MATH 1600 Summer 2026
MATH 1600 Assignment No. 3 Summer 2026
Section Assignment
650, SummerNo.
2026,
3 UWO
Section 650, Summer 2026, UWO
Instructions. (1) Submission of this assignment needs to be done through Gradescope. For in-
structions on how
Instructions. (1) to upload your
Submission assignment
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ments” tab on OWL.
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(2) Thetab
ments” solutions
on OWL.must be well written, clearly scanned, as covered in the lecture videos or the
textbook.
(2) The solutions must be well written, clearly scanned, as covered in the lecture videos or the
(3) We will only accept the hand written solutions. All the typed solutions will receive no marks
textbook.
for
(3) We whole
the assignment.
will only accept the hand written solutions. All the typed solutions will receive no marks
(4) Note that after uploading your files, you must make sure that you assign each page of your
for the whole assignment.
assignment to one of the questions
(4) Note that after uploading youroffiles,
the assignment on Gradescope.
you must make Youassign
sure that you may loose
each marks foryour
page of not
doing this step.
assignment to one of the questions of the assignment on Gradescope. You may loose marks for not
doing this step.
Questions.
(1) (6 marks) True or False? Justify your answer.
Questions.
Answers
(1) (6 marks) True without
or False? correct
Justify justification will receive no credit.
your answer.
Answers without correct justification will receive no credit.
a) If A is an invertible square matrix ∗ 2
then A A is also invertible.
a) If A can
is anchange
invertible square 1 2 ∗ 2−1 0 invertible.
b) One matrix A =matrix
3 4then to BA=A is also using elementary row operations.
1 2 2 01
−1
b) One can change matrix A = 1 1 to 1 B = 2 11 3using 1 elementary row operations.
c) One can change matrix A = 3 4 to B = using elementary row oper-
21 11 31 21 23 31
c) ations.
One can change matrix A = to B = using elementary row oper-
2 1 3 2 2 3
Assume A is a 2 × 3 matrix and B is a 1 × 3 matrix. The set of all vectors x ∈ R3 that
d) ations.
d) satisfies
Assume A AxBis a=2 ×
0 form a subspace.
3 matrix and B is a 1 × 3 matrix. The set of all vectors x ∈ R3 that
e) satisfies
If A andAxBB are=60×form
6 matrices of rank 6, then AB also has rank 6.
a subspace.
f) If
e) If A
A is
anda 5B×are6 matrix of rank 3,
6 × 6 matrices of then
rank the solutions
6, then of Ax
AB also has=rank
0 is has
6. two parameters.
f) If A is a 5 × 6 matrix of rank 3, then the solutions of Ax = 0 is has two parameters.
(1a) (2) (6 marks)
True. Since AConsider theboth
is invertible, following
A2 and matrices
its conjugate transpose A∗ are invertible:
(2) (6 marks) Consider 1 1the following
2 −1 2matrices
−11 −1 ∗ −13 0 −1 ∗ −1 −1
A= ,(A B) ==A A , ,, C(A = ) = (A , )D
. =
01 −11 12 11 13 20 −2
−1 −3
−1
The product of A = matrices is ,invertible,
invertible B = so ,, C = , D=
(a) Show that 0the−1 matrices A, B, 1 and
1 C are linearly −2
1 2 independent? −3
(b) Write that
(a) Show D asthelinear combination
matrices of A,
A, ∗B,2 and
−1 B,
areand
C −1 C. ∗ independent?
linearly
−1 −1
(A A ) =A A (A ) .
(b) Write D as linear combination of A, B, and C.
(3) (8 marks) Consider the following matrices
(1b) True. Starting with A, perform the following elementary
row operations:
(3) (8 marks) Consider the following
−1 2 3matrices
− 3i
=
A2 −3R −i 1 1 1 B= 1 + 2i −2
←− 2,R2 1 2 −i
1 2 R2 ←R
−−−−−−−−→−5 i1 −11 22 3− R2 3i
−8
−−−−−−→ 1 + 2i 1
−2
3 4 A = −i0 1−2 1 , 0B 1=
−i 1
(a) Find the inverse of the −5 i matrices
given 8 usingGauss-Jordan method?
R1 ←R1 −2R2 1 0 R1 ←−R1 −1 0 R2 ←R2 −2R1 −1 0
(b) Writethe
(a) Find theinverse
−inverse
−−−−−−of of
− thegiven
→the given
−−−matrices
−−−→ using
matrices as the−product
−−−−−−−→ of elementary
= Bmatrices
. (Find a
0 1 0 1 Gauss-Jordan method? 2 1
sequence of elementary matrices whose product in a correct order is
(b) Write the inverse of the given matrices as the product of elementary matrices (Findequal to inverse
a
matrix).
sequence of elementary matrices whose product in a correct order is equal to inverse
(1c) False. Elementary row operations preserve row-equivalence and reduced row echelon form. For the given two
matrix).
(4) (8 marks) Find a basis for the row space, column space and null space of the following
(4) matrices.
(8 marks) (SHOW
Find a YOUR
basis forWORK)
the row space, column space and null space of the following
matrices. (SHOW YOUR WORK)
, matrices:
1 1 1 R2 ←R2 −2R1 1 1 1 R2 ←−R2 1 1 1
A= −−−−−−−−→ −−−−−−→
2 1 3 0 −1 1 0 1 −1
R1 ←R1 −R2 1 0 2
−−−−−−−−→ rref(A) = ,
0 1 −1
1
1 3 1 R2 ←R2 −2R1 1 3 1 R2 ←− 4 R2 1 3 1
B= −−−−−−−−→ −−−−−−−→
2 2 3 0 −4 1 0 1 − 14
7
R1 ←R1 −3R2 1 0
−−−−−−−−→ rref(B) = 4 .
0 1 − 14
Since the reduced row echelon forms differ, A cannot be transformed into B using elementary row operations.
(1d) True. With x a 3 × 1 column vector, the product (Ax)B is a 2 × 3 matrix. Define
L(x) = (Ax)B.
If L(x) = 0 and L(y) = 0, then
L(x + y) = (Ax + Ay)B = (Ax)B + (Ay)B = 0.
For every scalar c, L(cx) = (cAx)B = cL(x) = 0, and L(0) = 0. Thus the solution set is closed under addition and
scalar multiplication and contains 0:
{x ∈ R3 : (Ax)B = 0} = ker L is a subspace.
(1e) True. Since rank(A) = rank(B) = 6 for 6 × 6 matrices, both are invertible. Hence
(AB)−1 = B −1 A−1 , rank(AB) = 6 .
(1f ) False. The rank-nullity theorem gives
dim(ker A) = # columns − rank(A) = 6 − 3 = 3.
Therefore the solutions of Ax = 0 have 3 free parameters , not two.
MATH 1600 Assignment No. 3 Summer 2026
Section Assignment
650, SummerNo.
2026,
3 UWO
Section 650, Summer 2026, UWO
Instructions. (1) Submission of this assignment needs to be done through Gradescope. For in-
structions on how
Instructions. (1) to upload your
Submission assignment
of this properly
assignment needstotogradescope, visit the
be done through “Written Assign-
Gradescope. For in-
ments” tab on OWL.
structions on how to upload your assignment properly to gradescope, visit the “Written Assign-
(2) Thetab
ments” solutions
on OWL.must be well written, clearly scanned, as covered in the lecture videos or the
textbook.
(2) The solutions must be well written, clearly scanned, as covered in the lecture videos or the
(3) We will only accept the hand written solutions. All the typed solutions will receive no marks
textbook.
for
(3) We whole
the assignment.
will only accept the hand written solutions. All the typed solutions will receive no marks
(4) Note that after uploading your files, you must make sure that you assign each page of your
for the whole assignment.
assignment to one of the questions
(4) Note that after uploading youroffiles,
the assignment on Gradescope.
you must make Youassign
sure that you may loose
each marks foryour
page of not
doing this step.
assignment to one of the questions of the assignment on Gradescope. You may loose marks for not
doing this step.
Questions.
(1) (6 marks) True or False? Justify your answer.
Questions.
Answers
(1) (6 marks) True without
or False? correct
Justify justification will receive no credit.
your answer.
Answers without correct justification will receive no credit.
a) If A is an invertible square matrix ∗ 2
then A A is also invertible.
a) If A can
is anchange
invertible square 1 2 ∗ 2−1 0 invertible.
b) One matrix A =matrix
3 4then to BA=A is also using elementary row operations.
1 2 2 01
−1
b) One can change matrix A = 1 1 to 1 B = 2 11 3using 1 elementary row operations.
c) One can change matrix A = 3 4 to B = using elementary row oper-
21 11 31 21 23 31
c) ations.
One can change matrix A = to B = using elementary row oper-
2 1 3 2 2 3
Assume A is a 2 × 3 matrix and B is a 1 × 3 matrix. The set of all vectors x ∈ R3 that
d) ations.
d) satisfies
Assume A AxBis a=2 ×
0 form a subspace.
3 matrix and B is a 1 × 3 matrix. The set of all vectors x ∈ R3 that
e) satisfies
If A andAxBB are=60×form
6 matrices of rank 6, then AB also has rank 6.
a subspace.
f) If
e) If A
A is
anda 5B×are6 matrix of rank 3,
6 × 6 matrices of then
rank the solutions
6, then of Ax
AB also has=rank
0 is has
6. two parameters.
f) If A is a 5 × 6 matrix of rank 3, then the solutions of Ax = 0 is has two parameters.
(1a) (2) (6 marks)
True. Since AConsider theboth
is invertible, following
A2 and matrices
its conjugate transpose A∗ are invertible:
(2) (6 marks) Consider 1 1the following
2 −1 2matrices
−11 −1 ∗ −13 0 −1 ∗ −1 −1
A= ,(A B) ==A A , ,, C(A = ) = (A , )D
. =
01 −11 12 11 13 20 −2
−1 −3
−1
The product of A = matrices is ,invertible,
invertible B = so ,, C = , D=
(a) Show that 0the−1 matrices A, B, 1 and
1 C are linearly −2
1 2 independent? −3
(b) Write that
(a) Show D asthelinear combination
matrices of A,
A, ∗B,2 and
−1 B,
areand
C −1 C. ∗ independent?
linearly
−1 −1
(A A ) =A A (A ) .
(b) Write D as linear combination of A, B, and C.
(3) (8 marks) Consider the following matrices
(1b) True. Starting with A, perform the following elementary
row operations:
(3) (8 marks) Consider the following
−1 2 3matrices
− 3i
=
A2 −3R −i 1 1 1 B= 1 + 2i −2
←− 2,R2 1 2 −i
1 2 R2 ←R
−−−−−−−−→−5 i1 −11 22 3− R2 3i
−8
−−−−−−→ 1 + 2i 1
−2
3 4 A = −i0 1−2 1 , 0B 1=
−i 1
(a) Find the inverse of the −5 i matrices
given 8 usingGauss-Jordan method?
R1 ←R1 −2R2 1 0 R1 ←−R1 −1 0 R2 ←R2 −2R1 −1 0
(b) Writethe
(a) Find theinverse
−inverse
−−−−−−of of
− thegiven
→the given
−−−matrices
−−−→ using
matrices as the−product
−−−−−−−→ of elementary
= Bmatrices
. (Find a
0 1 0 1 Gauss-Jordan method? 2 1
sequence of elementary matrices whose product in a correct order is
(b) Write the inverse of the given matrices as the product of elementary matrices (Findequal to inverse
a
matrix).
sequence of elementary matrices whose product in a correct order is equal to inverse
(1c) False. Elementary row operations preserve row-equivalence and reduced row echelon form. For the given two
matrix).
(4) (8 marks) Find a basis for the row space, column space and null space of the following
(4) matrices.
(8 marks) (SHOW
Find a YOUR
basis forWORK)
the row space, column space and null space of the following
matrices. (SHOW YOUR WORK)
, matrices:
1 1 1 R2 ←R2 −2R1 1 1 1 R2 ←−R2 1 1 1
A= −−−−−−−−→ −−−−−−→
2 1 3 0 −1 1 0 1 −1
R1 ←R1 −R2 1 0 2
−−−−−−−−→ rref(A) = ,
0 1 −1
1
1 3 1 R2 ←R2 −2R1 1 3 1 R2 ←− 4 R2 1 3 1
B= −−−−−−−−→ −−−−−−−→
2 2 3 0 −4 1 0 1 − 14
7
R1 ←R1 −3R2 1 0
−−−−−−−−→ rref(B) = 4 .
0 1 − 14
Since the reduced row echelon forms differ, A cannot be transformed into B using elementary row operations.
(1d) True. With x a 3 × 1 column vector, the product (Ax)B is a 2 × 3 matrix. Define
L(x) = (Ax)B.
If L(x) = 0 and L(y) = 0, then
L(x + y) = (Ax + Ay)B = (Ax)B + (Ay)B = 0.
For every scalar c, L(cx) = (cAx)B = cL(x) = 0, and L(0) = 0. Thus the solution set is closed under addition and
scalar multiplication and contains 0:
{x ∈ R3 : (Ax)B = 0} = ker L is a subspace.
(1e) True. Since rank(A) = rank(B) = 6 for 6 × 6 matrices, both are invertible. Hence
(AB)−1 = B −1 A−1 , rank(AB) = 6 .
(1f ) False. The rank-nullity theorem gives
dim(ker A) = # columns − rank(A) = 6 − 3 = 3.
Therefore the solutions of Ax = 0 have 3 free parameters , not two.
