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Parul University
Faculty of Engineering & Technology
Department of Applied Sciences and Humanities
1st Year B
...
These number or
functions are called entries or elements of the matrix
...
The trace of 𝐴 is undefined if𝐴 is not a square matrix
...

6

Symmetric matrix: - For any square matrix A, if 𝐴 = 𝐴𝑇 ,then it is known as symmetric matrix
...


Skew-symmetric matrix: - For any square matrix A, if 𝐴 = −𝐴𝑇 then it is known as Skew
symmetric matrix
...


−3
0
−7

4
7] = −𝐴𝑇
0

Singular and non-singular matrix: For any square matrix A, if|𝐴| ≠ 0, then it is known as non-singular matrix and if |𝐴| = 0 then it
is known as singular matrix
...
i
...
𝐴𝐴𝑇 = 𝐼
Example: Given𝐴 is an orthogonal matrix because
𝐴=[

−1
0

0
−1
] Then 𝐴𝑇 = [
1
0

0
−1
] and 𝐴𝐴𝑇 = [
1
0

0 −1
][
1 0

0
1 0
]=[
]=𝐼
1
0 1

System of linear equation
Linear Equations: Any straight line in the 𝑥𝑦-plane can be represented algebraically by
equation of the form 𝑎𝑥 + 𝑏𝑦 = 𝑐, where 𝑎&𝑏 are real numbers
...

A linear system of m linear equations in n variables: An arbitrary system of m linear equations in
n variables 𝑥1 , 𝑥2 , 𝑥3 , …
...
No solutions, or
2
...
Infinitely many solutions

Geometrical representation

Exactly one solution

No solution

Infinitely many solutions

Notes
(i) The system is said to be consistent if we get infinitely many solutions or unique solution
...


Augmented matrix
A system of 𝑚 equations in 𝑛 unknowns can be abbreviated by writing only the rectangular array
of numbers
...

For example: Find the augmented matrix for each of the following system of linear equations:
2𝑥1 + +2𝑥3 = 1
3𝑥1 − 𝑥2 + 4𝑥3 = 7
6𝑥1 + 𝑥2 − 𝑥3 = 0
2
Then, augmented matrix is given by[3
6

0
−1
1

2 |1
4 |7]
...

1 0 2| 1
For example: [0 1 4|7]
𝟎 𝟎 𝟎| 𝟒
(2) If at least one of the columns on the left of the augmentation bar has zero element pivot entry,
then the system has infinitely many solutions
...

1 0 2| 1
For example: [0 1 4|7]
𝟎 𝟎 𝟏| 𝟒

Row-Echelon (RE)form and Row-Reduced Echelon (RRE) formof a matrix
Definition: A rectangular matrix is in row-echelon form (or echelon form) if it has the following
three properties:
1
...

2
...

3
...

If the matrix satisfies the 4th property (i
...
, In each column except leading 1 if all entries are
zero) then row-echelon form (RE form) becomes row-reducedechelon form (RRE form)
...

𝑥+𝑦+𝑧 =6
𝑥 + 2𝑦 + 3𝑧 = 14
−2𝑏 + 3𝑐 = 1
2𝑥 + 4𝑦 + 7𝑧 = 30
3𝑎 + 6𝑏 − 3𝑐 = −2
Solution:
6𝑎 + 6𝑏 + 3𝑐 = 5

The matrix form of the given system is
1
[1
2

1 1 𝑥
6
2 3] [𝑦] = [14]
4 7 𝑧
30

The augmented matrix is
1 1
[𝐴|𝐵] = [1 2
2 4

1| 6
3|14]
7|30

Now, to convert the given augmented matrix in
row-echelon form we apply elementary
operations as following
...

𝑥 = 0, 𝑦 = 4, 𝑧 = 2 is unique solution of given

2
−2
6

1
[0
0

2
−2
−6

−1|−2/3
3| 1 ]
9| 9

𝑅3 → 𝑅3 − 3𝑅2

system
...

𝑎 + 2𝑏 − 𝑐 = −2/3
4𝑥 − 2𝑦 + 6𝑧 = 8
−2𝑏 + 3𝑐 = 1
𝑥 + 𝑦 − 3𝑧 = −1
15𝑥 − 3𝑦 + 9𝑧 = 21
0𝑎 + 0𝑏 + 0𝑐 = 6 is not possible
...

The matrix form of the given system is
4 −2 6 𝑥
8
𝑦
[1
]
[
]
[
=
1 −3
−1]
15 −3 9 𝑧
21

Example 4:
Solve the following system by gauss elimination
method
...

Since 𝑡 is arbitrary real number, The system has
infinitely many solutions
...


1 −3 −4|−30
[0 11 11 | 99 ]
0 5 10 | 70

𝑥+𝑦+𝑧 =6

Solution: The Augmented matrix is
1
[1
1

1
2
2

1| 6
3|10]
𝜆| 𝜇

𝑅2 → 𝑅2 − 𝑅1 , 𝑅3 → 𝑅3 − 𝑅1
1
[0
0

1
1 | 6
1
2 | 4 ]
1 𝜆 − 1|𝜇 − 6
𝑅3 → 𝑅3 − 𝑅2

𝑅2 → (

1
1
) 𝑅2 , 𝑅3 → ( ) 𝑅3
11
5

1 −3 −4|−30
[0 1
1| 9 ]
0 1
2 | 14
𝑅3 → 𝑅3 − 2𝑅2
1 −3 −4|−30
[0 1
1| 9 ]
0 0
1| 5

1
[0
0

1
1
0

1 | 6
2 | 4 ]
𝜆 − 3|𝜇 − 10

The corresponding system of equations is
𝑢 − 3𝑣 − 4𝑤 = −30
𝑣+𝑤 =9

(i) If 𝜆 − 3 = 0 and 𝜇 − 10 = 0, that is if 𝜆 =
3 and 𝜇 = 10 then the system has infinitely
many solutions
...
That is 𝜆 ≠ 3 and 𝜇 can possess any
real value
...

2
4
5
solution of the system
...

(1) 𝑥 + 𝑦 + 2𝑧 = 9

Ans:
𝑥 = 1, 𝑦 = 2, 𝑧 = 3

2𝑥 + 4𝑦 − 3𝑧 = 1
3𝑥 + 6𝑦 − 5𝑧 = 0

(2)3𝑥 + 𝑦 − 3𝑧 = 13
2𝑥 − 3𝑦 + 7𝑧 = 5

Ans: No solution as the
augmented matrix in
row-echelon form is

2𝑥 + 19𝑦 − 47𝑧 = 32

1 1/3
−1 |13/3
[0 1
−27/11| 1 ]
0 0
0 | 5

(3) 2𝑥 + 2𝑦 + 2𝑧 = 0

Ans: Infinitely many
solutions
...


−3𝑘−1 1−4𝑘
7

,

7

, 𝑘)/

Examples: Solve the following system using Gauss-Jordan Method
Case-1: Unique Solution
Case-2: Infinitely many Solutions
(1) 𝑥 + 𝑦 + 2𝑧 = 8

(2) 𝑥 + 2𝑦 − 3𝑧 = −2

−𝑥 − 2𝑦 + 3𝑧 = 1

3𝑥 − 𝑦 − 2𝑧 = 1

3𝑥 − 7𝑦 + 4𝑧 = 10

2𝑥 + 3𝑦 − 5𝑧 = −3

Solution: The matrix form of the system is
1
[−1
3

1
−2
−7

Solution:The matrix form of the system is

2 𝑥
8
3] [𝑦] = [ 1 ]
4 𝑧
10

The augmented matrix is
1
1 2| 8
[𝐴|𝐵] = [−1 −2 3| 1 ]
3 −7 4|10
𝑅2 → 𝑅2 + 𝑅1 , 𝑅3 → 𝑅3 − 3𝑅1
1
1
2| 8
[0 −1
5| 9 ]
0 −10 −2|−14
𝑅2 → (−1)𝑅2
1
1
2| 8
[0
1
−5| −9 ]
0 −10 −2|−14

1
[3
2

2
−1
3

−3 𝑥
−2
−2] [𝑦 ] = [ 1 ]
−5 𝑧
−3

The augmented matrix is
1
[𝐴|𝐵] = [3
2

2
−1
3

−3|−2
−2| 1 ]
−5|−3

𝑅2 → 𝑅2 − 3𝑅1 , 𝑅3 → 𝑅3 − 2𝑅1
1
[0
0

2
−7
−1

−3|−2
7|7]
1|1

𝑅2 → (−1/7)𝑅2
1
[0
2

2
1
−1

−3|−2
−1|−1]
1|1

𝑅3 → 𝑅3 + 10𝑅2
1 1
[0 1
0 0

𝑅3 → 𝑅3 + 𝑅2

2 | 8
−5 | −9 ]
−52|−104

1
[0
0

𝑅3 → (−1/52)𝑅3
1
[0
0

1 2|8
1 −5|−9]
0 1|2

2
1
0

−3|−2
−1|−1]
0|0

𝑅1 → 𝑅1 − 2𝑅2
1
[0
0

0
1
0

−1| 0
−1|−1]
0|0

𝑅2 → 𝑅2 + 5𝑅3 , 𝑅1 → 𝑅1 − 2𝑅3
1
[0
0

1
1
0

0|4
0|1]
1|2

The corresponding system of equations gives
𝑥−𝑧=0
𝑦 − 𝑧 = −1

𝑅1 → 𝑅1 − 𝑅2
1
[0
0

0
1
0

0|3
0|1]
1|2

Assigning the free variable 𝑧 an arbitrary value 𝑡,
𝑧=𝑡
𝑥=𝑧=𝑡

The corresponding system of equation is

𝑦=𝑧−1=𝑡−1
𝑥 = 3, 𝑦 = 1, 𝑧 = 2 which is a unique solution of
the given system of equations
...

3𝑥 − 𝑦 − 𝑧 = 4
𝑥 + 5𝑦 + 5𝑧 = −1

Exercise: Solve the following system
equations by using Gauss- Jordan method
...
The solution
set is {(1, 𝑘, 𝑘)/ k
R}
...

This shows that the system has no solution
...
Solve the system of linear equations using Gauss-Jordan method (Winter 2017)
𝑥 + 2𝑦 + 𝑧 = 5; −𝑥 − 𝑦 + 𝑧 = 2; 𝑦 + 3𝑧 = 1
Solution: Matrix form of the system is
1
2 1 𝑥
5
[−1 −1 1] [𝑦] = [2]
0
1 3 𝑧
1
1
2 1
Let 𝐴 = [−1 −1 1]
0
1 3

𝑥
5
𝑋 = [𝑦] 𝐵 = [2]
𝑧
1
1
2
[𝐴|𝐵] = [−1 −1
0
1
𝑅2 → 𝑅2 + 𝑅1

15
12]
31

1 2 15
~ [0 1 27]
0 1 31
𝑅1 → 𝑅1 − 2𝑅2
𝑅3 → 𝑅3 − 𝑅2
1
~ [0
0

0 −3−9
1 2 7]
0 1 −6

𝑅1 → 𝑅1 + 3𝑅3
𝑅2 → 𝑅2 − 2𝑅3
1 0 0−27
~ [0 1 0 19 ]
0 0 1 −6
∴ The solution of the system is 𝑥 = −27, 𝑦 = 19, 𝑧 = −6
2) Solve the system of linear equations 𝒙 − 𝟐𝒚 + 𝒛 = 𝟏; −𝒙 + 𝒚 − 𝒛 = 𝟎; 𝟐𝒙 − 𝒚 + 𝒛 = −𝟏 using
Gauss Elimination method
...

For example:(i) 𝑥 + 𝑦 + 𝑧 = 0
𝑥 + 2𝑦 − 𝑧 = 0
𝑥 + 3𝑦 + 2𝑧 = 0
(ii)𝑥 + 𝑦 = 0
𝑥 + 2𝑦 = 0
Homogeneous equations are never inconsistent
...
The solution (0, 0, …, 0) is often called the trivial solution
...


Example-1: Solve the following system:

Example-2: Solve the following system

4𝑥 + 3𝑦 − 𝑧 = 0
3𝑥 + 4𝑦 + 𝑧 = 0
5𝑥 + 𝑦 − 4𝑧 = 0
Solution:The matrix form of the system is
4
[3
5

3
4
1

Solution:

−1 𝑥
0
1 ] [𝑦 ] = [0]
−4 𝑧
0

The augmented matrix is
[𝐴|𝐵]
4 3
[3 4
5 1

=

−1|0
1 |0 ]
−4|0

=
=

=

=

=

=

The last equation does not give any information
about the equations
...


Exercise:Solve
equations
...


=

The solution set is
{(t/4, −7𝑡/4, 𝑡)/ t
R}
...


𝑦+𝑧 =0

Rank of a Matrix
The positive integer 𝑟 is said to be a rank of a matrix 𝐴 if it possesses the following properties:
(1) There is at least one minor of order 𝑟 which is non-zero
...



Notes:

1
...
The rank of matrix remains unchanged by elementary transformation
3
...
The rank of the product of two matrices always less than or equal to the rank of either
matrix(i
...
, 𝜌(𝐴𝐵) ≤ 𝜌(𝐴) or 𝜌(𝐴𝐵) ≤ 𝜌(𝐵))
...



Step-1: Find the determinant of 𝐴
...
Otherwise 𝜌(𝐴) < 𝑟
...
If any one of them is non-zero then
order is 𝑟 − 1, otherwise 𝜌(𝐴) < 𝑟 − 1
...


Example 1: Find the rank the following matrices by determinant method:
𝟐
(1) 𝑨 = [𝟒
𝟏

𝟑
𝟑
𝟐

𝟒
𝟏]
𝟒

2
Solution: Given, 𝐴 = [4
1
𝟏
(2) 𝑨 = [𝟐
𝟑

𝟐
𝟑
𝟓

4
1] then 𝑑𝑒𝑡(𝐴) ≠ 0
...
Hence, the rank of 𝐴 is less than 3
...
Hence, 𝜌(𝐴) = 2
...
Hence 𝜌(𝐴) < 3
...
e
...
There rank is less than 2
...


 1 2 1 4 
(4) A   2 4 3 5 
 1 2 6 7 
Solution: Here, the order of matrix 𝐴 is 3 × 4
...
Therefore, consider all the minors of order 3, i
...
,
1
|2
−1

2
4
−2

−1
2 −1 −4
1
3 | = 0, | 4
3
5 | = 0, | 2
6
−2 6 −7
−1

2
4
−2

−4
1
5 | = 0, | 2
−7
−1

−1
3
6

−4
5 | = −120
−7

Here, one minor of rank 3 is not equal to zero
...

 Method-2: Rank of a Matrix by Row Echelon Form
The Rank of a Matrix in Row Echelon Form is equal to the number of non-zero rows of
the matrix
...
Hence, rank of matrix 𝐴 is 2
...


1 2 3 1
0 3 3 3

R2  2 R1 , R3  R1 : 
0 2 2 2 


0 1 1 1
1 2 3 1
0 1 1 1

R24 : 
0 2 2 2 


0 3 3 3
1 2 3 1
0 1 1 1

R3  2 R2 , R4  3R2 : 
0 0 0 0 


0 0 0 0 

Number of non-zero rows = 3
...

Number

of

non-zero

rows

= 2
...

(2) If 𝜌(𝐴) = 𝜌(𝐴|𝐵) then the system is consistent
...

Example: Find the number of parameters in the general solution of 𝐴𝑋 = 𝑂 if 𝐴 is a 5 × 7
matrix of rank 3
...
Hence, number of parameters = 𝑛 − 𝜌(𝐴) = 7 − 3 =
4
...


Notes
1
...

2
...

3
...


4
...

5
...

6
...

7
...
Then the set𝐸𝜆 = {𝑋/𝐴𝑋 = 𝜆𝑋} is
called the eigen space of 𝜆
...
The eigen values of a diagonal matrix are its diagonal elements
...
The sum of eigen values of an 𝑛 × 𝑛 matrix is its trace and their product is |𝐴|
...
For the upper triangular (lower triangular) 𝑛 × 𝑛matrix 𝐴, the eigen values are its
diagonal elements
...

When 1  0

When 3  2

Therefore, we suppose

x  4 y  6 z  0,  2 z  0, y  k
 z  0, y  k , x  4k
Therefore, eigen vector space is

We suppose z  0, y  k , x  2 y  0
Therefore, eigen vector space for 1  0 is

When 2  1

 z  0, y  k , x  2 z  2k

Therefore, eigen vector space for 3  2

Algebraic multiplicity and Geometric multiplicity

Let 𝐴 be 𝑛 × 𝑛 matrix and 𝜆 be an eigen value for 𝐴
...

−2 2 −3
Example: Find eigen values and eigen vectors of the matrix
...
Also determine
−1 −2 0
algebraic and geometric multiplicity
...


Algebraic Multiplicity of 𝜆 = −3 is 2 and of 𝜆 = 5 is 1
...

We suppose

x2  k1 , x3  k2 , x1  2 x2  3x3  0  x1  2k1  3k2
=

Therefore, eigen space is for𝜆2 = −3, 𝜆3 = −3 is

k  2,1, 0   k  3, 0,1 / k , k
1

2

1

2

 R

Hence, Geometric multiplicity of 𝜆2 = −3 is 2 and
of𝜆 = 5 is 1
...

We suppose

x3  k , x2  2 x3  0  x2  2k ,
x1  x3  0  x1  k

Therefore, eigen space is for 𝜆1 = 5 is

k  1, 2,1 / k  R
0 1
Example: Find eigen values and eigen vectors of the matrix
...


1
1]
...


Algebraic Multiplicity of 𝜆 = −1 is 2 and of 𝜆 = 2 is 1
...

Therefore, eigen space is for𝜆2 = −1, 𝜆3 = −1
is k1  1, 0,1  k2  1,1, 0  / k1 , k2  R
Hence, Geometric Multiplicity of𝜆2 = −1 is 2 and
𝜆1 = 2 of is 1
...

Therefore, eigen space is for 𝜆1 = 2 is

k 1,1,1 / k  R

1
Example: Determine algebraic and geometric multiplicity ofmatrix 𝐴 = [ 0
−1

2
2
2

2
1]
...

For 𝜆 = 1 A
...
is 1 and G
...
is 1
...

Proof: Let 𝐴 be 𝑚 × 𝑛 matrix
...

Obviously, 𝐴 = 𝐵 + 𝐶
1

𝑇

1

1

1

2

2

2

Now,𝐵𝑇 = [ (𝐴 + 𝐴𝑇 )] = [(𝐴 + 𝐴𝑇 )]𝑇 = [𝐴𝑇 + (𝐴𝑇 )𝑇 ] = (𝐴𝑇 + 𝐴) = 𝐵
2

As 𝐵𝑇 = 𝐵 , 𝐵 is symmetric
...

Therefore, 𝐴 is a sum of symmetric and skew-symmetric matrices
...
e
...

1
Example (i): Verify Caley-Hamilton theorem and hence find the inverse of 𝐴 = [
2

4
] and 𝐴4
...


Now, by putting 𝜆 = 𝐴, we have

1 4  1 4 
1 4 1 0 9 16   4 16 5 0 0 0
A2  4 A  5I  
4



 5




O
 2 3  2 3
 2 3 0 1  8 17  8 12 0 5 0 0
Hence, Cayley-Hamilton theorem verified
...


Solution: The characteristics equation is

By Caley-Hamilton Theorem

1 1
1 0]and hence prove that
1 2


...

1
(2) Compute 𝐴9 − 6𝐴8 + 10𝐴7 − 3𝐴6 + 𝐴 + 𝐼, where 𝐴 = [−1
1

2
3
0

3
2 2
]
[
(Answer:
1
−1 4
2
1 0

3
1]
3

)

Diagonalization of a matrix:
An 𝑛 × 𝑛 matrix 𝐴 is diagonalizable if and only if 𝐴 has 𝑛 linearly independent eigenvectors
...
Here,𝑃 is the matrix with these eigenvectors as column
vectors
...

For   1
Solution (ii):
The characteristic equation is
|𝐴 − 𝜆𝐼𝑛 | = 0

=

𝑦 = 𝑘, 6𝑥 = 0 => 𝑥 = 0
∴ (𝑥, 𝑦) = {𝑘(0,1)/𝑘 ∈ 𝑅}

For   1

=

Now,

Suppose 𝑥 = 𝑘, 6𝑥 − 2𝑦 = 0
𝑥 = 𝑘, 𝑦 = 3𝑘
∴ (𝑥, 𝑦) = {𝑘(1,3)/𝑘 ∈ 𝑅}

Eigenvalues of A2 are: 12 = 1 and(−1)2 = 1
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For example:
(i)

𝑎𝑥 2 + 2ℎ𝑥𝑦 + 𝑏𝑦 2 is a quadratic form in the variables x and y

(ii)

2𝑥1 𝑥2 + 2𝑥2 𝑥3 + 2𝑥3 𝑥1 + 𝑥3 2 is a quadratic form in the variables 𝑥1 , 𝑥2 , 𝑥3
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Here, 𝐴 is known as the coefficient matrix
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Matrix Representation of Quadratic Forms
A quadratic form can be represented as a matrix product
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Solution: The coefficient matrix of 𝑄 is

So, C = symmetric matrix =

(ii) Express the following quadratic forms in matrix notation

Solution:

Transformation (Reduction) of Quadratic form to canonical form OR Diagonalizing
Quadratic Forms:
Procedure to Reduce Quadratic form to canonical form:
1
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2
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3
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(1)

where𝜆1 , 𝜆2 , …
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The matrix 𝑃 is said to
orthogonally diagonalize the quadratic form
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4
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Example: Reduce the quadratic form into canonical form

Solution:

4

Eigenvalues for𝐴 are3, − 3 , −1
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Positive definite if Q(x) > 0 for all x 0,
b
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Indefinite if Q(x) assumes both positive and negative values
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Positive semidefinite if Q(x) 0 for all x
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Negative semidefinite if Q(x) 0 for all x
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Positive definite if and only if the eigenvalues of A are positive,
b
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Indefinite if and only if A has both positive and negative eigenvalues
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Positive semi-definite if and only if A has only non-negative eigenvalues
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Indefinite if and only if A has only non-positive eigenvalues
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1
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