Dr. JaffeAdvanced Topics in MathematicsHomework due Tuesday 11/10

NAME: ___________________________________________________

1. Let A = [121101

143

367]

a) What size is A?3 x 4 (rows x columns)b) What is the third column of A?

[143 ]c) What is the second row of A?[2 0 4 6]d) What element of A is in the (3,2)th position?row 3, column 2 =1

2. Find A + B where

a) A = [ 1 0 4−1 2 20 −2 −3] B = [−1 3 5

2 2 −32 −3 0 ]

A + B = [0 3 91 4 −12 −5 −3]

b) A = [−1−4

0−3

55

6−2] B = [−3

09

−2−3−1

42]

A = [−4−4

9−5

24

100 ]

3. Let A be an m x n matrix and let 0 be the m x n matrix that has all entries equal to zero. Show that A = A + 0.

A = [ab cd

ef

gh]

[ab cd

ef

gh]+[000

000

00]=[ab c

def

gh ]

4. Show that matrix addition is commutative; that is, show that if A and B are both m x n matrices, then A + B = B + A.

A = [a bc d ]

B = [ e fg h ]

A + B = [a+e b+fc+g d+h ]

= B + A = [e+a f +bg+c h+d ]

This is because addition is commutative so a + e = e + a

b + f = f + b; c + g = g + c; d + h = h + d

Dr. JaffeAdvanced Topics in MathematicsHomework due Thursday 11/12

NAME: _________________________________________________________

1. Find the product AB, where

a) A = [ 1 0 10 −1 −1

−1 1 0 ] B = [ 0 1 −11 −1 0

−1 0 1 ]AB = [1+0+−1 1+0+0 −1+0+1

0+−1+1 0+1+0 0+0+−10+1+0 −1+−1+0 1+0+0 ]=[0 1 0

0 1 −11 −2 1 ]

b) A = [1 −3 01 2 22 1 −1] B = [ 1

−1−3

−10

−2

230

3−12 ]

AB = [ 1+3+−31+−2+−62+−1+3

−1+0+0−1+0+−4−2+0+2

2+−9+02+6+02+3+0

3+3+03+−2+4

6+−1+−2]=[ 1−74

−1−50

−785

653 ]

c) A = [ 0 −17 2

−4 −3 ] B = [ 4 −1−2 0

2 33 4

01 ]

AB = [ 0+2 0+0 0+−3 0+−4 0+−128+−4 −7+0 14+6 21+8 0+2−16+6 4+0 −8+−9 −12+−12 0+−3 ]=[ 2 0 −3 −4 −0

24 −7 20 29 2−10 4 −17 −24 −3 ]

2. Find a matrix A such that

[2 31 4] A=[3 0

1 2]Hint: You may need to solve a system of linear equations

Let A = [a bc d ]

2a + 3c = 3 2b + 3d = 01a + 4c = 1 1b + 4d = 2

2a + 3c = 3 2b + 3d = 02 *(1a + 4c = 1) 2 * (1b + 4d = 2)

2a + 3c = 3 2b + 3d = 0 Subtract these equations2a + 8c = 2 2b + 8d = 4 -5c = 1 -5d = -4

c = -1/5 d = 4/5Plug this back into the original equationsa + 4(-1/5) = 1 b + 4(4/5) = 2a - 4/5 = 1 b + 16/5 = 2a = 9/5 b = -6/5

3. Let A be a 3x4 matrix, B be a 4x5 matrix, and C be a 4x4 matrix. Determine which of the following products are defined and find the size of those that are defined:

a) AB(3 x4)(4x5) = (3x5)

b) BA(4x5)(3x4) = undefined

c) AC(3x4)(4x4) = 3x4

d) CA(4x4)(3x4) = undefined

e) BC(4x5)(4x4) = undefined

f) CB(4x4)(4x5) = 4x5

4. The nxn matrix A = [aij] is called a diagonal matrix if aij = 0 when i ≠ j. Show that the product of two nxn diagonal matrices is again a diagonal matrix. Give a simple rule for determining this product.

[a 00 b ] [c 0

0 d ]=[ac 00 bd ]

When you multiply two diagonal matrices, the product is a diagonal matrix in which the diagonal is the product of the corresponding diagonal elements.ABij = aij*bij

Dr. JaffeAdvanced Topics in MathematicsHomework due Monday 11/16

NAME: _________________________________________________________

1. If matrix B is the square matrix A after switching the columns and rows, what is the difference between AB and BA.

A = [a bc d ] , B = [

a cb d ]

B = [a cb d ] , A = [

a bc d ]

AB = [ a2+b2 ac+bdac+bd c2+d2 ]

BA = [ a2+c2 ab+cdab+cd b2+d2 ]

AB - BA = [ b2−c2 ac+bd−ab−cdac+bd−ab−cd c2−b2 ]

2. Give an example for which AB = BA, but neither A nor B is an identity matrix.

A = [1 22 1 ] B = [

1 22 1 ]

AB = [5 43 5 ] BA = [

5 43 5 ]

3. Prove AI = A, where I = identity matrix

[a bc d ][1 0

0 1 ]=[a bc d ]

Dr. JaffeAdvanced Topics in MathematicsHomework due Tuesday 11/17

NAME: ___________________________________________________________

1. Let A = [121101

143

367]. Find At.

At = [1 2 11 0 11 4 33 6 7 ]

2. What do we know about the sizes of the matrices A and B if AB and BA are both defined? Explain.

A and B must be the same size and square if both A and B are defined.

3. Let A and B be two n x n matrices. Show that

a) (A + B)t = At + Bt

A = [a bc d ] B = [

e fg h ]

At = [a cb d ] Bt =

[e gf h ] At + Bt = [

a+e c+gb+ f d+h ]

A + B = [a+e b+fc+g d+h ] (A + B)t =

[a+e c+gb+ f d+h ]

(A + B)t = At + Bt

b) (AB)t = At Bt

AB = [ae+bg af +bhce+dg cf +dh ]

(AB)t = [ae+bg ce+dgaf +bh cf +dh ]

AtBt = [ae+cf ag+chbe+df bg+dh ]

(AB)t ≠ At Bt

Dr. JaffeAdvanced Topics in MathHomework due Monday 11/23

NAME: ____________________________________________________

1. Show that [ 2 3 −11 2 1

−1 −1 3 ] is the inverse of [ 7 −8 5−4 5 −31 −1 1 ]

[ 7 −8 5−4 5 −31 −1 1 ] [ 2 3 −1

1 2 1−1 −1 3 ] =

[14+−8+−5 21−16−5 −7+−8+15−8+5+3 −12+10+3 4+5+−9

2+−1+−1 3+−2+−1 −1+−1+3 ]=[1 0 00 1 00 0 1 ]

2. Let A be a 2x2 matrix with A = [a bc d ]. Show that if ad – bc ≠ 0, then A-1 = [ d

ad−bc−b

ad−bc−c

ad−bca

ad−bc ]AA-1 = I

[a bc d ][ d

ad−bc−b

ad−bc−c

ad−bca

ad−bc ]=[ ad−bcad−bc

−ab+abad−bc

cd−cdad−bc

−bc+adad−bc ]=[1 0

0 1 ]

3. Let A = [−1 21 3]

a) Find A-1 (Hint: Use the previous problem to help you)

a = -1 b = 2 c = 1 d = 3

From the previous problem, we know A-1 = [ dad−bc

−bad−bc

−cad−bc

aad−bc ]

A-1 = [ 3−3−2

−2−3−2

−1−3−2

−1−3−2 ]=[−3 /5 2 /5

1/5 1 /5 ]

b) Find A3

A2 = AA = [−1 21 3] [−1 2

1 3] = [3 42 11 ]

A3 = A2A = [3 42 11 ][−1 2

1 3 ]=[1 189 37 ]

c) Find (A-1)3

(A-1)2 = [−3/5 2/51/5 1/5 ][−3/5 2/5

1/5 1/5 ]=[11/25 −4/25−2/25 3 /25 ]

(A-1)3 = (A-1)2A-1 = [11/25 −4/25−2/25 3 /25 ][−3/5 2/5

1/5 1/5 ] = [−37/125 18/125

9/125 −1/125 ]d) Use your answers to parts b and c to show that (A-1)3 is the inverse of A3

A3(A-1)3 = I

[1 189 37 ][−37/125 18/125

9/125 −1/125 ] = [(−37+162 )/125 (18−18 )/125(−333+333 )/125 (162−37 )/125 ]=[1 0

0 1 ]

4. Let A = [1 10 1] and B = [0 1

1 0]a) Find A ˅ B

[1 10 1]˅ [0 1

1 0] = [1∨0 1∨10∨1 1∨1 ]=[1 1

1 1 ]b) Find A ˄ B

[1 10 1]˄[0 1

1 0] = [1∧0 1∧10∧1 1∧0 ]=[0 1

0 0 ]

c) Find A ʘ B

[1 10 1] [0 1

1 0 ] = [(1∧0)∨(1∧1 ) (1∧1)∨(1∧0 )(0∧0 )∨(1∧1 ) (0∧1)∨(1∧0 ) ]=[0∨1 1∨0

0∨1 0∨0 ][1 11 0 ]

5. Find the Boolean product of A and B, where

A = [1 00 11 1

0 10 11 1] and B = [1 0

0 11 11 0 ]

A ʘ B =

[1∨0∨0∨1 0∨0∨0∨00∨0∨0∨1 0∨1∨0∨01∨0∨1∨1 0∨1∨1∨0 ]=[1 0

1 11 1 ]

Dr. JaffeAdvanced Topics in MathematicsHomework due Tuesday 11/24

NAME: _________________________________________________________

1. Draw the vectors v=[ 4

1 ] and

w=[−22 ]

and v + w and v - w in a single xy plane

You are drawing (4,1) and (-2,2), then v + w = (2,3) and v - w = (6, -1)

2. If v+w=[31 ]

and v−w=[13 ]

, compute and draw v and w.

We can do this using a system of equations:

v = [ xy ]

w = [wz ]

x + w = 3 y + z = 1 We can add the two equationsx - w = 1 y - z = 32x = 4 2y = 4x = 2 y = 2

Now we plug back in2 - w = 1 2 - z = 3

w = 1 z = -1

v = [22]

w = [ 1−1]

3. From v=[21 ]

and w=[12 ]

, find the components of 3v + w and v - 3w and cv + dw

3v + w = 3 [21] + [

12] = [

63 ]+ [

12] = [

75 ]

v - 3w = [21] - 3 [

12] = [

21] - [

36 ] = [

−1−1]

cv + dw = [2 c+dc+2d ]

4. Compute u + v, u + v + w, and 2u + 2v + w when

u=[123 ],

v=[−31

−2 ],

w=[ 2−3−1]

u + v = [123 ]

+[−3

1−2 ]

= [−2

31 ]

u + v + w = [123 ]

+[−3

1−2 ]

+[ 2−3−1 ]

= [000 ]

2u + 2v + w = 2[123 ]

+ 2[−3

1−2 ]

+ [ 2−3−1 ]

= [246 ]

+[−6

2−4 ]

+[ 2−3−1 ]

= [−2

31 ]

5. What combination of the vectors [12]

and [31 ]

produces [14

8 ]? Show how you found your answer.

Hint: use two equations for the coefficients c and d in a linear combination.

c + 3d = 14 2(c + 3d = 14) 2c + 6d = 282c + d = 8 2c + d = 8 2c + d = 8

5d = 20

d = 4Plug back in to get c: c + 3(4) = 14 c + 12 = 14 c = 2c = 2 and d = 4

Dr. JaffeAdvanced Topics in MathematicsHomework due Wednesday 11/25

NAME: ________________________________________________

1. What is a vector?

A vector has magnitude and direction

2. How do we represent a vector with a matrix?

v = [ xy ]

3. How can we use matrices to add vectors?

We add each of the corresponding components of the vectors.

4. What does the size of the matrix tell us about the size of the vector?

The size of the matrix tells us how many dimensions we are working in (2 or 3)

5. Consider the system2x – 5y = 10x + 6y = 13

a) Represent the system as a set of matrices

[2 −51 6 ][ x

y ]=[1013 ]

b) How would we solve this system using elimination?

We can multiply the second equation by and then subtract the two equations.

6. Consider the system:2x + 4y – 2z = 24x + 9y – 3z = 8-2x – 3y + 7z = 10

Represent the system as a set of matrices

[ 2 4 −24 9 −3

−2 −3 7 ] [ xyz ]=[ 2

810]

Dr. JaffeAdvanced Topics in MathematicsClass work Wednesday 11/25

NAME: _______________________________________

Given the following system of equations:

2w + 7x - 3y + z = 4-3w - 2x + 4y - 2z = 196w + 3x + 2y - 4z = -13-4w - 6x - 3y + 5z = -9

Solve the system using elimination. For each step, put your equations into a matrix. Do you notice anything about your matrix?

w = -5x = 3y = 2z = -1

Dr. JaffeAdvanced Topics in MathematicsHomework due Tuesday 12/1

NAME: _________________________________________________

1. Graph the vectors as specified

a) v = [67 ]

w = [ 3−2]

b) v = [ 4−5]

w = [−2−7 ]

c) v = [81 ]

w = [ 310]

2. On the same graph, show the resultant of the vectors from Question 1 - rewritten below:

a) [67 ]

+ [ 3−2]

= [95 ]

b) [ 4−5]

+ [−2−7 ]

= [ 2−12]

c) [81 ]

+ [ 310]

= [1111 ]

3. On a new graph, graph the difference of the two vectors from Question 1 - rewritten below:

a) [67 ]

- [ 3−2]

= [39 ]

b) [ 4−5]

- [−2−7 ]

= [62 ]

c) [81 ]

- [ 310]

= [ 5−9]

4. What is a resultant vector?The sum of two vectorsDr. JaffeAdvanced Topics in MathematicsHomework due Wednesday 12/2

NAME: ___________________________________________________

1. What multiple l of equation 1 should be subtracted from equation 2?2x + 3y = 1

10x + 9y = 11

l = 5. This is how we can make the values of x the same.

2. a) For which numbers a does elimination break down (1) permanently (2) temporarily?

ax + 3y = -34x + 6y = 6

If a = 2, it will break down permanently2x + 3y = -34x + 6y = 6

[2 34 6|−3

6 ]→[2 30 0|−3

12 ]If a = 0, it will break down temporarily because we would need to do a row switch:

0x + 3y = -34x + 6y = 6

[0 34 6|−3

6 ]→[ 4 60 3| 6

−3 ]b) Fix the temporary breakdown with a row exchange, and solve for x and y.

[0 34 6|−3

6 ]→[ 4 60 3| 6

−3 ]3y = -3 4x + 6y = 6y = -1 4x – 6 = 6

4x = 12x = 12

3. Choose a coefficient b that makes this system singular. Then choose a right side g that makes this system solvable.

2x + by = 134x + 8y = g

If b = 4, then the system will be singular because when you subtract, your second pivot is really zero.If b = 4, then they system will have infinitely many solutions when g = 26 because after elimination, you get 0y = 0

4. Choose a right side which gives no solution and another right side which gives infinitely many solutions.3x + 2y = 76x + 4y =

The system will give infinitely many solutions if the right side is 14 because after elimination, you will get 0y = 0The system will give no solution if the right side is any other number, such as 20 because then you get 0y=6.

5. Reduce this system to upper triangular form:2x + 3y + z = 14x + 7y + 5z = 7

-2y + 2z = 6

[2 3 14 7 50 −2 2|

176 ]→[2 3 1

0 1 30 −2 2|

156 ]→[2 3 1

0 1 30 0 8|

15

−4 ]6. Apply elimination and back substitution to solve

2x - 3y = 34x - 5y + z = 72x - y - 3z = 5

[2 −3 04 −5 12 −1 −3|

375 ]→[2 −3 0

0 1 12 −1 −3|

315 ]→[2 −3 0

0 1 10 2 −3|

312 ]→ [2 −3 0

0 1 10 0 −7|

310 ]

-7z = 0 y – z = 1 2x – 3y = 3z = 0 y – 0 = 1 2x – 3 = 3

y = 1 2x = 6x = 3

(x, y, z) = (3, 1, 0)

7. Apply elimination and back substitution to solvex + 4y - 2z = 1x + 7y - 6z = 63y + 11z = 6

[1 4 −21 7 −60 3 11 |166 ]→[1 4 −2

0 3 −40 3 11 |156 ]→[1 4 −2

0 3 −40 0 15 |151 ]

15z = 1 3y – 4z = 5 x + 4y – 2z = 1z = .067 3y – .27 = 5 x + 4(1.76) – 2(.067) = 1

3y = 5.27 x = -5.906y = 1.76

Dr. JaffeAdvanced Topics in MathematicsClasswork / Homework due Thursday 12/3

NAME: _________________________________________

1.[2 61 5 ]+[ 8 −5

10 −2]=[10 111 3 ]

2. [ 1 2−3 7 ]−[ 8 15

−4 −1] = [−7 −131 8 ]

3. [3 5 24 6 1][1 −1 6 1

2 9 5 −31 6 2 3 ] = [3+10+2 −3+45+12

4+12+4 −4+54+3618+25+4 3−15+624+30+2 4−18+3]=¿

[15 5420 86

47 −656 −11]

4. [3 12 6] [ 4 5 1 −3

−2 1 9 10 ] = [12−2 15+18−12 10+6

3+9 −9+102+54 −6+60 ] = [ 10 16

−4 1612 156 54]

5. [1 1 0 11 0 0 10 1 0 1]∧[0 1 1 0

0 1 0 11 1 0 1] = [0 1 0 0

0 0 0 10 1 0 1]

6.[1 0 1 0 10 1 0 1 01 0 1 1 0]Θ [1 0 1

0 1 01 1 00 1 11 0 1

] =

[1⋁0⋁ 1⋁ 0⋁ 1 0⋁0⋁1⋁ 0⋁ 0 1⋁ 0⋁ 0⋁0⋁ 10⋁ 0⋁ 0⋁ 0⋁0 0⋁1⋁0⋁ 1⋁ 0 0⋁ 0⋁0⋁1⋁ 01⋁ 0⋁1⋁0⋁ 0 0⋁0⋁ 1⋁ 1⋁ 0 1⋁ 0⋁ 0⋁1⋁0 ]=[1 1 1

0 1 11 1 1]

7. Find At if A = [ 3 1 0 710 2 3 −12 9 −5 0

−4 13 8 5 ]=[ 3 10 2 −41 2 9 132 9 −5 0

−4 13 8 5 ]8. Create a matrix A so that A = AT

Start with row 1 and column 1, then do row 2 and column 2 the same way. Make sure they are identical

9. How can we determine if two matrices are inverses of each other?You multiply them and see if you get the identity matrix.

10. Solve the following system:2x + 4y – 2z = 24x + 9y – 3z = 8

-2x – 3y + 7z = 10

[ 2 4 −24 9 −3

−2 −3 7 | 28

10 ]→[ 2 4 −20 1 1

−2 −3 7 | 24

10 ]→[2 4 −20 1 10 1 5 | 2

412]→[2 4 −2

0 1 10 0 4 |248 ]

4z = 8 y + z = 4 2x + 4y – 2z = 2 (x, y, z) = (-1, 2, 2)z = 2 y + 2 = 4 2x + 8 – 4 = 2

y = 2 2x + 4 = 22x = -2x = -1

Dr. JaffeAdvanced Topics in MathematicsHomework due Monday 12/7

NAME: ___________________________________________________

1. Find the pivots and the solution for these four equations:

2x + y = 0x + 2y + z = 0y + 2z + t = 0

z + 2t = 5

Switching the first two equations may make the elimination a little easier.

[1 2 1 02 1 0 00 1 2 10 0 1 2

|

0005 ]→[1 2 1 0

0 −3 −2 00 1 2 10 0 1 2

|

0005 ]→[1 2 1 0

0 −3 −2 00 1 2 10 0 1 2

|

0005 ]→[1 2 1 0

0 −3 −2 00 0 4/3 10 0 1 2

|

0005 ]→

[1 2 1 00 −3 −2 00 0 4 /3 10 0 0 5/4

|

0005 ]

5/4t = 5 4/3z + t = 0 -3y – 2z = 0 x + 2y + z = 0t = 4 4/3z + 4 = 0 -3y + 6 = 0 x + 4 - 3 = 0

4/3z = -4 -3y = -6 x + 1 = 0z = -3 y = 2 x = -1

(x, y, z, t) = (-1, 1, -1, 4)

2. Find a number, a, such that elimination will fail to give three pivots?

[a 2 3a a 4a a a]

a = 2

[2 2 32 2 42 2 2 ]→[2 2 3

0 0 12 2 2]→[2 2 3

0 0 10 0 −1]

3. Which number d forces a row exchange, and what is the triangular system for that d?2x + 5y + z = 04x + dy + z = 2y - z = 3

If d = 10 then the matrix becomes [2 5 14 11 10 1 −1|

023 ]→[2 5 1

0 0 −10 1 −1|

023 ] and you have to do a

row switch between the second and third rows. Then you can do backwards substitution.

[2 5 10 1 −10 0 −1|

032 ]

Which number d makes this system singular (no third pivot)?d = 11

[2 5 14 11 10 1 −1|

023 ]→[2 5 1

0 1 −10 1 −1|

023 ]→[2 5 1

0 1 −10 0 0 |021 ]

4. Solve the system:x + 2y + 3z = 62x + 5y + 2z = 46x - 3y + z = 2

[1 2 32 5 26 −3 1|

642 ]→[1 2 3

0 1 −46 −3 1 | 6

−82 ]→[1 2 3

0 1 −40 −15 −17|

6−8−34 ]→[1 2 3

0 1 −40 0 −77|

6−8

−154 ]-77z = -154 y – 4z = -8 x + 2y + 3z = 6 (x, y, z) = (0, 0, 2)z = 2 y – 8 = -8 x + 0 + 6 = 6

y = 0 x = 05. How do meet two matrices?

Aaaah! You were right and I got confused MEET is AND and JOIN is OR

We join two matrices by using the AND function1 AND 0 = 01 AND 1 = 10 AND 0 = 0

6. How do we join two matrices?We meet two matrices by using the OR function

1 OR 0 = 11 OR 1 = 10 OR 0 = 0

7. What is a linear combination?A linear combination is the form cv + dw = z, where v, w, and z are vectors and c and d are constants