This document contains notes from a Linear Algebra class covering systems of linear equations and matrices. Key topics discussed include linear systems, the matrix representation of linear systems using augmented matrices, Gaussian elimination to solve systems, elementary row operations, echelon forms, homogeneous systems, and issues with numerical stability in solving systems. Examples are provided to illustrate Gaussian elimination, Gauss-Jordan elimination, and determining the solution set of homogeneous systems from the reduced row echelon form.

1. Bachelor of Data
Science (Year 1)
Linear Algebra
Semester -2
MATHEMATICS (103)
8/6/2023
Linear Algebra-2 BDA SEM 2 Prof.A
Choudhury
1

2. Session 1,2
Systems of Linear Equations and Matrices
Introduction to Systems of Linear Equations
Gaussian Elimination
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3. Sections
O Linear Equation in one and many
variables
O Solution set
O Linear Systems
O Graphs of linear systems in one ,two and
three variables
O Matrix representation-augmented matrix
O Consistent and Inconsistent systems of
linear equations
O Homogeneous system of linear equations
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4. Linear systems(1)
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5. Linear systems (2)
(e)
Which of these figures represent (i)
no solution,(ii) infinitely many
solutions and iii) unique solution?
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6. Augmented matrix
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2x1+3x2=4
7x1+6x2=9
A=
2 3
7 6
b=
4
9
Augmented matrix
[A b]
𝐴 𝑏 =
2 3 4
7 6 9
Write the augmented
matrix :
x - y + z =9
y – x = 8
2x + z = 10

7. Gaussian elimination
Example: Three dozens oranges and a dozen bananas
cost $9 while a dozen oranges and a dozen bananas
cost $8.What is the price of a dozen oranges and a
dozen banana?
O Step 1:Formulation of the problem:
3𝑥 + 𝑦 = 9 ; 𝑥 + 𝑦 = 8
O Step 2:
3 1
1 1
𝑥
𝑦 =
9
8
← 𝐴𝑥 = 𝑏(Matrix form)
O Step 3:𝐴𝑢𝑔𝑚𝑒𝑛𝑡𝑒𝑑 𝑚𝑎𝑡𝑟𝑖𝑥 ∶
𝐴 𝑏 =
3 1 9
1 1 8
𝑨 𝒃 𝒉𝒂𝒔 𝒂𝒏 𝒂𝒅𝒅𝒆𝒅 𝒄𝒐𝒍𝒖𝒎𝒏 .
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8. Gaussian elimination
Step 4:Forward elimination:
[A b] =
3 1 9
1 1 8
→
3 1 9
0
2
3
5 R2-(1/3)R1
3𝑥 + 𝑦 = 9 → (𝑒𝑞. 1) Ax=b Ux=c
0𝑥 +
2
3
𝑦 = 5 → (𝑒𝑞. 2)
Step 5: Backward substitution:
Solving the equation(2) we get, 𝑦 = 7.5.Substituting in (1), we get
3𝑥 + 7.5 = 9
3𝑥 = 9 − 7.5
𝑥 = 0.5
Hence,(0.5,7.5) is the solution of the system of linear equations in
x and y.
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9. Elementary row operations
O The basic method for solving a system of linear equations is to
replace the given system by a new system that has the same
solution set but is easier to solve. This new system is generally
obtained in a series of steps by applying the following three
types of operations to eliminate unknowns systematically:
O Multiply an equation through by a nonzero constant.
𝑹𝒊→ 𝒌𝑹𝒊
O Interchange two equations. 𝑹𝒊 ↔ 𝑹𝒋
O Add a multiple of one equation to another. 𝑹𝒋→
𝑹𝒋 + 𝒌𝑹𝒊
Apply elementary row operations on the augmented matrix of
2𝑥 + 𝑦 − 𝑧 = 9
𝑥 + 𝑦 + 2𝑧 = 10
−𝑦 + 3𝑧 = 7
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10. Row echelon form of a matrix
Reduced row echelon form R
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Reduced row echelon form R
Row echelon form

11. Row echelon form
In each part, suppose that the row echelon form of the
augmented matrix of the linear systems are given. State which of
them have a) unique solution; b) no solution & c) infinitely
many solutions. Explain why.
1.
4 6 0
0 1 3
0 0 0
3
−5
0
infinitely
2.
1 −5 1
0 0 0
0 0 0
4
0
0
infinitely
3.
2 1 3
0 0 1
no solution
4.
2 1 −1
0 1 5
0 0 4
9
11
9
unique solution
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12. Elementary row operations
𝑥 + 𝑦 + 2𝑧 = 9 𝐴 𝑏 =
1 1 2
2 4 −3
3 6 −5
9
1
0
=
1 1 2
0 2 −7
0 3 −11
9
−17
−27
=
1 1 2
0 2 −7
0 0 −1/2
9
−17
−3/2
2𝑥 + 4𝑦 − 3𝑧 = 1
3𝑥 + 6𝑦 − 5𝑧 = 0 R3=R3-(3/2)R2 [U c]
Row echelon form:
1 1 2
0 1
−7
2
0 0
−1
6
9
−17
2
−1
2
-----------Gaussian elimination -1/2 z= -3/2 ; z=3
2y-7z=-17; 2y-21=-17 ;2y=4, y=2
Reduced row echelon form R:
1 0 0
0 1 0
0 0 1
1
2
3
-----Gauss-Jordan elimination
𝒙 = 𝟏; 𝒚 = 𝟐; 𝒛 = 𝟑
Gaussian elimination: In this procedure, there is a forward phase in which zeros are introduced below the leading
1’s.
Gauss-Jordan elimination: This algorithm has two phases; a forward phase in which zeros are introduced below
the leading 1’s and a backward phase in which zeros are introduced above the leading 1’s.
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13. Gauss- Jordan elimination
𝑥1 + 3𝑥2 − 2𝑥3 + +2𝑥5 = 0
2𝑥1 + 6𝑥2 − 5𝑥3 − 2𝑥4 + 4𝑥5 − 3𝑥6 = -1
5𝑥3 + 10𝑥4 + 15𝑥6 = 5
2𝑥1 + 6𝑥2 + 8𝑥4 + 4𝑥5 + 18𝑥6 = 6
Solution :
𝒙 =
𝑥1
𝑥2
𝑥3
𝑥4
𝑥5
𝑥6
= 𝑟
−3
1
0
0
0
0
+ 𝑠
−4
0
−2
1
0
0
+ 𝑡
−2
0
0
0
1
0
+
0
0
0
0
0
1/3
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15. Homogeneous Linear Systems
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16. Special case of a homogeneous
system of linear equations.
O a1x+ b1y = 0 ; a2x+b2y=0
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17. Theorems
Theorem 1: If a homogeneous system has n unknowns, and if
the reduced row echelon form R of its augmented matrix has r
nonzero rows, then the system has n-r free variables.
Theorem 2: A homogeneous linear system with more
unknowns than equations has infinitely many solutions.
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18. Homogeneous Linear Systems
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• What is the row reduced
echelon form of a square
invertible matrix?

19. Round Off Errors and
Instability
There is often a gap between mathematical theory and its
practical implementation.
O Gauss–Jordan elimination and Gaussian elimination are
good examples.
O Computers generally approximate numbers, thereby
introducing round off errors, so unless precautions are
taken, successive calculations may degrade an answer
to a degree that makes it useless.
O Algorithms (procedures) in which this happens are called
unstable.
O Gauss–Jordan elimination involves roughly 50% more
operations than Gaussian elimination.
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20. References:
1) Howard Anton (2014). Elementary Linear Algebra, 11 th Edition.
Wiley Publication.
2) Gilbert Strang (2016 ).Introduction to Linear Algebra , 5th Edition ,
Wellesley Publishers
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