. Also, complete explanations are provided for some writing exercises Practical Problems point out important exercises, give hints about what to study, and sometimes highlight potential exam questions. Frequent warnings identify common student errors. Don’t ever take an exam without reviewing these warnings! Good luck!
Andrzej Ma´ckiewicz Pozna´n, September 2014
Exercise 1.1 Simplify the imaginary numbers below
a)
b) ±
c) −
d) 11
e) ^9
f ) ^12
g) 24 ^20
h) 16 −
i) 16 −
j) ^ where is positive even number.
Exercise 1.2 Solve the following problems. Answers are to be in simplest + form.
a) 9 − 25 b) 25 c) 34
a) 15 − 12 b) 29 − 39 c) 83 + 39
8 1. Complex Numbers (Exercises)
a) 25 b) − 25 c) 12 − 12
a) 21 + 20 b) −21 + 20 c) 29 + 20
a) 7 + 8 b) 8 + 8 c) 9 + 8
a) 5 − 2 b) 3 + 2 c) 15 + 10
a) 35 37 + (1237) b) 35 + 12 c) 35 36 + (1236)
a) 5 + 3 b) − 5 − 3 c) 5 − 3
10 1. Complex Numbers (Exercises)
Exercise 1.10 a) Write ( ) + ( ) = ( ) and point out how it follows that the complex number 0 = (0 0) is unique as an additive identity.
b) Likewise, write ( )( ) = ( ) and show that the number 1 = (1 0) is a unique multiplicative identity.
Exercise 1.11 Solve the equation ^2 − 2 + 2 = 0, for = ( ) by writing
( )( ) − 2( ) + (2 0) = (0 0)
and then solving a pair of simultaneous equations in x and y. HINT: Use the fact that no real number satisfies the given equation to show that 6 = 0. Answer: Solution is: 1 + 1 − .
Exercise 1.12 Reduce each of these quantities to a real number:
a)
b)
c) (1 − )^8
Answer: a) −^25 b) 12 c) 16
Exercise 1.13 Show that
1 1
Exercise 1.14 Use the associative and commutative laws for multiplication to show that ( 1 2 )( 3 4 ) = ( 1 3 )( 2 4 )
Exercise 1.15 Prove that if 1 2 3 = 0, then at least one of the three factors is zero. HINT: Write ( 1 2 ) 3 = 0 and use a similar result involving two factors.
Exercise 1.16 Locate the numbers 1 + 2 and 1 − 2 vectorially when
a) 1 = 3 2 = 43 −
b) 1 =
c) 1 = (− 2 1) 2 =
d) 1 = 1 + 1 2 = 1 − 1
Exercise 1.17 Verify inequalities ??, involving Re(), Im(), and ||.
Exercise 1.18 Use established properties of moduli to show that when | 3 | 6 = | 4 |, Re( 1 + 2 ) | 3 + 4 |
Exercise 1.19 Verify that
2 || ≥ | Re | + | Im | HINT: Reduce this inequality to(|| − ||)^2 ≥ 0.
Exercise 1.20 In each case, sketch the set of points determined by the given condition:
a) | − 2 + | = 1;
b) | + | ≤ 2;
c) | − 4 | ≥ 3
Exercise 1.21 Using the fact that | 1 − 2 | is the distance between two points 1 and 2 , give a geometric argument that
a) | − 4 | + | + 4| = 10 represents an ellipse whose foci are (0 ±4);
b) | − 1 | = | + | represents the line through the origin whose slope is − 1.
Exercise 1.22 Use properties of conjugates and moduli to show that
a) ¯ + 4 = − 4 ;
b) = −;
c) (2 + )^2 = 3 − 4 ;
d)
a) is real if and only if ¯ = ;
b) is either real or pure imaginary if and only if ¯^2 = ^2
Exercise 1.31 Use mathematical induction to show that when = 2 3 ,
a) 1 + 2 + · · · + = ¯ 1 + ¯ 2 + · · · + ¯;
b) 1 2 · · · = ¯ 1 ¯ 2 · · · ¯.
Exercise 1.32 Let 0 1 2 ( ≥ 1) denote real numbers, and let be any complex number. With the aid of the results in previous, show that
0 + 1 + 2 ^2 + ··· + ^ = 0 + 1 ¯ + 2 ¯^2 + ··· + ¯
Exercise 1.33 Show that the equation | − 0 | = of a circle, centered at 0 with radius , can bewritten
||^2 − 2 Re( 0 ) + | 0 |^2 = ^2
Exercise 1.34 Find the principal argument Arg when
a) =
b) =
Answer: a) − 3 4 b)
Exercise 1.35 Show that a) || = 1; b) ^ = −.
Exercise 1.36 Use mathematical induction to show that
^1 ^2 · · · ^ = (^1 +^2 +···+)^ ( = 2 3 )
Exercise 1.37 Using the fact that the modulus |^ − 1 | is the distance between the points ^ and 1 give a geometric argument to find a value of in the interval 0 ≤ 2 that satisfies the equation |^ − 1 |. Answer:
Exercise 1.38 By writing the individual factors on the left in exponential form, performing the needed operations, and finally changing back to rectan- gular coordinates, show that
14 1. Complex Numbers (Exercises)
a)
b) 5 (2 + ) = 1 + 2;
c) (−1 + )^7 = − 8 − 8 = −8(1 + );
d)
Exercise 1.39 Show that if Re 1 0 and Re 2 0 , then
( 1 2 ) = ( 1 ) + ( 2 )
where principal arguments are used.
Exercise 1.40 Let be a nonzero complex number and a negative integer ( = − 1 − 2 ). Also, write = ^ and = − = 1 2 Using the expressions
^ = ^ and −^1 =
μ 1
verify that ()−^1 =
Exercise 1.41 Find the square roots of
a) 2 ;
b) 1 −
and express them in rectangular coordinates. Answer: a)± (1 + ) ; b) ±
√ √^3 − 2
Exercise 1.42 In each case, find all the roots in rectangular coordinates, ex- hibit them as vertices of certain squares, and point out which is the principal root:
a) (−16)^1 ^4 ;
b)
Answer: a) ±
2 (1 − ) b) ±
16 1. Complex Numbers (Exercises)
Get into the habit now of working the Practice Problems before you start the exercises. Probably, you should attempt all the Practice Problems before checking the solutions, because once you start reading the first solution, you might tend to read on through the other solutions and spoil your chance to benefit from those problems.
2.1 Practice Problems
Problem 1 Determine if the following system is consistent:
2 − 4 3 = 8 2 1 − 3 2 + 2 3 = 1 5 1 − 8 2 + 7 3 = 1
Solution: The augmented matrix is ⎡ ⎣
To obtain an 1 in the first equation, interchange rows 1 and 2 : ⎡ ⎣
To eliminate the 5 1 term in the third equation, add − 5 2 times row 1 to row 3 : (^) ⎡
⎣
Next, use the 2 term in the second equation to eliminate the − 1 2 2 term from the third equation. Add 1 2 times row 2 to row 3 : ⎡ ⎣
2.1 Practice Problems 19
Problem 2 State in words the next elementary row operation that should be performed on the system in order to solve it. [More than one answer is possi- ble.]
a) 1 + 2 − 4 3 + 6 4 = 8 2 + 2 3 − 3 4 = 1 7 3 + 4 = 1 3 − − 3 4 = 5
b) 1 + 2 − 4 3 + 6 4 = 8 2 + 2 3 = 1 2 3 = 1 4 = 5 Solution:
a) For “hand computation,” the best choice is to interchange equations 3 and 4. Another possibility is to multiply equation 3 by 1 7. Or, replace equation 4 by its sum with − 1 7 times row 3. (In any case, do not use the 2 in equation 2 to eliminate the 2 in equation 1 .)
b) The system is in triangular form. Further simplification begins with the 4 in the fourth equation. Use the 4 to eliminate all 4 terms above it. The appropriate step now is to add − 6 times equation 4 to equation 1. (After that, move to equation 3 , multiply it by 1 2 , and then use the equation to eliminate the 3 terms above it.)
Problem 3 The augmented matrix of a linear system has been transformed by row operations into the form below. Determine if the system is consistent. ⎡ ⎣
Solution: The system corresponding to the augmented matrix is 1 + 5 2 + 2 3 = − 6 4 2 − 7 3 = 2 5 3 = 0
The third equation makes 3 = 0, which is certainly an allowable value for 3. After eliminating the 3 terms in equations 1 and 2 , you could go on to solve for unique values for 2 and 1. Hence a solution exists, and it is unique (see Figure 2.2).
20 2. Systems of Linear Equations (Exercises)
Fig. 2.2. Each of the equations 2.8 determines a plane in three-dimensional space. The solution lies in all three planes.