MAT1503
2023
ASSIGNMENT 4
DUE DATE: 20 JULY 2023
, ASSIGNMENT 04
Due date: Thursday, 20 July 2023
Total Marks: 100
ONLY FOR YEAR MODULE
This assignment covers chapter 2 & 3 of the prescribed book as well as the
study guide, it is based on Study Units 2.1, 2.2, 2.3 & 3.1
Question 1:
Given a square matrix A:
(1.1) Performing the operation (3):
R2 ← R2 - 2R1
E1 =
[1 0 0]
[0 1 -2]
[0 0 1]
Performing E1A:
B = E1A =
[2 -1 1]
[1 3 -3]
[1 -3 k + 9]
To find values of k such that a33 = a233, we set them equal:
k + 14 = k + 9
5=0
Since there is no value of k that satisfies the equation, there are no values of k in the
range [-1, 0] that make a33 = a233.
(1.2) Performing the operation (3):
R1 ←→ R3
E2 =
[0 0 1]
[0 1 0]
[1 0 0]
Performing E2A:
C = E2A =
[1 -3 k + 14]
[3 1 -1]
[2 -1 1]
, Therefore, the matrices E2 and C such that A = E2C are:
E2 =
[0 0 1]
[0 1 0]
[1 0 0]
C=
[1 -3 k + 14]
[3 1 -1]
[2 -1 1]
Question 2:
(2.1) ⃗u = < 1, 3, -2 >, ⃗v = < -5, 3, 2 >
To determine if ⃗u and ⃗v are orthogonal, we calculate their dot product:
⃗u · ⃗v = 1*(-5) + 3*3 + (-2)*2 = -5 + 9 - 4 = 0
Since the dot product is zero, ⃗u and ⃗v are orthogonal vectors.
(2.2) ⃗u = < 1, -2, 4 >, ⃗v = < 5, 3, 7 >
⃗u · ⃗v = 1*5 + (-2)3 + 47 = 5 - 6 + 28 = 27
Since the dot product is positive (non-zero), ⃗u and ⃗v do not form an acute or obtuse
angle.
Question 3:
(3.1) ⃗u = < -2, 1, -3 >, ⃗a = < -2, 1, 2 >
To find the orthogonal projection of ⃗u on ⃗a, we can use the formula:
proj⃗a⃗u = ((⃗u · ⃗a) / (||⃗a||^2)) ⃗a
Calculating the dot product:
⃗u · ⃗a = (-2)(-2) + 11 + (-3)*2 = 4 + 1 - 6 = -1
Calculating the norm of ⃗a:
||⃗a|| = √((-2)^2 + 1^2 + 2^2) = √(4 + 1 + 4) = √9 = 3
Calculating the projection:
proj⃗a⃗u = ((-1) / (3^2)) < -2, 1, 2 > = (-1/9) < -2, 1, 2 >
2023
ASSIGNMENT 4
DUE DATE: 20 JULY 2023
, ASSIGNMENT 04
Due date: Thursday, 20 July 2023
Total Marks: 100
ONLY FOR YEAR MODULE
This assignment covers chapter 2 & 3 of the prescribed book as well as the
study guide, it is based on Study Units 2.1, 2.2, 2.3 & 3.1
Question 1:
Given a square matrix A:
(1.1) Performing the operation (3):
R2 ← R2 - 2R1
E1 =
[1 0 0]
[0 1 -2]
[0 0 1]
Performing E1A:
B = E1A =
[2 -1 1]
[1 3 -3]
[1 -3 k + 9]
To find values of k such that a33 = a233, we set them equal:
k + 14 = k + 9
5=0
Since there is no value of k that satisfies the equation, there are no values of k in the
range [-1, 0] that make a33 = a233.
(1.2) Performing the operation (3):
R1 ←→ R3
E2 =
[0 0 1]
[0 1 0]
[1 0 0]
Performing E2A:
C = E2A =
[1 -3 k + 14]
[3 1 -1]
[2 -1 1]
, Therefore, the matrices E2 and C such that A = E2C are:
E2 =
[0 0 1]
[0 1 0]
[1 0 0]
C=
[1 -3 k + 14]
[3 1 -1]
[2 -1 1]
Question 2:
(2.1) ⃗u = < 1, 3, -2 >, ⃗v = < -5, 3, 2 >
To determine if ⃗u and ⃗v are orthogonal, we calculate their dot product:
⃗u · ⃗v = 1*(-5) + 3*3 + (-2)*2 = -5 + 9 - 4 = 0
Since the dot product is zero, ⃗u and ⃗v are orthogonal vectors.
(2.2) ⃗u = < 1, -2, 4 >, ⃗v = < 5, 3, 7 >
⃗u · ⃗v = 1*5 + (-2)3 + 47 = 5 - 6 + 28 = 27
Since the dot product is positive (non-zero), ⃗u and ⃗v do not form an acute or obtuse
angle.
Question 3:
(3.1) ⃗u = < -2, 1, -3 >, ⃗a = < -2, 1, 2 >
To find the orthogonal projection of ⃗u on ⃗a, we can use the formula:
proj⃗a⃗u = ((⃗u · ⃗a) / (||⃗a||^2)) ⃗a
Calculating the dot product:
⃗u · ⃗a = (-2)(-2) + 11 + (-3)*2 = 4 + 1 - 6 = -1
Calculating the norm of ⃗a:
||⃗a|| = √((-2)^2 + 1^2 + 2^2) = √(4 + 1 + 4) = √9 = 3
Calculating the projection:
proj⃗a⃗u = ((-1) / (3^2)) < -2, 1, 2 > = (-1/9) < -2, 1, 2 >
