F18CF - Assignment 1
Student Name
Institution
Date: 6th October, 2023
Question 1
(a) Determine the solutions to the system below, or conclude that no solutions
exist.
4x + 4y + 2z + 4u = 3
2x + y − z = 5
−2x + y + 5z + 4u = 1
4 4 2 4 3
2 1 −1 0 5
−2 1 5 4 1
Perform row operations to get the row echelon form:
1 1 0.5 1 0.75
0 1 −1 −2 3.5
0 2 4
8 4.5
Performing further row operations to get the reduced row echelon form:
1 0 −2 −5 −2
0 1 −1 −2 3.5
0 0 3
6 −2.5
The system is inconsistent, so there are no solutions.
(b) Let α ∈ R and define the matrix


α
α2
0
A = α + 1 α α 2 
1
α −α2
Determine the set of values α such that det(A) ̸= 0.


α
α2
0
A = α + 1 α α 2 
1
α −α2
1
We want to determine the set of values of α such that det(A) ̸= 0.
Using the properties of determinants, we can calculate det(A) as follows:
det(A) = α · α · α − (−α2 · (α + 1)) − α2 · α · 1 − (−α2 · 1)
= α(α2 + α3 + α2 ) − α2 (α + α2 )
= α(2α3 + 2α2 ) − α2 (α + α2 )
= 2α4 + 2α3 − α3 − α4
= α4 + α3 .
Now, we want to find the set of values of α such that det(A) ̸= 0. So, we
need to solve the equation α4 + α3 ̸= 0. This equation holds true for all real
values of α except when α = 0 (since 04 + 03 = 0).
Therefore, the set of values of α for which det(A) ̸= 0 is α ∈ R such that
α ̸= 0.
Question 2
For this question, we will refer to the following matrices:
A = (⃗a1 |⃗a2 | . . . |⃗ak ) where the column vectors ⃗a1 , ⃗a2 , . . . , ⃗ak are linearly independent.
B = (⃗b1 |⃗b2 | . . . |⃗bm ) where the column vectors ⃗b1 , ⃗b2 , . . . , ⃗bm are linearly independent.
C = AB
(a) Show that A⃗x = 0 has a unique solution.
Let’s consider the system of equations A⃗x = 0, where A is a matrix with
linearly independent column vectors ⃗a1 , ⃗a2 , . . . , ⃗ak .
Suppose ⃗x is a solution to the system A⃗x = 0. Since ⃗a1 , ⃗a2 , . . . , ⃗ak are
linearly independent, the only solution to A⃗x = 0 is the trivial solution where
⃗x = ⃗0.
This can be explained as follows: Any nontrivial solution to A⃗x = 0 would
imply a nontrivial linear combination of the column vectors of A that gives
the zero vector, contradicting the assumption that ⃗a1 , ⃗a2 , . . . , ⃗ak are linearly
independent.
Therefore, the unique solution to A⃗x = 0 is ⃗x = ⃗0.
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(b) Show that the column vectors of the matrix C = AB are also linearly
independent.
Let A = (⃗a1 |⃗a2 | . . . |⃗ak ) and B = (⃗b1 |⃗b2 | . . . |⃗bm ). We want to show that
the column vectors of C, denoted as ⃗ci , are linearly independent.
Let’s express C in terms of A and B:
C = AB = (⃗a1 |⃗a2 | . . . |⃗ak )(⃗b1 |⃗b2 | . . . |⃗bm )
The i-th column of C, denoted as ⃗ci , is given by:
⃗ci = A⃗bi
Suppose we have a linear combination of the column vectors of C:
c1⃗c1 + c2⃗c2 + . . . + cm⃗cm = ⃗0
Substituting the expression for ⃗ci in terms of A and B, we get:
c1 (A⃗b1 ) + c2 (A⃗b2 ) + . . . + cm (A⃗bm ) = ⃗0
Now, we can factor out the A matrix:
A(c1⃗b1 + c2⃗b2 + . . . + cm⃗bm ) = ⃗0
Since A has linearly independent column vectors, the only solution to
A⃗x = ⃗0 is ⃗x = ⃗0. Therefore, the linear combination c1⃗b1 + c2⃗b2 + . . . + cm⃗bm
must be the zero vector:
c1⃗b1 + c2⃗b2 + . . . + cm⃗bm = ⃗0
Since ⃗b1 , ⃗b2 , . . . , ⃗bm are linearly independent, the only solution to this
equation is c1 = c2 = . . . = cm = 0. Hence, the column vectors of C are
linearly independent.
(c) Let n ∈ N be a fixed, but unknown positive integer. Determine (and
justify) the conditions of A and B such that the column vectors of C will
form the basis of Rn . Let n ∈ N be a fixed, but unknown positive integer.
We want to determine the conditions on matrices A and B such that the
column vectors of C = AB will form the basis of Rn .
For the column vectors of C to form the basis of Rn , the matrix C must
be full rank, which means its column vectors must be linearly independent
and span Rn .
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Matrix C = AB has column vectors given by C = (⃗c1 |⃗c2 | . . . |⃗cm ), where
⃗ci is the i-th column of C.
To ensure the column vectors of C are linearly independent, both matrices A and B must have linearly independent column vectors. Therefore,
⃗a1 , ⃗a2 , . . . , ⃗ak (the column vectors of A) and ⃗b1 , ⃗b2 , . . . , ⃗bm (the column vectors
of B) must be linearly independent.
Additionally, for the column vectors of C to span Rn , the number of
linearly independent column vectors of C must be n.
So, k × m = n, where k is the number of linearly independent column
vectors of A, and m is the number of linearly independent column vectors of
B.
Therefore, the conditions for matrices A and B such that the column
vectors of C will form the basis of Rn are: 1. The column vectors of A and
B must be linearly independent. 2. The product of the number of linearly
independent column vectors in A and B must be equal to n:
k×m=n
(d) Prove the following statement or provide an explicit counter-example:
Let P , Q, R be non-zero matrices satisfying the equation R = P Q. If the
column vectors of R are linearly dependent, then the column vectors of P
must form a linearly dependent set and the column vectors of Q must also
form a linearly dependent set.
Let P , Q, R be non-zero matrices satisfying the equation R = P Q. We
want to determine whether the following statement is true or provide a counterexample.
Statement: If the column vectors of R are linearly dependent, then the
column vectors of P must form a linearly dependent set, and the column
vectors of Q must also form a linearly dependent set.
Proof (if true): Suppose the column vectors of R are linearly dependent.
This means there exist scalars a1 , a2 , . . . , an , not all zero, such that:
a1 r1 + a2 r2 + . . . + an rn = 0
where ri is the i-th column vector of R.
Since R = P Q, we can express the column vectors of R in terms of the
column vectors of P and Q:
ri = P qi
where qi is the i-th column vector of Q.
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Substituting this into the linear dependence relation for R, we get:
a1 P q1 + a2 P q2 + . . . + an P qn = 0
Dividing both sides by P (assuming P is invertible), we obtain:
a 1 q1 + a 2 q2 + . . . + a n qn = 0
Since the qi ’s are the column vectors of Q, this shows that the column
vectors of Q are linearly dependent.
A similar argument can be made for P , showing that the column vectors
of P are linearly dependent.
Counterexample (if false):
Let’s provide a counterexample to show that the statement is not always
true.
Consider the following matrices:
1 0
0 1
0 1
P =
, Q=
, R = PQ =
0 0
0 1
0 0
The column vectors of R are linearly dependent (the second column is a
zero vector). However, the column vectors of P and Q are linearly independent.
This counterexample disproves the statement.
Question 3
Within the multiverse, each universe is completely determined by their fundamental forces that govern different essential aspect of existence. The combination of these forces create the reality experienced by individuals of that
universe.
Whilst studying secret scrolls in Kamar-Taj, Dr. Strange discovers that
the fundamental forces can be expressed as vectors in Rn and harness their
powers through linear combinations! In any Universe, the fundamental forces
can be expressed as the set F = {⃗r1 , ⃗r2 , . . . , ⃗rm } and the set of possible reality
in that universe is given by span(F ).
(a) With this epiphany, Dr. Strange attempts several spells that alter
the set of fundamental forces to create G. For each of the following spells,
show that the following spells do not alter the set of realities in this universe.
That is span(F ) = span(G).
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i. The interchange spell: Obtain G from F by swapping two forces ⃗ri ↔ ⃗rj
with i ̸= j.
ii. The amplify spell: Obtain G from F by amplifying one force by replacing ⃗ri with r⃗′ i = λ⃗ri with λ ̸= 0.
iii. The influence spell: Obtain G from F by allowing one force ⃗ri to be
influenced by a second so that ⃗ri is replaced by r⃗′ i = ⃗ri +λ⃗rj with λ ̸= 0
and i ̸= j.
i. Interchange Spell:
Swapping two forces ⃗ri ↔ ⃗rj with i ̸= j does not alter the span. The
span remains the same since swapping two vectors doesn’t change their
linear combinations.
ii. Amplify Spell:
Amplifying a force by replacing ⃗ri with r⃗′ i = λ⃗ri with λ ̸= 0 also does
not alter the span. This is because the amplified force is still a linear
combination of the original forces, hence still within the same span.
iii. Influence Spell:
Allowing one force ⃗ri to be influenced by a second by replacing ⃗ri with
r⃗′ i = ⃗ri + λ⃗rj with λ ̸= 0 and i ̸= j does not alter the span. The new
force is still a linear combination of the original forces, thus within the
same span.
(b) Upon sharing his mastery of this mystic art with the Sanctum, his
caretaker casually noted that this power originated from the long-forgotten
scroll of Elementary Row Operations (ERO). By considering the fundamental
forces as row vectors, express each reality ⃗v ∈ span(F ) as a matrix equation
⃗v = ⃗aP for some vector ⃗a and matrix P . Further, show that the span(F )
remains the same under the elementary row operations on P .
Let F = {⃗r1 , ⃗r2 , . . . , ⃗rm }, where each ⃗ri is a row vector. Let P be the
matrix formed by stacking these row vectors.
Each reality ⃗v ∈ span(F ) can be expressed as a linear combination of the
row vectors in F :
⃗v = ⃗aP
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where ⃗a is a vector of coefficients for the linear combination.
Now, let’s demonstrate how elementary row operations on P don’t change
the span.
• Elementary Row Operation 1: Swapping Rows
Swapping two rows of P corresponds to swapping the corresponding
forces in F . This doesn’t change the span because the linear combinations remain the same.
• Elementary Row Operation 2: Scaling a Row
Scaling a row of P corresponds to amplifying a force in F . This doesn’t
change the span because it’s still a linear combination of the original
forces.
• Elementary Row Operation 3: Adding a Multiple of One Row
to Another
Adding a multiple of one row to another corresponds to allowing one
force to influence another in F . This also doesn’t change the span
because the new force is still a linear combination of the original forces.
Therefore, elementary row operations on P don’t alter the span, i.e.,
span(F ) remains the same.
(c) As Dr. Strange progresses in his astral journey, he encounters two
universes U616 and U838, which seem to experience the same reality. From
his studies, he concludes that the fundamental forces of the two universes are
made up of the following:
Universe U616 consists of F = {⃗r1 , ⃗r2 , ⃗r3 } where:
 
 
1
2
2
4

 
⃗r1 = 
1 , ⃗r2 = 1 ,
3
4
 
3
6

⃗r3 = 
1 .
5
Universe U838 consists of G = {⃗s1 , ⃗s2 } where:
 
 
2
1
2
4

 
⃗s1 = 
4 , ⃗s2 =  4  .
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9
7
Using the new powers he developed in (b), show that these two universes
experience the same set of realities. That is, show that span(F ) = span(G).
Let the fundamental forces for U616 be denoted as F = {⃗r1 , ⃗r2 , ⃗r3 }, where:
 
 
 
1
2
3
2
4
6

 
 
⃗r1 = 
1 , ⃗r2 = 1 , ⃗r3 = 1 .
3
4
5
For U838, let the fundamental forces be denoted as G = {⃗s1 , ⃗s2 }, where:
 
 
1
2
2
4

 
⃗s1 = 
4 , ⃗s2 =  4  .
9
10
We express the forces of both universes in terms of the matrix P .


1 2 3
2 4 6

P =
1 1 1
3 4 5
Matrix P represents the transformation from forces in U616 to their corresponding forces in U838.
 
 
1
2
2
4

 
⃗s1 = ⃗r1 P = 
4 , ⃗s2 = ⃗r2 P =  4  .
9
10
It’s evident that ⃗s1 and ⃗s2 can be expressed as linear combinations of ⃗r1
and ⃗r2 respectively, thus showing that span(F ) = span(G), implying they
experience the same set of realities.
Question 4
The pilgrimage to savor global culinary delights is undertaken by many. In
this question, we model the transition of foodies across three famous culinary
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destinations: Malaysia (x1 ), Dubai (x2 ) and Edinburgh (x3 ), where each xi
denotes the number of foodies in that region. In particular, we want to
determine the stable distribution denoted by ⃗x = (x1 , x2 , x3 )T ∈ R3 such
that the number of foodies in each place remains unchanged between each
transition.
(a) Suppose that in each transition:
Malaysia-based foodies will move to Edinburgh with a probability of 0.4, otherwise they must stay
Dubai-based foodies will move to Malaysia with a probability of 0.4, otherwise they must stay in D
Edinburgh-based foodies will move to Dubai with a probability of 0.6, otherwise they must move t
We are given that the total number of foodies in this model is 500. Determine
the stable distribution for the transition problem above.
Malaysia-based foodies will move to Edinburgh with a probability of 0.4, otherwise they must stay
Dubai-based foodies will move to Malaysia with a probability of 0.4, otherwise they must stay in D
Edinburgh-based foodies will move to Dubai with a probability of 0.6, otherwise they must move t
We are given that the total number of foodies in this model is 500. Determine
the stable distribution for the transition problem above.
Stable distribution:
x1 = 0.6x3 + 0.4x2
x2 = 0.4x1 + 0.6x3
x3 = 0.4x2 + 0.6x1
x1 + x2 + x3 = 500
(b) Edinburgh proposes an International Food Fiesta as part of their
annual Cringe Festival to increase the Edinburgh (x3 ) component of the stable
distribution. This initiative is projected to change the transition of foodies
across the three places into the following new scheme, parametrized by three
probabilities, α, β, and γ:
The competition of Malaysia and Dubai remains strong. Malaysia-based foodies will move to Duba
International Food Fiesta results in Malaysia-based and Dubai-based foodies to have a probability
The problem can be formulated into the following system of linear equations:

   
1−α−β
α
0
x1
x1

α
1 − α − γ 0 x2  = x2 
β
γ
0
x3
x3
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with α + β ≤ 1 and α + γ ≤ 1, which can then be expressed as the augmented
matrix below:


−α − β
α
0 0
 α
−α − γ 0 0
β
γ
0 0
Denote the solution set of this matrix as S. Since it is a homogeneous
system, there exists a set V = {⃗v1 , ⃗v2 , . . . , ⃗vk } for some k ∈ Z+ such that
the set S = span(V ). In particular, we are interested in the scenario where
V = {⃗v }. That is, where S is spanned by a single vector ⃗v .
i. Without applying any row operations to the matrix, determine that if
S = span(⃗v ) for ⃗v = (v1 , v2 , v3 )T ∈ R3 , then v1 = v2 = 0.
Since S is spanned by a single vector ⃗v , it implies that the system
has a nontrivial solution. For ⃗v to be nontrivial, v1 , v2 , or v3
must be nonzero, violating the property of a homogeneous system.
Therefore, v1 = v2 = 0 to maintain a nontrivial solution.
ii. Show that the augmented matrix can be further simplified into:


−α − β
α
0 0
 α
−α − γ 0 0
0
0
0 0
To simplify the augmented matrix, we notice that the third row
corresponds to the homogeneous equation 0 = 0, thus it does not
affect the solution. We can omit it, leading to the given simplified
augmented matrix.
iii. Suppose that α + β ̸= 0. Determine the condition on α, β, and γ
that result in the matrix having exactly one free variable.
For the matrix to have exactly one free variable, the determinant
of the coefficient matrix must be zero. Thus:
−α − β
α
Det
=0
α
−α − γ
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Solving this, we get the condition:
α2 + α(β + γ) − βα = 0
which simplifies to:
α(α + β + γ) = 0
iv. For each case in (iii), express the solutions set S as the span of a
minimal number of vectors.
- **Case 1: α = 0**
In this case, the coefficient matrix reduces to:
0 0
0 −γ
The solutions can be expressed as the span of a single vector ⃗v :
 
1


0
S = span
0
- **Case 2: α + β + γ = 0**
In this case, the coefficient matrix reduces to:
−β 0
0 −β
The solutions can be expressed as the span of a single vector ⃗v :
 
1


0
S = span
0
v. For the case α + β = 0, validate or disprove that the matrix has
exactly one free variable.
In this case, we already solved it in part iv. The matrix reduces
to:
0 0
0 −γ
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The solutions can be expressed as the span of a single vector ⃗v :
 
1
S = span 0
0
Thus, the matrix has exactly one free variable, consistent with
part iv.
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