Unit 1
Matrices
- Definition:
An arrangement of m.n numbers in m rows and n columns is called a matrix of order mxn.
Generally, a matrix is denoted by capital letters. Like, A, B, C, ….. Etc.
2. Types of matrices:- (Review)
- Row matrix
- Column matrix
- Square matrix
- Diagonal matrix
- Trace of a matrix
- The determinant of a square matrix
- Singular matrix
- Non – singular matrix
- Zero/ null matrix
- Unit/ Identity matrix
- Scaler matrix
- Transpose of a matrix
- Triangular matrices
Upper triangular and lower triangular matrices,
14. Conjugate of a matrix
15. Symmetric matrix
16. Skew – symmetric matrix
3. Operations on matrices:
- Equality of two matrices
- Multiplication of A by a scalar k.
- Addition and subtraction of two matrices
- Product of two matrices
- The inverse of a matrix
4. Elementary transformations
a) Elementary row transformation
These are three elementary transformations
- Interchanging any two rows (Rij)
- Multiplying all elements in ist row by a non – zero constant k is denoted by KRi
- Adding to the elements in an ith row by the kth multiple of the jth row is denoted by
.
b) Elementary column transformations:
There are three elementary column transformations.
- Interchanging ith and jth column. Is denoted by Cij.
- Multiplying ith column by a non – zero constant k is denoted by kCj.
- Adding to the element of the ith column by the kth multiple of the jth column is denoted by Ci + kCj.
The rank of a matrix:
Let A be a given rectangular matrix ofa square matrix. From this matrix select, any r rows from these r rows select any r columns thus getting a square matrix of order r x r. The determinant of this matrix of order r x r is called a minor of order r.
e.g.
If
For example, select the 2nd and 3rd row. i.e.
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188325_2332735.png)
Now select any two columns. Suppose 1st and 2nd.
i.e.
Invariance of rank through elementary transformations.
- The rank of the matrix remains unchanged by elementary transformations. i.e. from a matrix. We get another matrix B by using some elementary transformation. Then
The rank of A = Rank of B
2. Equivalent matrices:
The matrix B is obtained from a matrix A by a sequence of a finite no. Of elementary transformations is said to be equivalent to A. And we write.
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188325_6782.png)
Normal form or canonical form:
Every mxn matrix of rank r can be reduced to the form
By a finite sequence of elementary transformation. This form is called the normal form or the first canonical form of the matrix A.
Ex. 1
Reduce the following matrix to the normal form of Hence find it’s rank,
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188325_7747207.png)
Solution:
We have,
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188325_8184652.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188325_8936388.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188325_9415293.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188325_9859753.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188326_0246463.png)
Apply
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188326_1114726.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188326_152682.png)
The rank of A = 1
Ex. 2
Find the rank of the matrix
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188326_2800007.png)
Solution:
We have,
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188326_3213856.png)
Apply R12
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188326_3687062.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188326_4169698.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188326_4981813.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188326_537839.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188326_5806425.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188326_6255221.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188326_6670911.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188326_7110384.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188326_7536085.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188326_8085644.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188326_887728.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188326_934782.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188326_9780362.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188327_038345.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188327_0910838.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188327_1449435.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188327_2165577.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188327_2608206.png)
The rank of A = 3
Ex. 3
Find the rank of the following matrices by reducing it to the normal form.
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188327_3643935.png)
Solution:
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188327_4087813.png)
Apply C14
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188327_4504259.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188327_4983368.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188327_5615077.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188327_6065881.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188327_6621404.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188327_7288382.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188327_775776.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188327_831524.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188327_8960426.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188327_959379.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188328_0070262.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188328_0479877.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188328_1160855.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188328_1743543.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188328_2175024.png)
H.W.
Reduce the follo9wing matrices into the normal form and hence find their ranks.
a)
b)
5. Reduction of a matrix a to normal form PAQ.
If A is a matrix of rank r, then there exists a non – singular matrices P & Q such that PAQ is in normal form.
i.e.
To obtain the matrices P and Q we use the following procedure.
Working rule:-
- If A is anmxn matrix, write A = Im A In.
- Apply row transformations on A on l.h.s. And the same row transformations on the prefactorIm.
- Apply column transformations on A on l.h.s and the column transformations on the postfactor In.
So that A on the l.h.s. Reduces to normal form.
Example 1
If Find Two
Matrices P and Q such that PAQ is in normal form.
Solution:
Here A is a square matrix of order 3 x 3. Hence, we write,
A = I3 A.I3
i.e.
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188328_5326338.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188328_576005.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188328_631789.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188328_673721.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188328_7337937.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188328_778885.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188328_8673975.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188328_9289427.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188328_976561.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188329_0221534.png)
i.e.
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188329_1154819.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188329_194032.png)
Example 2
Find a non – singular matrices p and Q such that P A Q is in a normal form where
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188329_2371583.png)
Solution:
Here A is a matrix of order 3 x 4. Hence, we write A as,
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188329_3219042.png)
i.e.
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188329_3975294.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188329_4558392.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188329_5065188.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188329_579795.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188329_6291857.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188329_7011943.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188329_7563212.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188329_8026786.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188329_8605204.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188329_9109917.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188329_966.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188330_0175955.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188330_0787823.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188330_1353157.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188330_2103152.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188330_2684946.png)
i.e.
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188330_3879895.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188330_434783.png)
A linear transformation is said to orthogonal if it transforms
.
The matrix of an orthogonal transfer motion is called on orthogonal matrix.
Definition:
A square matrix ‘A’ is said to be orthogonal if
Note:-
, for an orthogonal transformation
- If ‘A’ is orthogonal then A-1, AT are also orthogonal.
- If a is orthogonal,
Ex.
- Show that the transformation
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188330_8717442.png)
is orthogonal.
2. If is orthogonal, find a, b, c.
Ex. Determine the values of a, b, c when
is orthogonal.
Eigenvalues and Eigenvectors
Introduction:
In this chapter we are going to study a very important theorem viz first we have to study eigenvalues and eigenvector.
- Vector
An ordered n-tuple of numbers is called an n – vector. Thus the ‘n’ numbers x1, x2, ………… xn took in order denotes the vector x. i.e. x = (x1, x2, ……., xn).
Where the numbers x1, x2, ……….., xn are called component or coordinates of a vector x. A vector may be written as a row vector or a column vector.
If A be anmxn 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
Note:-
- 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 a 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 is said to be linearly independent then no vector can be expressed as a linear combination of the other vectors.
Example 1
Are the vectors ,
,
linearly dependent. If so, express x1 as a linear combination of the others.
Solution:
Consider a vector equation,
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188331_7831507.png)
i.e.
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188331_9934554.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188332_0425751.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188332_0935266.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188332_1415913.png)
Which can be written in matrix form as,
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188332_1906157.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188332_2398884.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188332_3178258.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188332_3655722.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188332_4104552.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188332_4586904.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188332_5038612.png)
Here & no. Of unknown 3. Hence the system has infinite solutions. Now rewrite the questions as,
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188332_6131957.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188332_6560926.png)
Put
and
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188332_8195887.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188332_8694527.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188332_9149222.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188332_995161.png)
Thus
i.e.
i.e.
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188333_2400384.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188333_2999322.png)
Since F11 k2, 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,
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188333_529472.png)
i.e. … (1)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188333_6457713.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188333_6907547.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188333_7602754.png)
Which can be written in matrix form as,
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188333_807051.png)
R12
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188333_8618972.png)
R2 – 3R1, R3 – R1
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188333_929263.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188334_0295506.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188334_0757813.png)
R3 + R2
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188334_1230881.png)
Here Rank of the coefficient matrix is equal to the no. Of unknowns. i.e. r = n = 3.
Hence the system has a unique trivial solution.
i.e.
i.e. vector equation (1) has an 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.
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188334_2447731.png)
Solution:
Consider the vector equation.
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188334_2977679.png)
i.e.
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188334_3664086.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188334_41386.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188334_4607751.png)
Which is a homogeneous system of three equations in 3 unknowns and has a unique trivial solution.
If and only if Determinant of the coefficient matrix is non zero.
consider
.
.
i.e.
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188334_674513.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188334_7324913.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188334_7795403.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188334_864104.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188334_922889.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188334_9928632.png)
Thus for the system has an 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 a linear combination of these vectors.
Characteristic equation:-
Let A be a square matrix, be any scaler then
is called a characteristic equation of a matrix A.
Note:
Let a be a square matrix and ‘’ be any scaler then,
1) is called a characteristic matrix
2) is called a characteristic polynomial.
The roots of characteristic equations are known as characteristic root or latent roots, eigenvalues, or proper values of a matrix A.
Eigenvector:-
Suppose be an eigenvalue of a matrix A. Then
a non – zero vector x1 such that.
… (1)
Such a vector ‘x1’ is called as eigenvector corresponding to the eigenvalue .
Properties of Eigenvalues:-
- Then the sum of the eigenvalues of a matrix A is equal to the sum of the diagonal elements of a matrix A.
- The product of all eigenvalues of a matrix A is equal to the value of the determinant.
- If
are n eigenvalues of square matrix A then
is m eigenvalues of a matrix A-1.
- The eigenvalues of a symmetric matrix are all real.
- If all eigenvalues are non – zen then A-1 exists and conversely.
- The eigenvalues of A and A’ are the same.
Properties of eigenvector:-
- The eigenvector corresponding to distinct eigenvalues is linearly independent.
- If two are more eigenvalues are identical then the corresponding eigenvectors may or may not be linearly independent.
- The eigenvectors corresponding to distinct eigenvalues of a real symmetric matrix are orthogonal.
Example 1
Determine the eigenvalues of the eigenvector of the matrix.
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188335_725891.png)
Solution:
Consider the characteristic equation as
i.e.
i.e.
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188335_9280512.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188335_9919426.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188336_0630128.png)
i.e.
Which is the required characteristic equation.
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188336_221235.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188336_2651887.png)
are the required eigenvalues.
Now consider the equation
… (1)
Case I:
If Equation (1) becomes
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188336_472706.png)
R1 + R2
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188336_5491667.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188336_5994313.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188336_6623383.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188336_7272768.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188336_7723346.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188336_8421326.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188336_8989463.png)
Thus
independent variable.
Now rewrite the equation as,
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188337_0476987.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188337_0988412.png)
Put x3 = t
&
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188337_2305005.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188337_279165.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188337_3392544.png)
Thus .
Is the eigenvector corresponding to .
Case II:
If equation (1) becomes,
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188337_5992882.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188337_6474855.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188337_7031584.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188337_7491639.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188337_8195868.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188337_8652449.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188337_917875.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188337_9681313.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188338_0329616.png)
Here
independent variables
Now rewrite the equations as,
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188338_1974618.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188338_2437925.png)
Put
&
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188338_4136713.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188338_4643788.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188338_5158644.png)
.
Is the eigenvector corresponding to .
Case III:
If equation (1) becomes,
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188338_7602782.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188338_820133.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188338_8658266.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188338_936832.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188338_980597.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188339_0467591.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188339_0931425.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188339_1751747.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188339_241075.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188339_2986372.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188339_3424134.png)
Here the rank of
independent variable.
Now rewrite the equations as,
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188339_4862733.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188339_5529804.png)
Put
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188339_6709762.png)
Thus .
Is the eigenvector for .
Example 2
Find the eigenvalues of an eigenvector for the matrix.
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188339_8220625.png)
Solution:
Consider the characteristic equation as
i.e.
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188339_9545403.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188340_007125.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188340_0739.png)
i.e.
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188340_1827016.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188340_2288878.png)
are the required eigenvalues.
Now consider the equation
… (1)
Case I:
Equation (1) becomes,
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188340_4247577.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188340_4984567.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188340_5478995.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188340_6137946.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188340_6624181.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188340_7189157.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188340_7643046.png)
Thus and n = 3
3 – 2 = 1 independent variables.
Now rewrite the equations as,
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188341_0045373.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188341_0503087.png)
Put
,
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188341_2794635.png)
i.e. the eigenvector for
Case II:
If equation (1) becomes,
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188341_4393075.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188341_488041.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188341_5333593.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188341_581639.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188341_6275868.png)
Thus
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188341_7402186.png)
Independent variables.
Now rewrite the equations as,
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188341_8082335.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188341_8517563.png)
Put
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188341_9824743.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188342_0504391.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188342_101038.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188342_1578703.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188342_2041063.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188342_2572355.png)
Is the eigenvector for
Now
Case II:-
If equation (1) gives,
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188342_4194667.png)
R1 – R2
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188342_4673796.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188342_5203898.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188342_599192.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188342_6509488.png)
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188342_693498.png)
Thus
independent variables
Now
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188342_913494.png)
Put
Thus
![](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1643188343_0299227.png)
Is the eigenvector for .
Reference Book:
1. G.B. Thomas and R.L. Finney, Calculus and Analytic geometry, 9th Edition, Pearson,Reprint, 2002.
2. Erwin kreyszig, Advanced Engineering Mathematics, 9th Edition, John Wiley &Sons,2006.
3. Veerarajan T., Engineering Mathematics for the first year, Tata McGraw-Hill, NewDelhi,2008.
4. Ramana B.V., Higher Engineering Mathematics, Tata McGraw Hill New Delhi, 11thReprint, 2010.
5. D. Poole, Linear Algebra: A Modern Introduction, 2nd Edition, Brooks/Cole, 2005.
6. B.S. Grewal, Higher Engineering Mathematics, Khanna Publishers, 36th Edition, 2010