UNIT-1

Matrices

Matrices: It is a rectangular arrangement of numbers formed by keeping them in m numbers of rows and n numbers of columns.

Types of matrices:

Row matrix:

A matrix with only one single row and many columns can be possible.

A =

Column matrix

A matrix with only one single column and many rows can be possible.

A =

Square matrix

A matrix in which number of rows is equal to number of columns is called a square matrix. Thus an matrix is square matrix then m=n and is said to be of order n.

The determinant having the same elements as the square matrix A is called the determinant of the matrix A. Denoted by |A|.

The diagonal elements of matrix A are 2, 2 and 1 is the leading and the principal diagonal.

The sum of the diagonal elements of square matrix A is called the trace of A.

A square matrix is said to be singular if its determinant is zero otherwise non-singular.

Hence the square matrix A is non-singular.

Diagonal matrix

A square matrix is said to be diagonal matrix if all its non diagonal elements are zero.

Scalar matrix:

A diagonal matrix is said to be scalar matrix if its diagonal elements are equal.

Identity matrix:

A square matrix in which elements in the diagonal are all 1 and rest are all zero is called an identity matrix or Unit matrix.

Null Matrix:

If all the elements of a matrix are zero, it is called a null or zero matrixes.

Ex: etc.

Symmetric matrix:

Any square matrix is said to be symmetric matrix if its transpose equals to the matrix itself.

For example:

and

Example: check whether the following matrix A is symmetric or not?

A =

Sol. As we know that if the transpose of the given matrix is same as the matrix itself then the matrix is called symmetric matrix.

So that, first we will find its transpose,

Transpose of matrix A ,

Here,

A =

So that, the matrix A is symmetric.

Example: Show that any square matrix can be expressed as the sum of symmetric matrix and anti- symmetric matrix.

Sol. Suppose A is any square matrix .

Then,

A =

Now,

(A + A’)’ = A’ + A

A+A’ is a symmetric matrix.

Also,

(A - A’)’ = A’ – A

Here A’ – A is an anti – symmetric matrix

So that,

Square matrix = symmetric matrix + anti-symmetric matrix

Hermitian matrix:

A square matrix A = is said to be hermitian matrix if every element of A is equal to conjugate complex j-ith element of A.

It means,

For example:

Necessary and sufficient condition for a matrix A to be hermitian –

A = (͞A)’

Skew-Hermitian matrix-

A square matrix A = is said to be hermitian matrix if every element of A is equal to negative conjugate complex j-ith element of A.

Note- all the diagonal elements of a skew hermitian matrix are either zero or pure imaginary.

For example:

The necessary and sufficient condition for a matrix A to be skew hermitian will be as follows-

- A = (͞A)’

Note: A Hermitian matrix is a generalization of a real symmetric matrix and also every real symmetric matrix is Hermitian.

Similarly a Skew- Hermitian matrix is a generalization of a Skew symmetric matrix and also every Skew- symmetric matrix is Skew –Hermitian.

Theorem: Every square complex matrix can be uniquely expressed as sum hermitian and skew-hermitian matrix.

Or If A is given square complex matrix then is hermitian and is skew-hermitian matrices.

Example1: Express the matrix A as sum of hermitian and skew-hermitian matrix where

Let A =

Therefore and

Let

Again

Hence P is a hermitian matrix.

Let

Again

Hence Q is a skew- hermitian matrix.

We Check

P +Q=

Hence proved.

Example2: If A = then show that

(i) is hermitian matrix.

(ii) is skew-hermitian matrix.

Given A =

Then

Let

Also

Hence P is a Hermitian matrix.

Let

Also

Hence Q is a skew-hermitian matrix.

Skew-symmetric matrix-

A square matrix A is said to be skew Symmetric matrix if –

1. A’ = -A, [ A’ is the transpose of A]

2.all the main diagonal elements will always be zero.

For example-

A =

This is skew symmetric matrix, because transpose of matrix A is equals to negative A.

Example: check whether the following matrix A is symmetric or not?

A =

Sol. This is not a skew symmetric matrix, because the transpose of matrix A is not equals to -A.

-A = A’

Orthogonal matrix-

Any square matrix A is said to be an orthogonal matrix if the product of the matrix A and its transpose is an identity matrix.

Such that,

A. A’ = I

Matrix × transpose of matrix = identity matrix

Note- if |A| = 1, then we can say that matrix A is proper.

Examples: and are the form of orthogonal matrices.

Unitary matrix-

A square matrix A is said to be unitary matrix if the product of the transpose of the conjugate of matrix A and matrix itself is an identity matrix.

Such that,

( ͞A)’ . A = I

For example:

A = and its ( ͞A)’ =

Then ( ͞A)’ . A = I

So that we can say that matrix A is said to be a unitary matrix.

Rank of a matrix by echelon form-

The rank of a matrix (r) can be defined as –

1. It has at least one non-zero minor of order r.

2. Every minor of A of order higher than r is zero.

Example: Find the rank of a matrix M by echelon form.

M =

Sol. First we will convert the matrix M into echelon form,

M =

Apply, , we get

M =

Apply , we get

M =

Apply

M =

We can see that, in this echelon form of matrix, the number of non – zero rows is 3.

So that the rank of matrix X will be 3.

Example: Find the rank of a matrix A by echelon form.

A =

Sol. Convert the matrix A into echelon form,

A =

Apply

A =

Apply , we get

A =

Apply , we get

A =

Apply ,

A =

Apply ,

A =

Therefore the rank of the matrix will be 2.

Example: Find the rank of a matrix A by echelon form.

A =

Sol. Transform the matrix A into echelon form, then find the rank,

We have,

A =

Apply,

A =

Apply ,

A =

Apply

A =

Apply

A =

Hence the rank of the matrix will be 2.

Example: Find the rank of the following matrices by echelon form?

Let A =

Applying

A

Applying

A

Applying

A

Applying

A

It is clear that minor of order 3 vanishes but minor of order 2 exists as

Hence rank of a given matrix A is 2 denoted by

2.

Let A =

Applying

Applying

Applying

The minor of order 3 vanishes but minor of order 2 non zero as

Hence the rank of matrix A is 2 denoted by

3.

Let A =

Apply

Apply

Apply

It is clear that the minor of order 3 vanishes where as the minor of order 2 is non zero as

Hence the rank of given matrix is 2 i.e.

Rank of a matrix by normal form-

Any matrix ‘A’ which is non-zero can be reduced to a normal form of ‘A’ by using elementary transformations.

There are 4 types of normal forms –

The number r obtained is known as rank of matrix A.

Both row and column transformations may be used in order to find the rank of the matrix.

Note-Normal form is also known as canonical form

Example: reduce the matrix A to its normal form and find rank as well.

Sol. We have,

We will apply elementary row operation,

We get,

Now apply column transformation,

We get,

Apply

, we get,

Apply and

Apply

Apply and

Apply and

As we can see this is required normal form of matrix A.

Therefore the rank of matrix A is 3.

Example: Find the rank of a matrix A by reducing into its normal form.

Sol. We are given,

Apply

Apply

This is the normal form of matrix A.

So that the rank of matrix A = 3

If A (matrix) reduced to I by elementary transformation then

PA = I

P = A‾¹ where P is elementary matrix.

Example: Find the inverse of the following matrix by using gauss Jordan method.

Sol. Here we are given,

A =

Example: Find the inverse of the following matrix by using gauss Jordan method.

Sol. Let the matrix is A, then

There are two types of linear equations-

1. Consistent 2. Inconsistent

Let’s understand about these two types of linear equations

Consistent –

If a system of equations has one or more than one solution, it is said be consistent.

There could be unique solution or infinite solution.

For example-

A system of linear equations-

2x + 4y = 9

x + y = 5

Has unique solution,

Whereas,

A system of linear equations-

2x + y = 6

4x + 2y = 12

Has infinite solutions.

Inconsistent-

If a system of equations has no solution, then it is called inconsistent.

Consistency of a system of linear equations-

Suppose that a system of linear equations is given as-

This is the format as AX = B

Its augmented matrix is-

[A:B] = C

(1) consistent equations-

If Rank of A = Rank of C

Here, Rank of A = Rank of C = n ( no. Of unknown) – unique solution

And Rank of A = Rank of C = r , where r<n - infinite solutions

(2) inconsistent equations-

If Rank of A ≠ Rank of C

Solution of homogeneous system of linear equations-

A system of linear equations of the form AX = O is said to be homogeneous, where A denotes the coefficients and of matrix and and O denotes the null vector.

Suppose the system of homogeneous linear equations is,

It means,

AX = O

Which can be written in the form of matrix as below,

Note- A system of homogeneous linear equations always has a solution if

1. r(A) = n then there will be trivial solution, where n is the number of unknown,

2. r(A) < n , then there will be an infinite number of solution.

Example: Find the solution of the following homogeneous system of linear equations,

Sol. The given system of linear equations can be written in the form of matrix as follows,

Apply the elementary row transformation,

, we get,

, we get

Here r(A) = 4, so that it has trivial solution,

Example: Find out the value of ‘b’ in the system of homogeneous equations-

2x + y + 2z = 0

x + y + 3z = 0

4x + 3y + bz = 0

Which has

(1) trivial solution

(2) non-trivial solution

Sol. (1)

For trivial solution, we already know that the values of x , y and z will be zerp, so that ‘b’ can have any value.

Now for non-trivial solution-

(2)

Convert the system of equations into matrix form-

AX = O

Apply respectively , we get the following resultant matrices

For non-trivial solutions, r(A) = 2 < n

b – 8 = 0

b = 8

Solution of non-homogeneous system of linear equations-

Example-1: check whether the following system of linear equations is consistent of not.

2x + 6y = -11

6x + 20y – 6z = -3

6y – 18z = -1

Sol. Write the above system of linear equations in augmented matrix form,

Apply , we get

Apply

Here the rank of C is 3 and the rank of A is 2

Therefore both ranks are not equal. So that the given system of linear equations is not consistent.

Example: Check the consistency and find the values of x , y and z of the following system of linear equations.

2x + 3y + 4z = 11

X + 5y + 7z = 15

3x + 11y + 13z = 25

Sol. Re-write the system of equations in augmented matrix form.

C = [A,B]

That will be,

Apply

Now apply ,

We get,

~ ~

Here rank of A = 3

And rank of C = 3, so that the system of equations is consistent,

So that we can can solve the equations as below,

That gives,

x + 5y + 7z = 15 ……………..(1)

y + 10z/7 = 19/7 ………………(2)

4z/7 = 16/7 ………………….(3)

From eq. (3)

z = 4,

From 2,

From eq.(1), we get

x + 5(-3) + 7(4) = 15

That gives,

x = 2

Therefore the values of x , y , z are 2 , -3 , 4 respectively.

Matrix form of the linear system can be written as follows,

Ax = b

Where,

A is called augmented matrix.

Gauss elimination method can be presented as ,

Suppose there is a linear system in triangular form ,

We can solve this as X2 = -26/13 = -2 and X1 = 6

Lets another set of eq. Is given as ,

Its augmented matrix is given by,

We eliminate X1 from the second eq.

To get the triangular system . We add twice the first equation to the second and we do same operation on rows of the augmented matrix. This gives,

-4X1 +4X1 +3X2 + 10X2 = -30 + 2.2

This is called Gauss elimination.

Example: find the solution of the following linear equations.

+ - = 1

- + = 2

- 2 + = 5

Sol. These equations can be converted into the form of matrix as below-

Apply the operation, and , we get

We get the following set of equations from the above matrix,

+ - = 1 …………………..(1)

+ 5 = -1 …………………(2)

= k

Put x = k in eq. (1)

We get,

+ 5k = -1

Put these values in eq. (1), we get

and

These equations have infinitely many solutions.

Step by step method to solve the system of linear equation by using Gauss Seidel iteration method-

Suppose,

This system can be written as after dividing it by suitable constants,

Step-1 Here put y = 0 and z = 0 and x = in first equation

Then in second equation we put this value of x that we get the value of y.

In the third eq. We get z by using the values of x and y

Step-2: we repeat the same procedure

Example: solve the following system of linear equations by using Gauss seidel method-

6x + y + z = 105

4x + 8y + 3z = 155

5x + 4y - 10z = 65

Sol. The above equations can be written as,

………………(1)

………………………(2)

………………………..(3)

Now put z = y = 0 in first eq.

We get

x = 35/2

Put x = 35/2 and z = 0 in eq. (2)

We have,

Put the values of x and y in eq. 3

Again start from eq.(1)

By putting the values of y and z

y = 85/8 and z = 13/2

We get

The process can be showed in the table format as below

At the fourth iteration , we get the values of x = 14.98 , y = 9.98 , z = 4.98

Which are approximately equal to the actual values,

As x = 15 , y = 10 and y = 5 ( which are the actual values)

Example: : solve the following system of linear equations by using Gauss seidel method-

5x + 2y + z = 12

x + 4y + 2z = 15

x + 2y + 5z = 20

Sol. These equations can be written as ,

………………(1)

………………………(2)

………………………..(3)

Put y and z equals to 0 in eq. 1

We get,

x = 2.4

Put x = 2.4, and z = 0 in eq. 2 , we get

Put x = 2.4 and y = 3.15 in eq.(3) , we get

Again start from eq.(1), put the values of y and z , we get

= 0.688

We repeat the process again and again ,

The following table can be obtained –

We see that the values are apprx. Equal to exact values.

Exact values are , x = 1 , y = 2, z = 3.