MATH 1003 Calculus and Linear Algebra (Lecture 6)
Albert Ku
HKUST Mathematics Department
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Outline
1 Matrices
2 Linear Systems and Augmented Matrices
3 Row Operations
4 Solving Linear Systems by Row Operations
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Matrices
Matrices
Definition
A matrix is a rectangular array of numbers written within brackets.
Example
3 7
1 −5 7
A= , B = 6.8 √
2.3 .
0 1/2 − 31 1
2 2
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Matrices
Some Definitions
Definition
Each number in a matrix is called an element of the matrix.
If a matrix has m rows and n columns, it is called an m × n matrix.
A matrix with n rows and n columns is called a square matrix of order
n.
The position of an element in a matrix is given by the row and the
column containing the element. The element on the i th row and j th
column is denoted by aij .
The principal diagonal of a matrix A consists of the element
a11 , a22 , a33 , . . ..
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Linear Systems and Augmented Matrices
Linear Systems and Augmented Matrices
Consider the following linear system:
(
3x1 + 4x2 =1
x1 − 2x2 =7
We will represent it as an augmented matrix as follows:
3 4 1
1 −2 7
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Linear Systems and Augmented Matrices
General Case
In general, associated with each linear system of the form
(
a11 x1 + a12 x2 = k1
a21 x1 + a22 x2 = k2
where x1 and x2 are variables, is the augmented matrix of the system:
a11 a12 k1
.
a21 a22 k2
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Row Operations
Row Operations
When using the method of elimination to solve a linear system, we usually
need to multiply an equation by a non-zero constant and then add the
equation to another one. Here we want to consider such operations on the
corresponding augmented matrix. The following are the three basic types
of operations:
Two rows are interchanged:
Ri ↔ Rj means interchanging i th and j th rows.
A row is multiplied by a non-zero constant:
kRi → Ri means multiplying i th row by a nonzero constant k.
A constant multiple of one row is added to another row:
Ri + kRj → Ri means that i th row is added by k multiple of j th row.
Notice that the above operations are called the row operations.
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Row Operations
An Example
3 4 1
Consider the augmented matrix :
1 −2 7
3 4 1 R1 ↔R2 1 −2 7
−→
1 −2 7 3 4 1
3 4 1 2R2 →R2 3 4 1
−→
1 −2 7 2 −4 14
3 4 1 R1 +2R2 →R1 5 0 15
−→
1 −2 7 1 −2 7
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Row Operations
Properties of Row Operations
All row operations are reversible.
If A and B are augmented matrices such that one can be transformed
to another by a sequence of row operation(s), then A and B are said
to be equivalent.
If two linear systems have equivalent augmented matrices, both
systems have the same set of solution(s).
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Solving Linear Systems by Row Operations
Solving Linear Systems by Row Operations
The aim of using row operations is to reduce the augmented matrix to one
of the following simple forms:
1 0 m
(A) , where m and n are real numbers.
0 1 n
1 m n
(B) , where m and n are real numbers.
0 0 0
1 m n
(C) , where p, m and n are real numbers and p 6= 0.
0 0 p
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Solving Linear Systems by Row Operations
Conclusion
1 A linear system has a unique solution if its augmented matrix reduces
to Form A.
2 A linear system has infinitely many solutions if its augmented matrix
reduces to Form B.
3 A linear system has no solution if its augmented matrix reduces to
Form C.
Why? (Hint: consider the systems corresponding such basic forms.)
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Solving Linear Systems by Row Operations
Examples
Example
Solve the following systems of linear equations:
(
3x1 + 4x2 = 1
(a)
x1 − 2x2 =7
(
2x1 − x2 =4
(b)
−6x1 + 3x2 = −12
(
2x1 + 6x2 = −3
(c)
x1 + 3x2 =2
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Solving Linear Systems by Row Operations
Solution to (a)
3 4 1 R1 ↔R2 1 −2 7 R2 +(−3)R1 →R2 1 −2 7
−→ −→
1 −2 7 3 4 1 0 10 −20
1
R →R2
10 2 1 −2 7 R1 +2R2 →R1 1 0 3
−→ −→
0 1 −2 0 1 −2
The solution is x1 = 3, x2 = −2. The system is independent consistent.
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Solving Linear Systems by Row Operations
Solution to (b)
1 − 21
1
2 −1 4 R2 +3R1 →R2 2 −1 4 R →R1
2 1 2
−→ −→
−6 3 −12 0 0 0 0 0 0
There are infinitely many solutions. The system is dependent consistent.
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Solving Linear Systems by Row Operations
Solution to (c)
2 6 −3 R1 ↔R2 1 3 2 R2 +(−2)R1 →R2 1 3 2
−→ −→
1 3 2 2 6 −3 0 0 −7
There is no solution. The system is inconsistent.
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