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MAST10007 Linear Algebra
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Linear Algebra

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MAST10007 Linear Algebra

These notes have been made in accordance with the provisions of Part VB of the copyright act for teaching purposes of the
University. These notes are for the use of students of the University of Melbourne enrolled in MAST10007 Linear Algebra.
1
Topic 1: Linear equations [AR 1.1 and 1.2]
1.1. Systems of linear equations. Row operations.
1.2. Reduction of systems to row-echelon form.
1.3. Reduction of systems to reduced row-echelon form.
1.4. Consistent and inconsistent systems.
2
1.1 Systems of linear equations. Row operations
Linear equations
Definition (Linear equation and linear system)
A linear equation in n variables, x1, x2, . . . , xn, is an equation of the form
a1x1 + a2x2 + · · ·+ anxn = b,
where a1, . . . , an and b are constants.
A finite collection of linear equations in the variables x1, x2, . . . , xn is
called a system of linear equations or a linear system.
Example
x1 + 5x2 + 6x3 = 100
x2 − x3 = −1
−x1 + x3 = 11
3
Definition (Homogeneous linear system)
A system of linear equations is called homogeneous if all the constants
on the right hand side are zero.
Example
x1 + 5x2 + 6x3 = 0
x2 − x3 = 0
−x1 + x3 = 0
Definition (Solution of a system of linear equations)
A solution to a system of linear equations in the variables x1, . . . , xn is a
set of values of these variables which satisfy every equation in the
system.
4
Example 1. Data fitting using a polynomial
Find the coefficients c , α, β, γ such that the cubic polynomial
y = c + αx + βx2 + γx3 passes through the points specified below.
x y
−0.1 0.90483
0 1
0.1 1.10517
0.2 1.2214

Solution:
Substituting the points (x , y) into the cubic polynomial gives:
0.90483 = c − 0.1α + 0.01β − 0.001γ
1 = c
1.10517 = c + 0.1α + 0.01β + 0.001γ
1.2214 = c + 0.2α + 0.04β + 0.008γ
5
So c = 1.
We can rewrite the system as 3 equations in 3 unknowns.
−0.1α+ 0.01β − 0.001γ = −0.09517
0.1α + 0.01β + 0.001γ = 0.10517
0.2α + 0.04β + 0.008γ = 0.2214
Note:
To solve these equations by hand is tedious, so we can use a computer
package such as MATLAB.
6
Example 2. (Revision)
Solve the following linear system.
2x − y = 3
x + y = 0
Solution:
Graphically
Solution is the intersection of the lines: (x , y) = (1,−1).
Note:
◮ Need accurate sketch.
◮ Not practical for three or more variables.
7
Elimination
2x − y = 3 (1)
x + y = 0 ⇒ y = −x (2)
Substituting (2) in (1) gives
2x + x = 3 ⇒ 3x = 3 ⇒ x = 1
Substituting into (2) gives y = −1.
Note:
Elimination of variables will always give a solution, but we need to do
this systematically and not in an adhoc manner, for three or more
variables.
8
Definition (Coefficient matrix and augmented matrix)
The coefficient matrix for a linear system is the matrix formed from the
coefficients in the equations.
The augmented matrix for a linear system is the matrix formed from the
coefficients in the equations and the constant terms. These are
separated by a vertical line.
Example 3.
Write down an augmented matrix for the following linear system:
2x − y = 3
x + y = 0
Solution: [
2 −1 3
1 1 0
]
9
Row Operations
To find the solutions of a linear system, we perform row operations to
simplify the augmented matrix. An essential condition is that whichever
row operations we perform, we can recover the solutions to the original
system from the new matrix.
Definition (Elementary row operations)
The elementary row operations are:
1. Interchanging two rows.
2. Multiplying a row by a non-zero constant.
3. Adding a multiple of one row to another row.
Note:
The matrices produced after each row operation are not equal but are
equivalent, meaning that the solution is the same for the system
represented by each augmented matrix. We use the symbol ∼ to denote
equivalence of matrices.
10
Example 4.
Solve the following system using elementary row operations:
2x − y = 3
x + y = 0
Solution:[
2 −1 3
1 1 0
]
R2 ↔ R1 ∼
[
1 1 0
2 −1 3
]
R2 → R2 − 2R1
∼
[
1 1 0
0 −3 3
]
R2 → −13R2
∼
[
1 1 0
0 1 −1
]
R1 → R1 − R2
∼
[
1 0 1
0 1 −1
]
So x = 1, y = −1.
11
1.2 Reduction of systems to row-echelon form
Definition (Leading entry)
The leftmost non-zero entry in each row of the matrix is called the
leading entry.
Definition (Row-echelon form)
A matrix is in row-echelon form if:
1. For any row with a leading entry, all elements below that entry and
in the same column as it, are zero.
2. For any non-zero two rows, the leading entry of the lower row is
further to the right than the leading entry in the higher row.
3. Any row that consists solely of zeros is lower than any row with
non-zero entries.
12
Examples[
1 −2 3 4 5 ] row-echelon form

 1 0 0 30 4 1 2
0 0 0 3

 row-echelon form


0 0 0 2 4
0 0 3 1 6
0 0 0 0 0
2 −3 6 −4 9

 not row-echelon form
13
Gaussian elimination
Gaussian elimination is a systematic way to reduce a matrix to
row-echelon form using row operations.
Note:
The row-echelon form obtained is not unique.
Gaussian elimination
1. Interchange rows, if necessary, to bring a non-zero number to the
top of the first column with a non-zero entry.
2. Add suitable multiples of the top row to lower rows so that all
entries below the leading entry are zero. (Multiplying a row by a
constant is also allowed and is often useful.)
3. Start again at Step 1 applied to the matrix without the first row.
To solve a linear system we read off the equations from the row-echelon
matrix and then solve the equations to find the unknowns, starting with
the last equation. This final step is called back substitution.
14
Example 5.
Solve the following system of linear equations using Gaussian
elimination:
x − 3y + 2z = 11
2x − 3y − 2z = 13
4x − 2y + 5z = 31
Solution:
Step 1. Write the system in augmented matrix form.
 1 −3 2 112 −3 −2 13
4 −2 5 31


15
Step 2. Use elementary row operations to reduce the augmented matrix
to row-echelon form.
 1 −3 2 112 −3 −2 13
4 −2 5 31

 R2 → R2 − 2R1
R3 → R3 − 4R1
∼

 1 −3 2 110 3 −6 −9
0 10 −3 −13

 R2 → 13R2
∼

 1 −3 2 110 1 −2 −3
0 10 −3 −13


R3 → R3 − 10R2
∼

 1 −3 2 110 1 −2 −3
0 0 17 17


16
Step 3. Read off the equations from the augmented matrix and use back
substitution to solve for the unknowns, starting with the last equation.
This gives us three equations:
17z = 17 ⇒ z = 1
y − 2z = −3 ⇒ y = 2− 3 = −1
x − 3y + 2z = 11 ⇒ x = −3− 2 + 11 = 6
17
1.3 Reduction of systems to reduced row-echelon form
Definition (Reduced row-echelon form)
A matrix is in reduced row-echelon form if the following three conditions
are satisfied:
1. It is in row-echelon form.
2. Each leading entry is equal to 1 (called a leading 1).
3. In each column containing a leading 1, all other entries are zero.
18
Examples[
1 −2 3 −4 5 ] reduced row-echelon form
[
1 2 0
0 0 1
]
reduced row-echelon form

 1 0 0 30 1 1 2
0 0 0 3

 is not in reduced row-echelon form


1 0 0 2 4
0 1 0 1 6
0 0 0 0 0
0 0 1 −4 9

 is not in reduced row-echelon form
19
Gauss-Jordan elimination
Gauss-Jordan elimination is a systematic way to reduce a matrix to
reduced row-echelon form using row operations.
Note:
The reduced row-echelon form obtained is unique.
Gauss-Jordan elimination
1. Use Gaussian elimination to reduce matrix to row-echelon form.
2. Multiply each non-zero row by an appropriate number to create a
leading 1 (type 2 row operations).
3. Use row operations (of type 3) to create zeros above the leading
entries.
20
Example 6.
Solve the following system of linear equations using Gauss-Jordan
elimination:
x − 3y + 2z = 11
2x − 3y − 2z = 13
4x − 2y + 5z = 31
Solution:
Step 1. Use Gaussian elimination to reduce the augmented matrix to
row-echelon form.
From example 5, we have:
 1 −3 2 112 −3 −2 13
4 −2 5 31

∼

 1 −3 2 110 1 −2 −3
0 0 17 17


21
Step 2. Divide rows by their leading entry.
 1 −3 2 110 1 −2 −3
0 0 17 17


R3 → 117R3
∼

 1 −3 2 110 1 −2 −3
0 0 1 1


Step 3. Add multiples of the last row to the rows above to make the
entries above its leading 1 equal to zero.

 1 −3 2 110 1 −2 −3
0 0 1 1

 R1 → R1 − 2R3R2 → R2 + 2R3 ∼

 1 −3 0 90 1 0 −1
0 0 1 1


22
Step 4. Add multiples of the penultimate (2nd last) row to the rows
above to make the entries above its leading 1 equal to zero.
 1 −3 0 90 1 0 −1
0 0 1 1

 R1 → R1 + 3R2 ∼

 1 0 0 60 1 0 −1
0 0 1 1


Step 5. Read off the answers.
The solution is x = 6, y = −1, z = 1.
23
1.4 Consistent and inconsistent systems
Theorem
A system of linear equations has zero solutions, one solution, or
infinitely many solutions. There are no other possibilities.
We say that the system is:
◮ inconsistent if it has no solutions.
◮ consistent if it has at least one solution.
Note:
Every homogeneous linear system is consistent, since it always has at
least one solution: x1 = 0, . . . , xn = 0.
24
Inconsistent systems
We can determine whether a system is consistent or inconsistent by
reducing its augmented matrix to row-echelon form.
Theorem
The system is inconsistent if and only if there is at least one row in the
row-echelon matrix having all entries equal to zero except for a non-zero
final entry.
Example 7.
Solve the system with row-echelon form:
 1 0 1 −20 2 2 4
0 0 0 −3


Solution:
The third equation says 0x1 + 0x2 + 0x3 = −3.
So the system is inconsistent with no solution.
25
Geometrically, an inconsistent system in 3 variables has no common
point of intersection for the planes determined by the system.
26
Consistent systems
Unique solution
For a consistent system with n variables, a unique solution exists when
the row reduced coefficient matrix and augmented matrix has n
non-zero rows. We can read off the solution from the reduced
row-echelon form of the matrix.
Example 8.
Solve the system with reduced row-echelon form:
 1 0 0 20 1 0 −4
0 0 1 15


Solution:
Reading off the answers, we get
x1 = 2, x2 = −4, x3 = 15.
27
Infinitely many solutions
Example 9. Calculating flows in networks
At each node • we require
sum of flows in = sum of flows out.
Determine a, b, c and d in the following network.
Case category
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COMP2013 IB9X60 INVP001 EMATM0051 COMP220 BLAW1002 E20 TCP8001 ECON61001 EART40005 ECONM1022 MATH 365 MATH 367 ELEC6218 COMP 5812M EDUC 5061M ELEC ENG 1101 equation MATH6174 ST334 C4 EC9570 A33648 EBUS603 ECON47815 stata WFMS0001 PEM102 INF6060 ionically ELECTRICAL MATH 375 MMW 12 MKTG860 ECON60401 ALY6010 SOST20131 Introduction COM3524 MME2 GGR305 JNB 538 EECS101 COMM5030 IEOR E4601 2B CSC336 EN2315 LAW 432 CSC263 MATH2001 OLET1143 CSCB63 SIT772 COMM1170 STAT 441 MS930 INFT3050 GLBH0030 Stat 120B MAST10007 MGEC61 Machine GGR 252H1S ECON7030 Financial Marketing MGMT30019 ECON5120 SDSC 8009 CSC165H1S DSCI 551 3081W ACM41000 ECA 5304 MATH 1064 ECSE444 DATA 604 COMP7105 MIE1615 ECON4004 ENV200H1S EC481 MKT 306 EC1B1 MATH08051 ALY6140 CS260C CSE30 MSIN0226 CS4103 CSC3064 AM16 COMP11212 CSE4101 PSTAT 174 Math 380W CNT 5106C STAT 310 FIN2028 CS5014 STAT 425 COMP 4102A ELEC2140 MANF9544 QRM 9 QRM 8 ECON 457 COMPSCI 4091 MATH 523 DNSC-4211 PEAS 6000 MSIN0180 COMP2119 A2A HT 2022 IB93F0 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COMSM0093 COMP281 AFM 333 IBUS1001 CP2501 Math 105A VE320 TEC101 SOCY2166 COM240 EFIM20033 ENGI4467 EFIM20034 EC 979 BE432 7EJ502 CSSE6400 COMP90025 Econ 312 ENG2031 ECO0006 EXERSCI 207 CAN209 ECO 137W FT312 ECO 2199 MATH32052 BDW4213 APS 360 SAS 6 CHME0017 PO92Q Economics 01AV MATH39512 MATH 5486 COMP8620 ISE102 ECON235 ECON 206 FI 345 PHAY0020 MATH0099 MATH0085 EFB106 PHY 134 MAE115 circuits Chemistry CSE12 MATH-UA 233 PHAS0038 COMP2140 CONMGNT 7049 COMP151101 ECON 485 ACX5903 ENGN6536 ARCH1162 Law PHAS0042 CCF1C03 MATH 351 Phys 139 AG942 MGT001371 APH106 ECON1051 MA4AM MA3Z7 NUMBER THEORY REST0006 SESS0026 STAT0023 TABL 2741 CMP4008Y EFIM10015 Acct 560 PSYC735 GEOG 5 ACCT3104 CSEC5616 PHIL1012 ITLS6111 ECON8022 5CCM226A compX123 MA4XJ R001 EWB LLP120 COMP643 ENG 103 ECON 23950 ACCT 207 FINC 616 Bio99 Math 137A Psychology BCPM0054 MATH20212 EC3450 CENV6141 PSY 205 CE335 CS365 Calculus ACSD INFT6800 MCIT INFO6030 PHYS 111LV H63EEJ ECON2151 ENGI 2181 MTRX1701 PLIT08009 CS240 Math 270 CMP7000A CMP-7038B Math 263 LAW3016 CS371 STAT 334 FIN0008 ACC3004 STAT 371 ​COMP1921 MTH201 ACCT2000 UN3412 CIV6000 MATHS 3021 MATE7014 ECON1 CIBC6035 LAW121G COMM 321 CSC8204R TTM2033 CSCI 2600 MAE 115 LNG302 CENV6175W1 ECON30019 CPS2015 BST969 DDES9015 AMATH242 CS 1151 N1592 ACCT600 EG503G SMM519 EC318 Ecstatistics EEE 471 MARK2051 Topology COM00037H ACCT3000 MTHM501 EG501V COM00055H BUSN9085 GGR273 MATH 2B ITD102 MA214 ECON0060 ECE3114 ECMM447 Mechanics CCT361H5S BSAN3210 WELF2001 Fin3001 RESP5001 STA 137 Linguistics 101 CSCI368 CMPSC497 LUBS3655 OMBA 5355 BUSM2562 ​BFF5902 CYBR7001 CULT2005 ECON 7720 MTRX3760 FSE598 ​LUE1001 MGSC 7100 CS3240 COSC7500 ACCT3091 INFOGR TRAN 2080 DESIGN CIVE215001 CIVE2360 CPSC 8430 FNCE2003 Stoichiometry INFS6023 Quantitative Method Math 21D ECE5179 Design ​ECON8022 Chemical DMS 213 ADM1370 BFF5902 BFF2701 AGRI10051 AMN400 ECON7322 ​FINC5090 ​MATH1052 ​SciComp Elec4622 ​FINS5516 ENG 101 MMM143 CS431CA2 BSAN2201 EBSC7070 ES9v9-10 PLIN0009 DATA 8 MATH 101 BCPM0097 ​MAT 237 BISM2201 EEE8099 Physics 11 ECE2238B Biostat 234 ​ECON30009 INF10025 CHEM252 CCMA4008 CIVE 3007 STAT0031 ELEC ENG 3108 MGMT90140 CHEM 252 STAT3016 MMM095 ​MMGT6016 Legal STAT 311 ELEC9764 COMP 370 TCS 3053 EECS 1021 FIT1051 C/C++ EE3C COMP2003J COM398 COMP51915 Simulator FIN42330 8PRO102 PGD ELEC5160 XJCO1921 COMP5123M INFS 2042 CHE1148H ECA5313 IN3329 Microsoft Modeling B15 2TT COMP3687 COMP2432 DMS2030 KAN4TSF CFKG CSE 5322 CSE 445 STATS 3DA3 CHC5223 COMP1048 EL3105 EECS 3221 CS231 ME 4434 ITP4510 ITP4501 CSCI 5708 Controller EECE 5699 CPEN 231L CPN programs CSC8016 MATU9D2 ENGN4213 2. SVM. (10 points) Consider an SVM classifier using RBF kernel, and finish the following python code to search for the optimum kernel parameter γ. You are required to calculate the RBF function and search yourself, instead of using functions in Scikit-Learn (sklearn). def parameter search(gamma list ,Xtr,ytr,Xts,yts ,...): """ Parameters gamma list: A list of RBF kernel parameter gamma to search for Xtr: Training data, shape (ntr,nrow,ncol) ytr: Training labels, shape (ntr,) Xts: Test data, shape (nts,nrow,ncol) yts: Test labels, shape (nts,) ... Add other parameters as needed Returns: gamma optimum: The gamma with minimum error on Xts, yts """ ... return gamma optimum
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