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YORK UNIVERSITY Faculty of Science Department of Mathematics and Statistics MATH 1025 3.00 B Test #3 SOLUTIONS November 28, 2018 Surname (print): Student No:

Given Name: Signature:

INSTRUCTIONS: 1. Please write your name, student number and final answers in ink. 2. This is a closed-book, closed-notes test, duration – 50 minutes. 3. CALCULATOR, LAPTOP, INTERNET CONNECTED DEVICES OR OTHER AIDS ARE NOT PERMITTED. 4. There are six questions on seven pages. Answer all questions. Fill in final answers in designated spaces. Your work must justify the answer you give. Show your work on the space provided. If you need more space, use the back of a page and clearly indicate this fact on an original page each time when you use the back of a page for your work. 5. If you use pencil for your solutions, you may not submit your paper for regrading. 6. Remain seated until we collect all the test papers. 7. Do the easiest questions first. GOOD LUCK!

c I. Raguimov Copyright 2018 

Question

Points

1 2 3 4 5 6 Total:

12 8 6 7 4 13 50

Scored

Name:

Student No:

1. (2 + 4 + 6 pts) (a) Given z = 3 − 2i. Find z −1 and express it in standard (Cartesian) form. ANSWER: ANSWER:

z −1 = =

1 z 3 + 2i = = z (3 − 2i)(3 + 2i) zz 3 2 3 + 2i i. = + 32 + 22 13 13

(b) Solve the equation z(2 − i) = (z + 1)(1 + i) for the complex number z. Hint: z is the conjugate of z and (zz) ∈ R. ANSWER: ANSWER: Let z = a + bi. Then z = a − bi, and we need to solve for a and b the equation (a + bi)(2 − i) = [(a − bi) + 1](1 + i). =⇒ 2a − ai + 2bi − bi2 = a + ai − bi − bi2 + 1 + i =⇒ 2a − (a − 2b)i = (a + 1) + (a + 1 − b)i. Hence, 

2a = a+1 −a + 2b = a + 1 − b  a = 1 ⇐⇒ 2a − 3b = −1,

that is, a = b = 1. ∴ z = 1 + i.

Continued... c I. Raguimov Copyright 2018  1

Name:

Student No: 15

(c) Compute (1 + i) . Express your answer in standard (Cartesian) form. ANSWER: ANSWER: √ 1+i = 2 So,

√  √     √ πi √ 2 π π 2 i] = 2e 4 . i = 2[cos + sin + 4 2 4 2

√ √ 15πi √ πi 15πi (1 + i)15 = ( 2e 4 )15 = ( 2)15e 4 = (2)7 2e 4 .

Now,

16π π π 15π = − + 2π(2). =− + 4 4 4 4

Hence, e

15πi 4

− πi 4πi 4

=e

e

0 − πi 4

=e

∴ (1 + i)

15

− πi 4

e =e

√ =2 2 7

= cos



π − 4



+ sin



√ √  π 2 2 i. − − i= 2 2 4

√  √ 2 2 i = 27 − 27 i = 128 − 128i. − 2 2

Continued... c I. Raguimov Copyright 2018  2

Name:

Student No:

2. (8 pts) Find the standard matrix of the transformation T : R2 → R2 if T reflects a vector about the x-axis, then rotates the vector counterclockwise through π6 radians around the ori√ gin. Determine the image of the vector ~u = [ 3 − 1]T under this linear transformation. ANSWER: T = R π6 ◦ Q0 , where Q0 is the reflection about the x-axis and R π6 is the counterclockwise π rotation through the angle θ = around the origin. 6 T (~e1 ) = R π ◦ Q0 (~e1 ) = R π (Q0 (~e1 )) = R π (~e1 ) 6 6 6 

π π sin = R ([1 0] ) = cos 6 6 T

π 6

T

=

√ 3 2

1 2

T

.

T (~e2 ) = R 6π ◦ Q0 (~e2 ) = R π6 (Q0 (~e2 )) √ T  T  3 π 1 π T − . = − cos = R π6 ([0 − 1] ) = sin 2 2 6 6 Therefore, the standard matrix of the transformation T, #    " √3 1  2 √2 . [T ] = T (~e1 )  T (~e2 ) = 1 3 − 2 2

Or equivalently,

cos π6 sin 6π

  π 1 0 − sin 6 [T ] = [R π ][Q0 ] = 6 0 −1 cos π6 " √ " # # √  1 3 3 1 − 1 0 2 2 √2 √2 = = . 1 1 3 3 0 −1 − 2 2 2 2 

Finally, "

√ 3 2 1 2

Therefore, the image of ~u, T (~u) = [1

1 √2 − 23

# √    1 3 √ = . 3 −1

√ T 3] .

Continued... c I. Raguimov Copyright 2018  3

Name:

Student No:

3. (3 + 3 pts) Let T : R3 → R2 be a linear transformation such that       x1   x1 x1 + x2 + x3 , for all  x2  ∈ R3 T  x2  = x1 − x2 − x3 x3 x3 (a) Is T one to one? Answer “YES” or “NO”, and justify your answer.

ANSWER: ANSWER: NO. T is one-to-one, whenever ∀~u ∈ R3 T (~u) = ~0 =⇒ ~u = ~0. Let t 6= 0. Then ~u = [0 − t t]T = 6 ~0, while         0 0−t+t 0   = . = −t T (~u) = T 0 − (−t) − t 0 t

∴ T is not one-to-one.

(b) Is T onto? Answer “YES” or “NO”, and justify your answer. ANSWER: ANSWER: YES, T is onto. The system of linear equations with the augmented coefficient matrix   1 1 1 | a 1 −1 −1 | b is consistent ∀[a b]T ∈ R2 , because    1 1 1 | a 1 0 0 | −→ · · · −→ 1 −1 −1 | b 0 1 1 |

a+b 2 a−b 2



.

Continued... c I. Raguimov Copyright 2018  4

Name:

Student No: 







5 −3 1 3 . and P = 1 1 1 1 Find P −1 AP and determine a formula for Ak , where k is any positive integer greater than 1.

4. (7 pts) Let A =

ANSWER: ANSWER: The quick way to go is to suspect that ~v1 = [1 1]T and ~v2 = [5 2]T (the columns of P ) might be eigenvectors of A. Direct computation shows that               5 −3 5 −3 3 12 3 1 2 1 A~v1 = , =4 = =2 = and A~v2 = 1 4 1 1 1 1 2 1 1 1 which means that ~v1 is an eigenvector associated to the eigenvalue λ1 = 2, while ~v2 is an eigenvector associated to the eigenvalue λ2 = 4. Moreover,   2 0 −1 P AP = D = . 0 4 Otherwise, compute first det P = 1 · 1 − 3 · 1 = −2 6= 0, hence P is invertible, and     1 1 −3 1 −1 3 −1 P =− = . 2 −1 1 2 1 −1 Then P

−1

        1 −1 3 2 0 2 0 1 3 5 −3 = = D. = AP = 0 22 0 4 1 1 2 1 −1 1 1

Therefore, k

k

A = PD P

−1

          1 3 2k 0 1 1 −1 3 1 3 2k 0 −1 3 = = 1 −1 1 −1 1 1 0 22k 1 1 0 22k 2 2   k k k−1 3 · 2 − 1 3(1 − 2 ) . =2 2k − 1 3 − 2k

for all k ∈ Z, k ≥ 2.

Continues... c I. Raguimov Copyright 2018  5

Name:

Student No:

5. (4 pts) Let  4 0 1 A = −2 1 0 , −2 0 1 

  −1  ~v = 1  . 1

Is ~v an eigenvector of A? Answer “YES” or “NO”, then justify your answer. If ~v is an eigenvector of A, find the corresponding eigenvalue. ANSWER:

ANSWER: We have



      −1 −3 −1 4 0 1 A~v = −2 1 0  1  =  3  = 3  1  1 3 1 −2 0 1

Therefore, YES, ~v = [−1 1 1]T is an eigenvector of the matrix A and the corresponding eigenvalue λ = 3.

 7 0 −4 0 . 6. (4 + 6 + 3 pts) Let A =  0 5 5 0 −2 

(a) Find the characteristic polynomial and eigenvalues of A. ANSWER: 

 λ−7 0 4 0 λ−5 0  = (λ − 5)[(λ − 7)(λ + 2) + 20] CA (λ) = det(λI3 − A) = det  −5 0 λ+2 = (λ − 5)(λ2 − 5λ + 6) = (λ − 5)(λ − 3)(λ − 2).

So, eigenvalues of A are: λ1 = 2, λ2 = 3 and λ3 = 5.

Continued... 6

Name:

Student No:

(b) Find basic eigenvectors of A corresponding to the eigenvalues from part(a). ANSWER: For λ1 = 2, the general solution of the linear homogeneous system (2I3 − A)~v = ~0 will be  4  5

~v = t  0  1

(∀t ∈ R).

For simplicity take t = 5 to get the basic eigenvector corresponding to λ1 = 2 as   4 ~p1 =  0  . 5

For λ2 = 3, the general solution of the corresponding homogeneous system will be   1 ~v = t  0  (∀t ∈ R). 1

So, the basic eigenvector corresponding to λ1 = 3 will be   1 ~p2 =  0  . 1

Similarly, for λ3 = 5, the general solution of the corresponding system will be   0 ~v = t  1  (∀t ∈ R). 0

Hence, the basic eigenvector corresponding to λ3 = 5 will be   0 ~p3 =  1  . 0

(c) Determine an invertible matrix P and a diagonal matrix D such that P −1 AP = D. ANSWER: P = [~p1 | ~p2



   4 1 0 2 0 0 | p~3 ] =  0 0 1  and P −1 AP = D = diag (2, 3, 5) =  0 3 0  . 5 1 0 0 0 5 END.

c I. Raguimov Copyright 2018  7...