Unit 1

Matrix operations and solving systems of linear equations

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

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 whereas the minor of order 2 is non zero as

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

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.

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.

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.

i.e.

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.

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

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-

(7-

Which gives,

Or

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

Or

Or

By Cayley-Hamilton theorem

L.H.S.

=

By Cayley-Hamilton theorem we have

Multiply both side by

.

Or

=

=

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,

11

=

=

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,

Or

We know that by Cayley-Hamilton theorem,

…………………….(1)t,

Multiply equation (1) by , we get

Or

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

Or

Or

Or

By Cayley-Hamilton theorem

L.H.S:

= =0=R.H.S

Multiply both side by on

Or

Or [

Or

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}

Or

Or

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

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

Or

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 =

Q-1 =

Q-1 = … (2)

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

Therefore, Q is an orthogonal matrix.

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.

Solution:

Characteristic eq, of matrix A is

For , Eigen Vector,

For , Eigen Vector,

For , Eigen Vector,

Length of eigen vector,

Length of eigen vector,

Length of eigen vector,

Normalised eigen vectors are:

Modal matrix P (Normalised Eigen vector as its column vectors)

Hence, the canonical form is

Rank (r) = 3 (No. Of non-zero eigen values)

Index = 3 (no. Of positive eigen values)

Example 2: Reduce the quadratic form to the canonical form through an orthogonal transformation.

Solution:

To find the characteristic equation.,

= 0 ,

To find the eigenvectors,

Case-1,

If

(A-)X=0

Case(2),

If

(A-)X = 0.

From equation ….(1)

Put in equation …(2)

Case (3),

If

(A-)X = 0

D =

Textbooks:

1. Erwin Kreyszig, Advanced Engineering Mathematics, 10/e, John Wiley & Sons, 2011.

2. B. S. Grewal, Higher Engineering Mathematics, 44/e, Khanna Publishers, 2017.

References:

1. R. K. Jain and S. R. K. Iyengar, Advanced Engineering Mathematics, 3/e, Alpha Science

International Ltd., 2002.

2. George B. Thomas, Maurice D. Weir and Joel Hass, Thomas Calculus, 13/e, Pearson

Publishers, 2013.

3. Glyn James, Advanced Modern Engineering Mathematics, 4/e, Pearson publishers, 201.