Unit 1 | unit 1 matrix operations and solving systems of linear equations

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Unit 1

Matrix operations and solving systems of linear equations

1.1 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 1: Find the rank of a matrix M by echelon form.

M =

Solution: 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 2: Find the rank of a matrix A by echelon form.

A =

Solution: 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 3: Find the rank of a matrix A by echelon form.

A =

Solution: 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 4: Find the rank of the following matrices by echelon form?

Solution: Let A =

Applying

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

Let A =

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

Let A =

Apply

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

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

1.2 Solving system of homogeneous and non – homogenous equations Linear equations

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

Solution: 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 2: 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

Solution: (1) For trivial solution, we already know that the values of x , y and z will be zero, 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-3: check whether the following system of linear equations is consistent of not.

2x + 6y = -11

6x + 20y – 6z = -3

6y – 18z = -1

Solution: 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 4: 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

Solution. 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 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.

Gauss-Seidel iteration method-

Step by step method to solve the system of linear equation by using Gauss Seidal 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 5: solve the following system of linear equations by using Guassseidel method-

6x + y + z = 105

4x + 8y + 3z = 155

5x + 4y - 10z = 65

Solution: 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

Iterations	1	2	3	4
	32/2=17.5	14.64	15.12	14.98
	85/8=10.6	9.62	10.06	9.98
	13/2=6.5	4.67	5.084	4.98

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)

Example6: Solve the following system of linear equations by using Guassseidel method-

5x + 2y + z = 12

x + 4y + 2z = 15

x + 2y + 5z = 20

Solution: 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 –

Iterations	1	2	3	4	5
	2.4	0.688	0.84416	0.962612	0.99426864
	3.15	2.448	2.09736	2.013237	2.00034144
	2.26	2.8832	2.99222	3.0021828	3.001009696

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

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

1.3 Eigen values and eigen vectors and their properties

First, we will go through some important definitions before studying Eigen values and Eigen vectors.

1. Vector-

An ordered n – couple of numbers is called an n – vector. Thus the ‘n’ numbers x1, x2, …………xn taken in order denote the vector x. i.e. x = (x1, x2, ……., xn).

Where the numbers x1, x2, ………..,xn are called component or co – ordinates of a vector x. A vector may be written as row vector or a column vector.

If A be an mxn matrix then each row will be an n – vector & each column will be an m – vector.

2. Linear dependence-

A set of n – vectors. x1, x2, …….., xr is said to be linearly dependent if there exist scalars. k1, k2, …….,kr not all zero such that

k1 + x2k2 + …………….. + xrkr = 0 … (1)

3. Linear independence-

A set of r vectors x1, x2, ………….,xr is said to be linearly independent if there exist scalars k1, k2, …………, kr all zero such that

x1 k1 + x2 k2 + …….. + xrkr = 0

Important notes-

Equation (1) is known as a vector equation.
If the vector equation has non – zero solution i.e. k1, k2, ….,kr not all zero. Then the vector x1, x2, ……….xr are said to be linearly dependent.
If the vector equation has only trivial solution i.e.

k1 = k2 = …….=kr = 0. Then the vector x1, x2, ……,xr are said to linearly independent.

4. Linear combination-

A vector x can be written in the form.

x = x1 k1 + x2 k2 + ……….+xrkr

Where k1, k2, ………….., kr are scalars, then X is called linear combination of x1, x2, ……, xr.

Results:

A set of two or more vectors are said to be linearly dependent if at least one vector can be written as a linear combination of the other vectors.
A set of two or more vector are said to be linearly independent then no vector can be expressed as linear combination of the other vectors.

Example1: Are the vectors , , linearly dependent. If so, express x1 as a linear combination of the others.

Solution: Consider a vector equation,

i.e.

Which can be written in matrix form as,

Here & no. Of unknown 3. Hence the system has infinite solutions. Now rewrite the questions as,

Put

and

Thus

i.e.

Since F11k2, k3 not all zero. Hence are linearly dependent.

Example 2: Examine whether the following vectors are linearly independent or not.

and .

Solution: Consider the vector equation,

i.e. … (1)

Which can be written in matrix form as,

R12

R2 – 3R1, R3 – R1

R3 + R2

Here Rank of coefficient matrix is equal to the no. Of unknowns. i.e. r = n = 3.

Hence the system has unique trivial solution.

i.e.

i.e. vector equation (1) has only trivial solution. Hence the given vectors x1, x2, x3 are linearly independent.

Example 3: At what value of P the following vectors are linearly independent.

Solution:

Consider the vector equation.

i.e.

This is a homogeneous system of three equations in 3 unknowns and has a unique trivial solution.

If and only if Determinant of coefficient matrix is non zero.

consider .

i.e.

Thus for the system has only trivial solution and Hence the vectors are linearly independent.

Note:-

If the rank of the coefficient matrix is r, it contains r linearly independent variables & the remaining vectors (if any) can be expressed as linear combination of these vectors.

Characteristic equation:-

Let A he a square matrix, be any scalar then is called characteristic equation of a matrix A.

Note:

Let a be a square matrix and ‘’ be any scalar then,

1) is called characteristic matrix

2) is called characteristic polynomial.

The roots of a characteristic equations are known as characteristic root or latent roots, Eigen values or proper values of a matrix A.

Eigen vector:-

Suppose be an Eigen value of a matrix A. Then a non – zero vector x1 such that.

… (1)

Such a vector ‘x1’ is called as Eigen vector corresponding to the Eigen value .

Properties of Eigen values:-

Then sum of the Eigen values of a matrix A is equal to sum of the diagonal elements of a matrix A.
The product of all Eigen values of a matrix A is equal to the value of the determinant.
If are n Eigen values of square matrix A then are m Eigen values of a matrix A-1.
The Eigen values of a symmetric matrix are all real.
If all Eigen values are non –zero then A-1 exist and conversely.
The Eigen values of A and A’ are same.

Properties of Eigen vector:-

Eigen vector corresponding to distinct Eigen values are linearly independent.
If two are more Eigen values are identical then the corresponding Eigen vectors may or may not be linearly independent.
The Eigen vectors corresponding to distinct Eigen values of a real symmetric matrix are orthogonal.

Example-4: Determine the Eigen values of Eigen vector of the matrix.

Solution: Consider the characteristic equation as,

i.e.

Which is the required characteristic equation.

are the required Eigen values.

Now consider the equation

… (1)

Case I:

If Equation (1) becomes

R1 + R2

Thus

independent variable.

Now rewrite equation as,

Put x3 = t

Thus .

Is the eigen vector corresponding to .

Case II:

If equation (1) becomes,

Here

Independent variables

Now rewrite the equations as,

Put

Is the eigen vector corresponding to .

Case III:

If equation (1) becomes,

Here rank of

independent variable.

Now rewrite the equations as,

Put

Thus .

Is the eigen vector for .

Example 5: Find the Eigen values of Eigen vector for the matrix.

Solution: Consider the characteristic equation as

i.e.

are the required eigen values.

Now consider the equation

… (1)

Case I:

Equation (1) becomes,

Thus and n = 3

3 – 2 = 1 independent variables.

Now rewrite the equations as,

Put

I.e.

The Eigen vector for

Case II:

If equation (1) becomes,

Thus

Independent variables.

Now rewrite the equations as,

Put

Is the Eigen vector for

Now

Case III:-

If equation (1) gives,

R1 – R2

Thus

Independent variables

Now

Put

Thus

Is the Eigen vector for

1.4 Cayley – Hamilton theorem (without proof)

Statement-

Every square matrix satisfies its characteristic equation, that means for every square matrix of order n,

|A - | =

Then the matrix equation-

Is satisfied by X = A

That means

Example-1: Find the characteristic equation of the matrix A = andVerify cayley-Hamlton theorem.

Solution: Characteristic equation of the matrix, we can be find as follows-

Which is,

( 2 - , which gives

According to cayley-Hamilton theorem,

2 …………(1)

Now we will verify equation (1),

Put the required values in equation (1) , we get

Hence the cayley-Hamilton theorem is verified.

Example-2: Find the characteristic equation of the the matrix A and verify Cayley-Hamilton theorem as well.

A =

Solution: Characteristic equation will be-

= 0

( 7 -

(7-

Which gives,

According to cayley-Hamilton theorem,

…………………….(1)

In order to verify cayley-Hamilton theorem , we will find the values of

So that,

Now

Put these values in equation(1), we get

= 0

Hence the cayley-hamilton theorem is verified.

Example-3: Using Cayley-Hamilton theorem, find , if A = ?

Solution: Let A =

The characteristics equation of A is

By Cayley-Hamilton theorem

L.H.S.

By Cayley-Hamilton theorem we have

Multiply both side by

1.5 Finding inverse and power of a matrix by Cayley – Hamilton theorem

Inverse of a matrix by Cayley-Hamilton theorem-

We can find the inverse of any matrix by multiplying the characteristic equation With .

For example, suppose we have a characteristic equation then multiply this by , then it becomes

Then we can find by solving the above equation.

Example-1: Find the inverse of matrix A by using Cayley-Hamilton theorem.

A =

Solution: The characteristic equation will be,

|A - | = 0

Which gives,

(4-

According to Cayley-Hamilton theorem,

Multiplying by

That means

On solving,

So that,

Example-2: Find the inverse of matrix A by using Cayley-Hamilton theorem.

A =

Solution: The characteristic equation will be,

|A - | = 0

= (2-

= (2 -

That is,

We know that by Cayley-Hamilton theorem,

…………………….(1)t,

Multiply equation (1) by , we get

Now we will find

Hence the inverse of matrix A is,

Example-3: Verify the Cayley-Hamilton theorem and find the inverse.

Solution: Let A =

The characteristics equation of A is

By Cayley-Hamilton theorem

L.H.S:

= =0=R.H.S

Multiply both side by on

Or [

1.6 Diagonalisation of a matrix

Two square matrixes and A of same order n are said to be similar if and only if

for some non singular matrix P.

Such transformation of the matrix A into with the help of non singular matrix P is known as similarity transformation.

Similar matrices have the same Eigen values.

If X is an Eigen vector of matrix A then is Eigen vector of the matrix

Reduction to Diagonal Form:

Let A be a square matrix of order n has n linearly independent Eigen vectors Which form the matrix P such that

Where P is called the modal matrix and D is known as spectral matrix.

Procedure: let A be a square matrix of order 3.

Let three Eigen vectors of A are Corresponding to Eigen values

Let

{by characteristics equation of A}

Note: The method of diagonalization is helpful in calculating power of a matrix.

.Then for an integer n we have

We are using the example of 1.6*

Example1: Diagonalise the matrix

Solution: Let A=

The three Eigen vectors obtained are (-1,1,0), (-1,0,1) and (3,3,3) corresponding to Eigen values .

Then and

Also, we know that

Example2: Diagonalise the matrix

Solution: Let A =

The Eigen vectors are (4,1),(1,-1) corresponding to Eigen values .

Then and also

Also, we know that

1.7 Quadratic forms and nature of the quadratic forms

Quadratic Forms: Definition : An expression of the form, where constants is called quadratic form in n variables

If the constants are real numbers, it is called quadratic form.

The second order homogenous expression in n variables is called quadratic form.

Procedure for solving Quadratic form:

Step 1: Write the coefficient matrix A associated with the given quadratic form.

Step 2: Find the eigen values of A.

Step 3: Write the canonical form using

Step 4: Form a matrix P containing the normalized eigen vectors of A. Then X=PY gives the required orthogonal transformation, which reduces Quadratic form to canonical form.

Example1: Consider the function .Determine whether q(0,0) is the global minimum.

Solution: Based on matrix technique

Rewrite the above equation

Note that we split the contribution -4 equally among the two components.

More succinctly, we can write

The matrix A is symmetric by construction. By the spectral theorem, there is an orthonormal eigen basis

With associated eigenvalues .

Let can express the value of the function as follows:

Therefore, is the global minimum if the function.

Example 2: The matrix A= has row vectors.

Solution: ,

Orthogonal matrix: An orthogonal matrix is the real specialization of a unitary matrix, and thus always a normal matrix. Although we consider only real matrices here, the definition can be used for matrices with entries from any field.

Suppose A is a square matrix with real elements and of n x n order and AT is the transpose of A. Then according to the definition, if, AT = A-1 is satisfied, then,

A AT = I

Where ‘I’ is the identity matrix, A-1 is the inverse of matrix A, and ‘n’ denotes the number of rows and columns.

Example 3: prove Q= is an orthogonal matrix

Solution: Given Q =

So, QT = …..(1)

Now, we have to prove QT = Q-1

Now we find Q-1

Q-1 =

Q-1 = … (2)

Now, compare (1) and (2) we get QT = Q-1

Therefore, Q is an orthogonal matrix.

1.8 Reduction of quadratic form to canonical forms by orthogonal transformation

Steps:

Convert quadratic form to matrix form.
Find eigen values and eigen vectors.
Find model matrix(P)
Find Normalised matrix (N)
Find
Find D (D =

Example 1: Reduce the following quadratic form to canonical form by orthogonal transformation. Also, find rank, index, signature and nature of quadratic form.