MATH1013 — Mathematics and Applications 1 Algebra
Mathematics and Applications 1 Algebra
Hello, dear friend, you can consult us at any time if you have any questions, add WeChat: THEend8_
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.