MTH 223 Problem bank - KEY

1. Coordinate systems. Operations with vectors and points.

Local and global coordinates

x’ = 3x +2 y’ = 2y + 1

Aspect Ratio

a) 32’’ x24‘’

b) 640√337

337 ’’ x

360√337

337 ’’

Geometrically meaningful operations

a) Yes

b) Yes

c) Yes

d) No

e) Yes (result is a vector)

f) Yes

g) No

h) Yes, it is always a midpoint between the two given points

Geometrically meaningful operations

a) No

b) Yes

c) No

d) Yes

e) Yes (result is a point B + (A-C) + (B-C))

f) Yes

It is the midpoint of line segment AB .

Vector addition

a) 0⃗

b) −2𝑆𝐶⃗⃗⃗⃗

c) −𝑆𝐶⃗⃗⃗⃗

d) 2𝑆𝐶⃗⃗ ⃗⃗

e) 0⃗

Vector addition

a) 0⃗

b) 2𝑆𝐴⃗⃗⃗⃗ ⃗

c) 𝑆𝐷⃗⃗⃗⃗ ⃗

d) 4𝑆𝐴⃗⃗⃗⃗ ⃗

e) 0⃗

f) 0⃗

g) 2𝑆𝐵⃗⃗⃗⃗ ⃗

h) 𝑆𝐵⃗⃗⃗⃗ ⃗

Vector addition

2. Vector magnitude. Dot product.

Magnitudes and angles

Find the lengths of sides and interior angles of a triangle ABC if

a) √5, √5, √10, 45°, 45°, 90°

b) 3√2, 3, 3, 45°, 45°, 90°

c) 3√2, 3√2, 3√2, 60°, 60°, 60°

𝑐 = (7,2)

Angle of two vectors

Find angles formed by these vectors:

a) 45°

b) 60°

c) 90°

Angle of two vectors

a) cos 𝜃 =4

5 , 𝜃 ≈ 36.87°

b) cos 𝜃 =2√5

5 , 𝜃 ≈ 26.57°

c) D = (-1,1,1) , cos𝜃 = −1

2 , 𝜃 = 120°

Perpendicular (orthogonal) vectors

Find a (non-zero) perpendicular vector to given vectors and explain why they are perpendicular:

a) (-4,-3) or (4,3)

b) for example (2, 1, 0)

c) If we disregard scaling, there are two perpendicular “directions” in the plane. There are infinitely

many in the space.

Hint: Expand (�⃗⃗� + �⃗� )2 using properties of vector algebra. Answer: 2 + √3

a) (2,1)

b) (1, -2) (or (-1, 2))

c) (1, -2) (or (-1, 2))

d) (2,1) (or (-2, -1))

e) X = (4, 2) + t(-1, 2) (one example)

f) (3,4)

g) -x + 2y - 5 = 0

h) When entering line equation in parametric form into Geogebra, make sure to insert a

multiplication sign (*) or space between the parameter and vector: (1,3) + t*(2,1)

a) x=3-2t, y=-7+5t, tR

b) x=5, y=2t, tR

c) x=4t, y=0, tR

d) For example: x=4-2t, y=3t, tR

e) For example: x=-3+4t, y=2, t ≥ 0

f) For example: x=3t, y=-5+2t, t[0,1]

g) For example x=2+t,y=-5+3t, tR

h) For example x=-3, y=2t, tR

a) x=-2+t, y=-3, tR

b) x = -2, y=3+t, tR

c) x=-2+t, y=3+t, tR

d) -y-3 =0, x+2 = 0, x+y-5=0

e) Line parallel to y axis does not satisfy the definition of a function (review your Calc knowledge

for explanation)

There are two such points: D1=(8,9) and D2=(-10,-15).

a) x=1+2t, y=2+3t, t≥0 b) (31, 47) c) 3 sec

a) x=5-6t, y=3+4t, z=6-8t, tR

b) No

c) x=5-10s, y=3-2s, z=6-5s, tR

a) x+2y+5=0

b) 12x+5y+11=0

c) x-6=0

d) 3x-4y-40=0

e) x+5=0

x+2y-10=0

7x-3y-6=0

(3,5), (-1,-2), (-3,2)

a) (-7,16) b) (0,7,-11)

a) (-4,0)

b) Yes, because the parameter t for the intersection point is ¾, which is in the interval [0,1].

Distance of a point from a line

a) 8

b) 3

c) 2.8

d) 0

e) 1.4

Distance of a point from a line

Hint: There is several ways to show that these lines are parallel. One fast method is by looking at their

normal vectors. When finding the distance of the two lines, pick any point on one of them and calculate

its distance to the other line. Answer: √13

2

Distance of a point from a line

√10

Linear combinations

a) 4.5 𝑎 + 2.5 �⃗�

b) 11.5 𝑎 − 4.5 �⃗�

c) not possible, vectors 𝑎 𝑎𝑛𝑑 �⃗� are linearly dependent

d) −8.8 𝑎 − 7.2 �⃗�

e) 3 𝑎 − 5 �⃗� + 3 𝑐

Normalized vector

a) -

b) [

4

53

5

] c) [

2

5√5

1

5√5

]

a) √2

2𝑥 +

√2

2𝑦 −

1

18√2 = 0 b)

2

5√5 𝑥 −

1

5√5 𝑦 +

4

5√5 = 0

*1

Exercise 2.2.11

e) [

√3

2−

1

2

1

2

√3

2

] f) [−

√2

2

√2

2

−√2

2−

√2

2

]

*

Exercise 2.2.12

1 Solutions to some of the problems marked with asterisk can be found at the end of the 2019 edition of open textbook “Linear Algebra with Applications” by W. Keith Nicholson. Hints are provided in student solution manual. The textbook and student solution manual are on my website.

*

Exercise 2.2.13

g) [0 −1 0 10

0 00 1

]

Orthogonal Projection

[53]

Orthogonal projection - applications

15√2

2

Orthogonal Projection

Exercise 4.2.10 on page240

Orthogonal Decomposition

Exercise 4.2.11 on page 240

Matrix operations*

Exercise 2.1.1 through Exercise 2.1.4 on pages 44, 45. Careful, there is a typo in the answer for 2.1.4 b) in

2019 Edition of the book.

Matrix multiplication*

Exercise 2.3.1 through Exercise 2.3.5 on pages 76, 77.

a) No, it does not have to be a zero matrix. Find a non-zero one.

b) Hint: you may want to start with a matrix A that has all 1s. Try to figure out what entries must

be in the matrix B for it to work. (A possible option for B: [1 1

−1 −1] Find another pair.)

Matrix multiplication – applications*

Exercise 2.3.25

Exercise 2.6.12 on page 116

Determinants

Exercises 3.1.1 through 3.1.3. For 3.1.3 keep in mind that only square matrices have determinants.

The value of the determinant will be multiplied by the same number.

Determinants

a) 5 units squared b) 5 units cubed

a) 2, full rank

b) 1, deficient rank

c) 2; full row rank (but

not full column rank)

a) Write the system in the most general form (system with 2 variables will do) and use the

elimination of variables method to show where the determinants |A| and |A2| are coming

from.

b) The system has a unique solution only if |A| is different from 0. Otherwise there are no

solutions or infinitely many solutions.

c) (5, -4) d) (0,2) e) a ≠ 0; (4/a ; 1)

a) Point (5,6,10)

b) Point (-1,0,1)

c) Line (5k, -11k, -7k) or (0,0,0) + (5, -11, -7)t

d) Line (2

3+

5

3𝑡,

5

3−

7

3𝑡, 𝑡) or (

2

3,5

3, 0) +

(5

3𝑡, −

7

3𝑡, 𝑡)

e) Line (-t, 13t, 5t) or (0,0,0) + (-1, 13, 5)t

f) Line (t, 7 – 3t, 18 – 7t) or (0,7,18) + (1, -3, -7)t

g) no solution

h) no solution

i) Point (2, 4 6, 8)

j) Point (10,8,6,4,2)

k) Plane 2x – 4y – z = 2; (1 + 2t + 0.5s, t, s)

or (1,0,0) + (2,1,0)t + (1/2, 0, 1)s

Linear systems – Elimination **

a) consistent, not unique

b) consistent, unique

c) consistent, not unique

d) depends on * in the last row. it can be

inconsistent (when?) or consistent but

not unique (when?)

Linear systems

a) Can have infinitely many solutions. Ambiguity is in the fact that we don’t know what equations

are in the system. If one of the equations is in the form 0y = 5 then even such system will not

have a solution. If no equations are in such form, then the system will have infinitely many

solutions. (provide an example of that)

b) It will have a solution only if some of the equations are linearly dependent. Otherwise there will

be no solution. (provide an example for both cases).

a) 4 b) any c) any

a) If h 2 there will be a unique solution for any k. If h= 2 and k 4, there are no solutions. If h= 2

and k = 4, then there are infinitely many solutions.

b) If h 4, then there is exactly one solution. If h = 4 and k 4, then there are no solutions. If h = 4

and k = 4, then there are infinitely many solutions.

200, 90, 60, 50

Keep in mind that for many of the problems, your solution may be different (yet correct) depending on

how you chose the parameters or which points you pick for generating vectors etc.

a) x = 4 + 2t – 1/3s, y = t, z =s, or (4, 0 ,0) + (2, 1, 0)t + (-1/3, 0, 1)s

b) 3x + 6y – 5z – 2 = 0

c) x=1 + t – 3s , y = 3 – 8s, z = -1 + 4t – 6s ; t, s R

d) These points are colinear, they determine a line, not a plane

e) x + 4y – 2z = 2 (note that the vector N – M must be the normal vector)

f) Yes, the coordinates of L satisfy the parametric equation of the plane ABC: x = 1 – 3t – 4s, y = 3t

– 2s, z = 3 – 3t + s ; t, s R

g) 2x + y = 0 or parametric equation X = (0,0,0) + (0,0,1)t + (2,-4,3)s

Inverse matrix**

a) - d)

e) – g)

Inverse Matrix – Solving linear systems**

Matrix Rank

a) 2 b) 2 c) 3

Linear dependence

a) Independent; Rank 3 - three independent vectors

b) Dependent; Rank 2 - two independent vectors

c) Independent; Rank 4 - four independent vectors

d) Dependent; Rank 3- three independent vectors

e) Dependent; Rank 2 – two independent vectors.

f) In each case they are dependent.

Matrix rank

Rank of a matrix is the number of linearly independent rows (or columns, it is the same number). mxn

can have at most min(m,n) linearly independent rows/colums, hence its rank cannot be higher than min

(m,n).

a) Solution will be a plane (two free variables).

b) Cannot happen (why?)

c) It is impossible, rank of A cannot be greater than 2; rank of the augmented matrix cannot be

greater than 3 (why?)

d) Rank of the augmented matrix is larger that the rank of the coefficient matrix. One row must be

in the form of “0 0 0 0 nonzero#”

e) Unique solution.

Basis of a vectors space

a) No

b) Yes

c) No

d) Yes

e) No

f) Yes

M’M-1 = A

a) [−1 02 −1

]

b)

Change of basis

a) [1 1

−1 2]

b) [1

−4]

c) [1 1

−1 2]−1

= [

2

3

−1

31

3

1

3

]

d) [−30

]

Eigen values and vectors

a) 5, -1; (1, -1), (1,1) b) 4, -1; (1, -3), (1, 2) c) 2 ; (1,0)

d) 4, 2, -1; (2,0,1),

(5,1,0), (3,0,-1)