MATH2111 Tutorial 1

MATH2111 Tutorial 1
General Information
Teaching Assistant
Leung Ho Ming
Office: Rm 3489
Tel: 3469 2017
Email: [email protected]
Tutorial Homepage
All materials used in the tutorials will be found at http://ihome.ust.hk/~malhm
More problems can be found at Dr. WL Lee’s webpage: http://www.math.ust.hk/~malung/2111.html
Consultation Hours
You will find me outside class in math support center or by appointment (preferably through email).
Math support center hours: http://www.math.ust.hk/~support/mschours_fall_2014.html
Time and Venue
T3A Mo 10:30AM - 11:20AM Rm 4504, Lift 25-26
T5C Tu 12:00PM - 12:50PM Rm 4505, Lift 25-26
T3C We 01:30PM - 02:20PM Rm 1033, LSK Bldg
T5A Fr 09:30AM - 10:20AM Rm 2304, Lift 17-18
T3B Fr 02:00PM - 02:50PM Rm 6591, Lift 31-32
T5B Fr 05:00PM - 05:50PM Rm 4502, Lift 25-26
System of Linear Equations
A system of m linear equations in n variables is in the form


a11 x1 + a12 x2 + · · · + a1n xn


a21 x1 + a22 x2 + · · · + a2n xn
..


.



an1 x1 + an2 x2 + · · · + amn xn
= b1
= b2
,
..
.
= bm
(1)
each of the unknown xi has its power exactly equal to 1 and there is no product xi xj . (x1 , . . . , xn ) = (s1 , . . . , sn ) is
a solution to the system if it is a solution to all m equations simultaneously. A system is said to be inconsistent if it
has no solution, the system is consistent if it has at least one solution.
If solutions can be taken from any n-tuples of real numbers, exactly one of the following is true for a system:
• There is no solution.
• There is exactly one unique solution.
• There are infinitely many solutions.
Prepared by Leung Ho Ming
Homepage: http://ihome.ust.hk/~malhm
1
Augmented Matrices
Given a system of linear equation (1), we can use the coefficients and constant terms to form the augmented matrix


a11 a12 · · · a1n b1
 a21 a22 · · · a2n b2 


 ..
..
..  .
..
..
 .
.
.
. 
.
am1 am2 · · · amn bm

a11
 a21

 ..
 .
a12
a22
..
.
···
···
..
.
am1
am2
···
1. Write down


x 1 + x 5
3x2 − x4


x1 + 2x3
 

b1
a1n
 b2 
a2n 
 

..  is called the coefficient matrix and  ..  is called the constant matrix of the system.
 . 
. 
bm
amn
the augmented matrix of the following system:
=1
=4
= −3
2. Write down the linear system with the following augmented matrix:
2 1 2 3 5
0 −2 0 4 1
Elementary Row Operations
Three kinds of elementary row operations (EROs) can be applied to an augmented matrix without changing the
solutions to the system represented by the augmented matrix:
• Interchange two rows. (ri ↔ rj )
• Multiply one row by a non-zero number. (cri )
• Add a multiple of onw row to a different row. (crj + ri )
A matrix A is row-equivalent to matrix B if we can apply a sequence of EROs to transform A into B. Since each
ERO has a reverse operation, namely ri ↔ rj , 1c ri and −crj + ri respectively, B is also row-equivalent to A if A is
row-equivalent to B.
3. Perform the EROs in the given sequence to the matrix A.


−1 −1 −1
2
3 , 1st: r1 + r2 , 2nd: r1 + r3 , 3rd: −2r2 + r3
A= 1
1
3
5
2
Reduced Row-echelon Form and Gauss-Jordan Elimination
A matrix is in row-echelon form if it satisfies the following three conditions:
• All zero rows are at the bottom.
• The first non-zero entry in each non-zero row (called leading entry) is to the right of other leading 1’s in the
rows above.
Intuitively, row-echelon matrices have their non-zero entries in a staircase shape. An example of row-echelon matrix is


0 1 ∗ ∗ ∗ ∗ ∗
0 0 0 1 ∗ ∗ ∗


0 0 0 0 1 ∗ ∗ ,


0 0 0 0 0 0 1
0 0 0 0 0 0 0
where the *’s represent arbitrary numbers. In addition, a matrix is in reduced row-echelon form (RREF) if it also
satisfies:
• Each leading entry is equal to 1.
• Each leading 1 is the only non-zero entry in its column.
To solve a system of linear equations, our main goal is to apply EROs to transform the augmented matrix into its
RREF. This can be done by the Gaussian Algorithm:
Step 1
If the matrix is the zero matrix, stop.
Step 2
Find the first non-zero column from the left and move a row containing a non-zero entry to the top.
Step 3
Multiply that row by a the reciprocal of the leading entry to create a leading 1.
Step 4
Subtract multiples of that row from other rows, to make the other entries in the same column become zero.
Step 5
Repeat steps 1 − 4 on the remaining columns.
Note that since the RREF of a given matrix is unique, it is not necessary to follow the algorithm strictly. We will end
up in the same RREF regardless of the sequence of EROs we use. The columns containing a leading entry in RREF
are called pivot columns. The variables corresponding to pivot columns are called basic variables, while the others are
called free variables. We can give the solutions of a system by observing the RREF of the augmented matrix:
• If there is a row 0 0 · · · 0 1 , the system is inconsistent and there is no solution.
• Otherwise if the number of basic variables equals to the number of unknowns, we can solve for a unique solution.
• Or if the number of basic variables is less than that of unknowns, there are infinitely many solutions. We assign
parameters to the free variables and solve for the basic variables in terms of the parameters.
4. Is the following system consistent?


=4
x1 + 3x2 − 2x3 + 5x4
2x1 + 8x2 − x3 + 9x4
=9


3x1 + 5x2 − 12x3 + 17x4 = 7
3
5. Solve the following system of equations.


=1
x1 − 2x2 − x3 + 3x4
2x1 − 4x2 + x3
=5


x1 − 2x2 + 2x3 − 3x4 = 4
6. A school has three clubs and each student is required to belong to exactly one club. One year the students
switched club membership as follows:
Club A.
Club B.
Club C.
4
10
7
10
6
10
remain in A,
remain in B,
remain in C,
1
10
2
10
2
10
switch to B,
switch to A,
switch to A,
5
10
1
10
2
10
switch to C.
switch to C.
switch to B.
If the fraction of the student population in each club is unchanged, find each of these fractions.
7. An amusement park charges $7 for adults, $2 for youths, and $0.50 for children. If 150 people enter and pay a
total of $100, find the numbers of adults, youths, and children. [Hint: These numbers are nonnegative integers.]
4