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MATH1013-无代写
时间：2023-02-20
Australian
National
University
Mathematical Sciences Institute
EXAMINATION: Semester 2 — Final 2021
MATH1013 — Mathematics and Applications 1
Algebra
Exam Duration: 90 minutes.
Reading Time: 15 minutes.
Materials Permitted During The Exam:
• One A4 page with hand written notes on both sides.
(This A4 page is to cover both Calculus and Linear Algebra.)
• Trig cheat sheet.
• No calculators or books are permitted, with the exception of paper dictionaries.
Instructions To Students:
• This exam is run as a Wattle quiz.
• The exam must be your own work. During the exam you must not use any resources
on the internet or have any examination help from any source.
• The Algebra exam is worth a total of 50 points, with the value of each question as
shown.
• A good strategy is not to spend too much time on any question. Read them through
rst and attempt them in the order that allows you to make the most progress.
• You must justify your answers. Do not expect credit for a correct answer with no
justication. Write clearly and legibly.
• You have 15 minutes reading time for the exam. You can make notes on scrap paper
during this time (but must not start writing your nal answers.)
• At the end of the 90 minutes exam time you will be given additional time to make a
single PDF and upload to Wattle.
Question 1 6 pts
Consider the system of linear equations
x1 + x2 = 0,
− x2 + 2x3 = −1,
−x1 − 2x2 + x3 = 1.
(a) Form the associated augmented matrix and then reduce the augmented matrix to row
echelon form. 3 pts
Solution
The augmented matrix for the linear system is
M =
1 1 0 0
0 −1 2 −1
−1 −2 1 1
.
Convert the matrix to row echelon form:
1 1 0 0
0 −1 2 −1
−1 −2 1 1
(R3 + R1→ R3)
−→
1 1 0 0
0 −1 2 −1
0 −1 1 1
(R3 − R2→ R3)
−→
1 1 0 0
0 −1 2 −1
0 0 −1 2
(b) Is the system consistent or inconsistent? If the system is consistent then nd the
solution of the system of linear equations. 3 pts
Solution
Every column/row has a pivot position. The system is consistent and has the unique
solution
x3 = −2, x2 = 1 + 2x3 = −3, x1 = 3.
MATH1013 Final Exam, Semester 2, 2021, Page 2 of 12
Question 2 6 pts
(a) Find the LU factorisation of the matrix
A =
−2 −1 2
−6 0 −2
8 −1 5
.
3 pts
Solution
Convert A to row echelon form:
−2 −1 2
−6 0 −2
8 −1 5
(R2 − 3R1→ R2)
(R3 + 4R1→ R3)
−→
−2 −1 2
0 3 −7
0 −5 13
:
E1 =
1 0 0
−3 1 0
4 0 1
−2 −1 2
0 3 −8
0 −5 13
(R3 + 5/3R2→ R3)
−→
−2 −1 2
0 3 −8
0 0 4/3
:
E2 =
1 0 0
0 1 0
0 5/3 1
L = E−11 E
−1
2 =
1 0 0
3 1 0
−4 −5/3 1
, U =
−2 −1 2
0 3 −8
0 0 −1/3
.
(b) Given
A =
1 0 0
−1 1 0
2 0 1
4 3 −5
0 −2 2
0 0 2
, b =
2
−4
6
.
Solve Ax = b . 3 pts
Solution
Forward substitution:
Ly = b =⇒ y1 = 2, y2 = −4 + y1 = −2, y3 = 6 − 2y1 = 2
MATH1013 Final Exam, Semester 2, 2021, Page 3 of 12
Backward substitution:
Ux = y =⇒ x3 = y3/2 = 1, x2 = −(y2−2x3)/2 = 2, x1 = (y1−3x2+5x3)/4 = (2−6+5)/4 = 1/4
MATH1013 Final Exam, Semester 2, 2021, Page 4 of 12
Question 3 6 pts
Consider the matrix A and its row echelon form R
A =
−2 0 −1 1
0 2 2 0
0 2 3 3
1 0 0 2
, R =
−2 1 −1 1
0 2 2 0
0 0 1 3
0 0 0 0
.
(a) Find a basis for the null space of A . 3 pts
Solution
Ax = 0 ⇐⇒ Rx = 0,
x1 = −(−3x4 − 3x4 − x4)/2 = 7/2x4
x2 = −x3 = 3x4
x3 = −3x4
x4 = x4
7/2
3
−3
1
(b) Find a basis for the column space of A 3 pts
Solution
A and R are row equivalent. Thus pivot columns will span the column space of A .
−2
0
0
1
,
0
2
2
0
,
−1
2
3
0
MATH1013 Final Exam, Semester 2, 2021, Page 5 of 12
Question 4 6 pts
(a) Let T : R2 → R2 , S : R2 → R3 , R : R3 → R2 be transformations such that
T (x1,x2) = *,x1 − 2x23|x1 | +- , S (x1,x2) =
*...,
3x1 − 2x2
−x1 + 3
6x2
+///- , R (x1,x2,x3) =
*,3x1 − 2x2 + x3−x1 + 3x3 +- .
Which of the transformations T , S,R is linear? Justify your answers. 3 pts
Solution
T and S are not linear, since T (αu) , αT (u) and S (αu) , αS (u)
R is linear since it satises R (u +v ) = R (u) + R (v ) and R (αu) = αR (u) .
(b) Let T : R2 → R3 be a transformation such that
T (x1,x2) =
*...,
4x1 − 7x2
x1 − 2x2
4x1 − 4x2
+///- .
Is b =
1
0
4
in the range of T (x1,x2)? 3 pts
Justify your answer.
Solution
First, formulate as matrix vector operation T (x ) := Ax = b
T (x1,x2) =
4 −7
1 −2
4 −4
xx2
=
1
0
4
Consider the augmented matrix
4 −7 1
1 −2 0
4 −4 4
MATH1013 Final Exam, Semester 2, 2021, Page 6 of 12
Convert the matrix to row echelon form:
4 −7 1
1 −2 0
4 −4 4
( 14 R1→ R1)
( 14 R3→ R3)
−→
1 −74
1
4
1 −2 0
1 −1 1
(R2 − R1→ R2)
(R3 − R1→ R3)
−→
1 −74
1
4
0 −14
−1
4
0 34
3
4
(-4R2→ R2)
((4/3)R3→ R3)
−→
1 −74
1
4
0 1 1
0 1 1
( R3−R2→ R3)
−→
1 −74
1
4
0 1 1
0 0 0
Having
x2 = 1, x1 =
1
4 +
7
4x2 = 2.
MATH1013 Final Exam, Semester 2, 2021, Page 7 of 12
Question 5 10 pts
Suppose that A is a 3×3 matrix which can be reduce to the identity matrix I3 by performing
the following row operations in order
1st R3 is replaced by R3 + α1R2 , α1 , 0.
2nd R1 is replaced by R1 + α2R3 , α2 , 0.
3rd R3 is replaced by α3R3 , α3 , 0.
(a) Determine the elementary matrices E1 , E2 , E3 corresponding to each of the row
operations. 3 pt
Solution
E1 =
1 0 0
0 1 0
0 α1 1
, E2 =
1 0 α2
0 1 0
0 0 1
, E3 =
1 0 0
0 1 0
0 0 α3
.
(b) Show that A−1 = E3E2E1 . 3 pt
Solution
The elementary row operations will correspond to mathrmE3E2E1A = I3 .
A−1 is an (left) inverse of A if A−1A = I3 .
Note that
E3E2E1A = I3 ⇐⇒ A−1A = I3.
We must have A−1 = E3E2E1
(c) Determine the matrix A. 2 pt
Solution
Note that
E3E2E1A = I3 ⇐⇒ A = E−11 E−12 E−13 =
1 0 0
0 1 0
0 −α1 1
1 0 −α2
0 1 0
0 0 1
1 0 0
0 1 0
0 0 1α3
=
1 0 −α2α3
0 1 0
0 −α1 1α3
(d) Determine det(A) and det(A−1) . 2 pt
Solution
MATH1013 Final Exam, Semester 2, 2021, Page 8 of 12
det(A) = det
(
E−11 E−12 E−13
)
= det(E−11 ) det(E−12 ) det(E−13 ) =
1
α3
det(A−1) = det (E3E2E1) = det(E3) det(E2) det(E1) = α3
MATH1013 Final Exam, Semester 2, 2021, Page 9 of 12
Question 6 6 pts
(a) Consider the 3 × 3 matrix
A =
1 1 1
3 0 −1
1 −1 1
.
Use the determinant to show whether Ax = b has a solution for all b ∈ R3 . 3 pts
Solution
det (A) = −3
1 1−1 1
+
1 11 −1
= −8
Since det (A) , 0 A is invertible with the solution x = A−1b for all b ∈ R3 .
(b) Consider a matrix A with the row echelon form
R =
3 −1 0 1
0 2 3 −1
0 0 2 −3
.
i. Are the columns of A linearly in dependent?
ii. Does the system Ax = 0 have nontrivial solutions?
Justify your answers. 3 pts
Solution
The matrices A and R are row equivalent. Since there are 4 columns with only 3
pivots, then the columns of A are linearly dependent.
Correct answers with good reasons gets 1.5marks
Otherwise award partial marks
Since the columns of A are linearly dependent the system Ax = 0 has nontrivial
solutions.
MATH1013 Final Exam, Semester 2, 2021, Page 10 of 12
Question 7 10 pts
(a) Given the complex numbers z1 = 2 + i , z2 = 1 − 3i, express the following expressions
as complex numbers in standard form.
i. z1z2
ii. z1/z2
3 pts
Solution
z1z2 = (2 + i ) (1 − 3i ) = (2 + 3) + i (1 − 6) = 5 − 5i
z1/z2 = (2 + i )/(1 − 3i ) = (2 + i ) (1 + 3i )10 = −
1
10 +
7
10i
(b) Find the 4th roots of z = −i , and express them in polar form. 4 pts
Solution
z = −i = cis (3/2pi )
zk = cis (
(3/2)pi + 2kpi
4 ) = cis (
(3 + 4k )pi
8 ), k = 0, 1, 2, 3
z0 = cis (
3pi
8 ) = cos(
3pi
8 ) + i sin(
3pi
8 ), z1 = cis (
7pi
8 ) = cos(
7pi
8 ) + i sin(
7pi
8 )
z2 = cis (
11pi
8 ) = cos(
11pi
8 ) + i sin(
11pi
8 ), z3 = cis (
15pi
8 ) = cos(
15pi
8 ) + i sin(
15pi
8 )
(c) Consider the fourth degree polynomial f (z) = z4 − 2z3 + 3z2 − 2z + 2.
i. Two roots of f (z) are z0 = −i and z1 = 1 + i . What are the other roots of f (z) ?
ii. Factorise f (z) as the product of two quadratic polynomials.
3 pts
Solution
The roots are conjugate pairs. Therefore with z0 = −i and z1 = 1 + i we must have
z2 = z¯0 = i, z3 = z¯1 = 1 − i
MATH1013 Final Exam, Semester 2, 2021, Page 11 of 12
f (z) = (z − z0) (z − z¯0)︸ ︷︷ ︸
(z2+1)
(z − z1) (z − z¯1)︸ ︷︷ ︸
(z2−2z+2)
= (z2 + 1) (z2 − 2z + 2)
MATH1013 Final Exam, Semester 2, 2021, Page 12 of 12
National
University
Mathematical Sciences Institute
EXAMINATION: Semester 2 — Final 2021
MATH1013 — Mathematics and Applications 1
Algebra
Exam Duration: 90 minutes.
Reading Time: 15 minutes.
Materials Permitted During The Exam:
• One A4 page with hand written notes on both sides.
(This A4 page is to cover both Calculus and Linear Algebra.)
• Trig cheat sheet.
• No calculators or books are permitted, with the exception of paper dictionaries.
Instructions To Students:
• This exam is run as a Wattle quiz.
• The exam must be your own work. During the exam you must not use any resources
on the internet or have any examination help from any source.
• The Algebra exam is worth a total of 50 points, with the value of each question as
shown.
• A good strategy is not to spend too much time on any question. Read them through
rst and attempt them in the order that allows you to make the most progress.
• You must justify your answers. Do not expect credit for a correct answer with no
justication. Write clearly and legibly.
• You have 15 minutes reading time for the exam. You can make notes on scrap paper
during this time (but must not start writing your nal answers.)
• At the end of the 90 minutes exam time you will be given additional time to make a
single PDF and upload to Wattle.
Question 1 6 pts
Consider the system of linear equations
x1 + x2 = 0,
− x2 + 2x3 = −1,
−x1 − 2x2 + x3 = 1.
(a) Form the associated augmented matrix and then reduce the augmented matrix to row
echelon form. 3 pts
Solution
The augmented matrix for the linear system is
M =
1 1 0 0
0 −1 2 −1
−1 −2 1 1
.
Convert the matrix to row echelon form:
1 1 0 0
0 −1 2 −1
−1 −2 1 1
(R3 + R1→ R3)
−→
1 1 0 0
0 −1 2 −1
0 −1 1 1
(R3 − R2→ R3)
−→
1 1 0 0
0 −1 2 −1
0 0 −1 2
(b) Is the system consistent or inconsistent? If the system is consistent then nd the
solution of the system of linear equations. 3 pts
Solution
Every column/row has a pivot position. The system is consistent and has the unique
solution
x3 = −2, x2 = 1 + 2x3 = −3, x1 = 3.
MATH1013 Final Exam, Semester 2, 2021, Page 2 of 12
Question 2 6 pts
(a) Find the LU factorisation of the matrix
A =
−2 −1 2
−6 0 −2
8 −1 5
.
3 pts
Solution
Convert A to row echelon form:
−2 −1 2
−6 0 −2
8 −1 5
(R2 − 3R1→ R2)
(R3 + 4R1→ R3)
−→
−2 −1 2
0 3 −7
0 −5 13
:
E1 =
1 0 0
−3 1 0
4 0 1
−2 −1 2
0 3 −8
0 −5 13
(R3 + 5/3R2→ R3)
−→
−2 −1 2
0 3 −8
0 0 4/3
:
E2 =
1 0 0
0 1 0
0 5/3 1
L = E−11 E
−1
2 =
1 0 0
3 1 0
−4 −5/3 1
, U =
−2 −1 2
0 3 −8
0 0 −1/3
.
(b) Given
A =
1 0 0
−1 1 0
2 0 1
4 3 −5
0 −2 2
0 0 2
, b =
2
−4
6
.
Solve Ax = b . 3 pts
Solution
Forward substitution:
Ly = b =⇒ y1 = 2, y2 = −4 + y1 = −2, y3 = 6 − 2y1 = 2
MATH1013 Final Exam, Semester 2, 2021, Page 3 of 12
Backward substitution:
Ux = y =⇒ x3 = y3/2 = 1, x2 = −(y2−2x3)/2 = 2, x1 = (y1−3x2+5x3)/4 = (2−6+5)/4 = 1/4
MATH1013 Final Exam, Semester 2, 2021, Page 4 of 12
Question 3 6 pts
Consider the matrix A and its row echelon form R
A =
−2 0 −1 1
0 2 2 0
0 2 3 3
1 0 0 2
, R =
−2 1 −1 1
0 2 2 0
0 0 1 3
0 0 0 0
.
(a) Find a basis for the null space of A . 3 pts
Solution
Ax = 0 ⇐⇒ Rx = 0,
x1 = −(−3x4 − 3x4 − x4)/2 = 7/2x4
x2 = −x3 = 3x4
x3 = −3x4
x4 = x4
7/2
3
−3
1
(b) Find a basis for the column space of A 3 pts
Solution
A and R are row equivalent. Thus pivot columns will span the column space of A .
−2
0
0
1
,
0
2
2
0
,
−1
2
3
0
MATH1013 Final Exam, Semester 2, 2021, Page 5 of 12
Question 4 6 pts
(a) Let T : R2 → R2 , S : R2 → R3 , R : R3 → R2 be transformations such that
T (x1,x2) = *,x1 − 2x23|x1 | +- , S (x1,x2) =
*...,
3x1 − 2x2
−x1 + 3
6x2
+///- , R (x1,x2,x3) =
*,3x1 − 2x2 + x3−x1 + 3x3 +- .
Which of the transformations T , S,R is linear? Justify your answers. 3 pts
Solution
T and S are not linear, since T (αu) , αT (u) and S (αu) , αS (u)
R is linear since it satises R (u +v ) = R (u) + R (v ) and R (αu) = αR (u) .
(b) Let T : R2 → R3 be a transformation such that
T (x1,x2) =
*...,
4x1 − 7x2
x1 − 2x2
4x1 − 4x2
+///- .
Is b =
1
0
4
in the range of T (x1,x2)? 3 pts
Justify your answer.
Solution
First, formulate as matrix vector operation T (x ) := Ax = b
T (x1,x2) =
4 −7
1 −2
4 −4
xx2
=
1
0
4
Consider the augmented matrix
4 −7 1
1 −2 0
4 −4 4
MATH1013 Final Exam, Semester 2, 2021, Page 6 of 12
Convert the matrix to row echelon form:
4 −7 1
1 −2 0
4 −4 4
( 14 R1→ R1)
( 14 R3→ R3)
−→
1 −74
1
4
1 −2 0
1 −1 1
(R2 − R1→ R2)
(R3 − R1→ R3)
−→
1 −74
1
4
0 −14
−1
4
0 34
3
4
(-4R2→ R2)
((4/3)R3→ R3)
−→
1 −74
1
4
0 1 1
0 1 1
( R3−R2→ R3)
−→
1 −74
1
4
0 1 1
0 0 0
Having
x2 = 1, x1 =
1
4 +
7
4x2 = 2.
MATH1013 Final Exam, Semester 2, 2021, Page 7 of 12
Question 5 10 pts
Suppose that A is a 3×3 matrix which can be reduce to the identity matrix I3 by performing
the following row operations in order
1st R3 is replaced by R3 + α1R2 , α1 , 0.
2nd R1 is replaced by R1 + α2R3 , α2 , 0.
3rd R3 is replaced by α3R3 , α3 , 0.
(a) Determine the elementary matrices E1 , E2 , E3 corresponding to each of the row
operations. 3 pt
Solution
E1 =
1 0 0
0 1 0
0 α1 1
, E2 =
1 0 α2
0 1 0
0 0 1
, E3 =
1 0 0
0 1 0
0 0 α3
.
(b) Show that A−1 = E3E2E1 . 3 pt
Solution
The elementary row operations will correspond to mathrmE3E2E1A = I3 .
A−1 is an (left) inverse of A if A−1A = I3 .
Note that
E3E2E1A = I3 ⇐⇒ A−1A = I3.
We must have A−1 = E3E2E1
(c) Determine the matrix A. 2 pt
Solution
Note that
E3E2E1A = I3 ⇐⇒ A = E−11 E−12 E−13 =
1 0 0
0 1 0
0 −α1 1
1 0 −α2
0 1 0
0 0 1
1 0 0
0 1 0
0 0 1α3
=
1 0 −α2α3
0 1 0
0 −α1 1α3
(d) Determine det(A) and det(A−1) . 2 pt
Solution
MATH1013 Final Exam, Semester 2, 2021, Page 8 of 12
det(A) = det
(
E−11 E−12 E−13
)
= det(E−11 ) det(E−12 ) det(E−13 ) =
1
α3
det(A−1) = det (E3E2E1) = det(E3) det(E2) det(E1) = α3
MATH1013 Final Exam, Semester 2, 2021, Page 9 of 12
Question 6 6 pts
(a) Consider the 3 × 3 matrix
A =
1 1 1
3 0 −1
1 −1 1
.
Use the determinant to show whether Ax = b has a solution for all b ∈ R3 . 3 pts
Solution
det (A) = −3
1 1−1 1
+
1 11 −1
= −8
Since det (A) , 0 A is invertible with the solution x = A−1b for all b ∈ R3 .
(b) Consider a matrix A with the row echelon form
R =
3 −1 0 1
0 2 3 −1
0 0 2 −3
.
i. Are the columns of A linearly in dependent?
ii. Does the system Ax = 0 have nontrivial solutions?
Justify your answers. 3 pts
Solution
The matrices A and R are row equivalent. Since there are 4 columns with only 3
pivots, then the columns of A are linearly dependent.
Correct answers with good reasons gets 1.5marks
Otherwise award partial marks
Since the columns of A are linearly dependent the system Ax = 0 has nontrivial
solutions.
MATH1013 Final Exam, Semester 2, 2021, Page 10 of 12
Question 7 10 pts
(a) Given the complex numbers z1 = 2 + i , z2 = 1 − 3i, express the following expressions
as complex numbers in standard form.
i. z1z2
ii. z1/z2
3 pts
Solution
z1z2 = (2 + i ) (1 − 3i ) = (2 + 3) + i (1 − 6) = 5 − 5i
z1/z2 = (2 + i )/(1 − 3i ) = (2 + i ) (1 + 3i )10 = −
1
10 +
7
10i
(b) Find the 4th roots of z = −i , and express them in polar form. 4 pts
Solution
z = −i = cis (3/2pi )
zk = cis (
(3/2)pi + 2kpi
4 ) = cis (
(3 + 4k )pi
8 ), k = 0, 1, 2, 3
z0 = cis (
3pi
8 ) = cos(
3pi
8 ) + i sin(
3pi
8 ), z1 = cis (
7pi
8 ) = cos(
7pi
8 ) + i sin(
7pi
8 )
z2 = cis (
11pi
8 ) = cos(
11pi
8 ) + i sin(
11pi
8 ), z3 = cis (
15pi
8 ) = cos(
15pi
8 ) + i sin(
15pi
8 )
(c) Consider the fourth degree polynomial f (z) = z4 − 2z3 + 3z2 − 2z + 2.
i. Two roots of f (z) are z0 = −i and z1 = 1 + i . What are the other roots of f (z) ?
ii. Factorise f (z) as the product of two quadratic polynomials.
3 pts
Solution
The roots are conjugate pairs. Therefore with z0 = −i and z1 = 1 + i we must have
z2 = z¯0 = i, z3 = z¯1 = 1 − i
MATH1013 Final Exam, Semester 2, 2021, Page 11 of 12
f (z) = (z − z0) (z − z¯0)︸ ︷︷ ︸
(z2+1)
(z − z1) (z − z¯1)︸ ︷︷ ︸
(z2−2z+2)
= (z2 + 1) (z2 − 2z + 2)
MATH1013 Final Exam, Semester 2, 2021, Page 12 of 12
