Subject Expert: Dr. Bijumon
Contend Editor : Prof. MADHUSOODANAN K.
Presenter: RASHMI
Algebra 22
Solution of Non-Homogeneous
System of Linear Equations
Objectives
From this unit a learner is expected to achieve the following
- Familiarize with system of linear equations and augmented matrix.
- Understand the concept of Consistent System of Linear Equations.
- Study thenature of the General Solution of Nonhomogeneous system of equations.
- Learn the method to find Solution of a Nonhomogeneous System of Linear Equations.
Sections
1. Introduction
2. System of Linear Equations and Augmented Matrix
3. Consistent System of Linear Equations
4. Nature of the General Solution
5. Method of Finding Solution to a Non-Homogeneous System of Linear Equations
1. Introduction We shall begin this session by reviewing the concept of system of linear equations and augmented matrix.We will also present the definition and a discussion of consistency of system of linear equations. Nature of the General Solution of a nonhomogeneous system of equations will be analysed. Method to find solution of a Nonhomogeneous System of Linear Equations will be illustrated through examples.
2. System of Linear Equations and Augmented Matrix
A system of m linear equations in n unknowns x1, x2, . . . , xn is a set of equations of the following form:
. . . (1)
The above system of equations can be expressed as the single matrix equation
. . . (2)
where A, X and B are as follows:
, and
If the zero column matrix, then the system (2) is called a nonhomogeneous system. In that case at least one of b j is different from 0 and the system of linear of equations (1) is a non- homogeneous system.
A solution of the system (1) is any set of values of which satisfy simultaneously the equations. Equivalently, any column matrix X that satisfies the matrix equation (2) is a solution of the system (2). When the system has a solution, it is said to be consistent; otherwise, the system is said to be inconsistent. A consistent system has either a unique solution or infinitely many solutions.
It may be noted thatthe zero column matrix
is not a solution of any non-homogeneous system of m equations.
Definition(Augmented matrix) The matrix [AB] obtained by placing the column matrix B to the right of the matrix A is called the augmented matrix. Thus the augmented matrix of the system AX = B is
i.e., is the matrix whose first n columns are corresponding ncolumns of A and the th column is the column matrix B.
3. Consistent System of Linear Equations
We now we characterize a consistent system of linear equations.
Theorem1(Necessary and sufficient condition for a system of linear equations to be consistent)
The system AX = B is consistent if and only if the matrix A and the augmented matrix [AB] have the same rank.
Equivalently, the system (1) of linear equations has a solution if and only if A and [AB] have the same rank.
Proof
Let the rank of the matrix A be r. Then any linearly independent set of column vectors of A contains at most r elements ; if a set consists of r linearly independent columns, then that set is a basis for the vector space generated by columns of A. Also wemay write
and, without loss of generality, assume that the first r columns form a basis for the column space of A and hence each column of A is a linear combination of these columns.
The given equation AX = B is equivalent to
…(3)
Suppose, now, that the rank of the matrix is the same as that of matrix A, i.e., suppose rank of is equal to r.
Now, since rank of is also equal to r, the maximum number of linearly independent columns of the matrix is r and its columns form such a system. Hence the column B can be expressed as a linear combination of as follows:
… (4)
where are real numbers.
We rewrite (4) as
…(5)
Comparing (3) and (5),
so that there exist a solution whose first r components are and each of remaining components is zero.
Thus the given system of equations is consistent if the matrices A and have the same rank.
Conversely, suppose the system of equations is consistent so that there exists n scalars
such that
… (6)
As the rank of the matrix A is r, each of the columns of the matrix A is a linear combination of the columns and hence we deduce from (6) that B alsoisa linear combination of Thus the maximum number of linearly independent columns in the matrix is r and, as such, its rank is r which is also the rank of A.
4. Nature of the General Solution
Suppose that the matrices A and are of the same rank r so that the system of equations
…(1)
is consistent.
Since r is the rank, the corresponding homogeneous system
…(2)
has linearly independent solutions.
Let
be a set of linearly independent solutions of the homogeneous system (2).
Let be any solution of the non-homogeneous system (1).
Claim:
where is the general solution of the homogeneous system (2) (Here being any set of scalars.)
We have
as
Hence is a solution of the given equation
Conversely, if Z be a solution of the equation (1), then we have
…(3)
and hence
so that is a solution of the homogeneous equation (2).
Writing
we see that every solution Z of the equation (1) is expressible as the sum of a solution and some solution of the homogeneous equation. This completes the proof.
Corollary If A is an n-rowed non-singular square matrix, then the system (1) is consistent and hasaunique solution.
Proof
In this case the rank of A and is n. Thus the equation is consistent. Since rank is equal to the number of unknowns, the solution is unique (In fact is the unique solution).
Corollary Suppose that A, B are n-rowed square matrices and X is an n-rowed square matrix of unknowns such that
The matrix equation has a solution if and only if each of the matrices A and have the same rank.
Proof
Suppose that A, B are n-rowed square matrices and X is ann-rowed square matrix of unknowns such that
Let and .
so that are columns of the matrix B and are columns of the matrix X of unknowns. Then the given equation is equivalent to the system of equations
Thus we see that the equation is consistent if and only if each of the matrices
has same rank, i.e., if and only if each of the columns of the matrix B is linearly dependent upon the set of columns of A.
These conditions are equivalent to saying that the matrices A and have the same rank.
Thus the matrix equation has a solution if and only if each of the matrices A and have the same rank.
Remarks
With the same assumption as in the theorem and that has a solution, we have the following particular cases.
- In particular, if A is an n-rowed non-singular matrix, then the matrices A and are of the same rank viz., n. In that case the equation possesses the unique solution .
- In case A is singular and B is the zero column matrix, then obviously the condition for consistency is satisfied. In this case the solution will not be unique as the solution of the auxiliary equation is also not unique in this case. But we bring the attention of the student to note that if A is singular and B is not the zero column matrix, then the condition for consistency may not besatisfied.
Remark In a consistent system of linear equations with n unknowns whose coefficient matrix Ahas rank r <n , we can assign arbitrary values to the (n r) unknowns and determine the remaining r unknowns uniquelyin terms of these arbitrary values. In this case the system have infinite number of solutions.
5. Method of Finding Solution of a Non-Homogeneous System of Linear Equations
If we are given with a system of m equations in n unknowns, proceed as follows:
- Write down the corresponding matrix equation AX = B.
- By elementary row transformations obtain row reduced echelon form of the augmented matrix [AB].
- Examine whether the rank of A and the rank of [AB] are the same or not.
Case 1 If
rank of A = rank of [AB] ,
then the system is consistent.
Case 1a If rank of A = rank of [AB] = n = number of unknowns,
then the system has aunique solution.
Case 1b If
r = rank of A = rank of [AB] < n = number of unknowns,
then the system has infinitely many solutions. We assign arbitrary values to (n r) unknowns and determine the remaining r unknowns uniquely.
Case 2 If rank of A rank of [AB] ,
then the system is inconsistentand has no solution.
Example 1 Show that the system of equations
is consistent and hence solve them.
Solution The matrix equation corresponding to the given system of equations is
. . . (4)
The augmented matrix [AB] is given by
The matrix given just above is the row echelon form of the augmented matrix [AB] obtained by a finite sequence of elementary row transformations to the matrix [AB]. Since the echelon matrix has 3 non-zero rows with leading coefficients 1 and columns containing that 1 has zeros elsewhere we can see that rank of [AB] is 3.
Also, by the same sequence of row transformations, we have
,
so that rank of A = 3.
Hence
rank of A = 3 = rank of [AB].
Therefore by the theorem, the system is consistent.
Also since
rank of A = 3 = the number of unknowns,
the system has unique solution.
Now to find the solution we see that the given system of equations (3) or the matrix equation (4) is equivalent to the matrix equation
x = 1; y = 0 ; z = 1.
So the solution to the system of equations is x = 1; y = 0; z = 1.
is the solution vector.
Example2 Show that the system of equations given below is consistent hence solve the same.
x1 + x2 + x3 = 4
2 x1 + 5 x2 2 x3 = 3
Solution
The augmented matrix corresponding to the given system of equations is given by
The matrix given just above is the row echelon form of [AB]. Since the echelon matrix has 2 non- zero rows, rank of [AB] is 2.
Also, by the same sequence of elementary transformations we have
,
so that rank of A = 2.
Hence
rank of [AB] = 2 = rank of A,
so that by the theorem, the system is consistent.
Also since the rank of A = 2 < 3 = the number of unknowns, the system possesses infinitely many solutions. We can assign arbitrary values to n r = 3 2 = 1 unknown.
Now to find the solution we see that the given system of equation is equivalent to the matrix equation
The matrix equation is equivalent to the system of equations
Choose x3 = a, where a is an arbitrary constant that can take any value.
Then we obtain
i.e.,
i.e.,
Hence the solution is given by
Remarks
- In the solution of Example 2, a is arbitrary and if we put particular values to a, we get particular solutions to the given equation. If we put a = 1 the solution will be . As we can put infinitely many values to a, the number of solutions to the given system of linear equations is infinite.
- In Example 2, one can assign arbitrary value to any one of the variables x1 , x2and x3 ; in the solution abovewe have put the arbitrary value to the variable x3 . If we put arbitrary valuea to the variable x1 , the solution vector will be
Similarly, if we put arbitrary valuea to the variable x2 , the solution vector will be
Example3 Apply the rank test to examine whether the following system of equations is consistent and if consistent, find the complete solutions:
x + y + z = 3
3 x + y 2 z = 2
2 x + 4 y + 7 z = 7.
Solution
By performing row elementary transformations to the augmented matrix five times, we obtain its row echelon form
.
Since the echelon form has three non-zero rows, the rank of = 3.
Also, by the same elementary transformation it can be seen that the echelon form of A is
.
Since the echelon form has two non-zero rows, the rank of A = 2.
Hence it can be seen that rank A = 2 3 = rank [AB].
Hence the system is inconsistent and has no solution.
Example 11 Investigate for what values of a, b the system of equations
x + y + 2 z = 2
2 x y + 3 z = 10
5 x y + a z = b
have
(i)no solution ;
(ii) unique solution ; and
(iii) an infinite number of solutions.
Solution Here the augmented matrix is
(i) the given system of equations have no solution if and only if
rank of A rank of the augmented matrix [AB].
From the last matrix (echelon form of the augmented matrix), it can be seen that this is possible if and only if
a 8 = 0 and b 22 0,
i.e., if and only if a = 8 and b 22.
i.e., the system has no solution if a = 8 and b 22.
(ii) the given system of equations have unique solution if
rank of A = rank of the augmented matrix [AB] = 3, the number of unknowns.
From the last matrix obtained by elementary row transformations on [AB], it can be seen that this is possible if and only if
a 8 0 and with b 22 has any value.
i.e., if and only if a 8 and b take any value.
i.e. the system has unique solution if a 8 and b take any value.
(iii) the given system of equations have infinitely many solutions if
rank of A = rank of the augmented matrix [AB] < 3, the number of unknowns.
From the last matrix obtained by elementary row transformations on [AB], it can be seen that this is possible if and only if
a 8 = 0 and with b 22 = 0.
i.e., the system has infinitely many solutions if a = 8 and b = 22.
Summary
In this session we have reviewed the concept of system of linear equations and augmented matrix. The definition of consistency of system of linear equations has been given. Nature of the General Solution nonhomogeneous system of equations are considered. Method of finding solution to a Non-Homogeneous System of Linear Equations is illustrated through examples.
Assignments
Test for consistency and if consistent solve (Exercises 1 to 5):
1.
2.
3.
4.
5.
6. Show that the equations:
x + y + z = a,
3x + 4y + 5z = b,
2x + 3y + 4z = c
(i) have no solution if
and (ii) have many solutions if
7. Find the values of a and b for which the system of equations
x + y + z = 3,
x + 2 y + 2 z = 6,
x + a y + 3 z = b have
(i) no solution ;
(ii) unique solution ; and
(iii) more than one solution.
8. Prove that the equations
3x + 4y + 5z = a ;
4x + 5y + 6z = b;
5x + 6y + 7z = c
are consistent only when a + c = 2 b.
9. Find the values of a for which the equations
x + y + z = 3;
x + 2 y + 2 z = 6;
x + a y + 3 z = 2 have
(i)no solution and
(ii) unique solution .
10. Given , find a matrix B of rank 2 such that .
11. Show that a square matrix is singular if and only if its rows (columns) are linearly dependent.
12. Let be a system of homogeneous equations in n unknowns and suppose of rank . Show that any non-zero vector of cofactors of a row of is a solution of .
Quiz
1. Vector space generated by columns of a matrix A is called
(a)column space of A
(b)row space of A
(c)null space of A
(d) none of the above.
Ans. (a)
2. Which one of the following system of equations is not consistent.
(a)
2x + 3 y + z = 9
x + 2 y + 3 z = 6
3x + y +2 z = 8
(b)
(c)
(d)
x + 2 y = 3
2 x + 4 y = 7
Ans. (d)
FAQ
1. Is there any geometrical significance to solutions of a system of linear equations?
Ans. Yes.
For a system involving two variables (x and y), each linear equation determines a line on the xy-plane. Because a solution to a linear system must satisfy all of the equations, the solution set is the intersection of these lines, and is hence either a line, a single point, or the empty set.
For three variables, each linear equation determines a plane in three-dimensional space, and the solution set is the intersection of these planes. Thus the solution set may be a plane, a line, a single point, or the empty set.
For n variables, each linear equation determines a hyperplane in n-dimensional space. The solution set is the intersection of these hyperplanes, which may be a flat of any dimension.
2. Is there any method other than the one described in this session to find solution of nonhomogeneous system of linear equations?
Ans. Yes. Cramer's ruleis an explicit formula for the solution of a system of linear equations, in which value of each unknown variable is given as a quotient of two determinants. Using this method, to find a solution of a consistent system of non-homogeneous equations in unknowns , we proceed as follows:
Denote by the matrix obtained from A by replacing its ith column with the column of constants (the b’s). Then, if , the system has the unique solution given by
As an example we solve the system
using Cramer’s Rule. First of all we determine the determinants of and and the values of them are:
;
;
and
Then
and
Though Cramer's rule is important theoretically, it has little practical value for large matrices, since the computation of large determinants is somewhat cumbersome. (Indeed, large determinants are most easily computed using row reduction.) Further, Cramer's rule has very poor numerical properties, making it unsuitable for solving even small systems reliably, unless the operations are performed in rational arithmetic with unbounded precision.
3. Describe one more method that is not described so far in this session.
Ans. Given a system of n linear equations in n unknowns with the determinant of the coefficient matrix is different from zero. Then the system is obviously consistent. One can use (i.e., the inverse of the coefficient matrix) to find a solution of the system. The method is as follows:
If exists and the solution of the system is given by
To describe this method we solve the system
The coefficient matrix of the system is
and its inverse is obtained to be
.
Then
The solution of the system is or in matrix form
Glossary
System of linear equations:A system of m linear equations in n unknowns x1, x2, . . . , xn is a set of equations of the following form, where the coefficients a j k ’s and b j ’s are real numbers:
. . . (1)
Homogeneous and Non-Homogeneous System of linear equations:The system of equations (1) is said to be homogeneous if all the are zero; otherwise, it is said to be non-homogeneous.
Solution of System of Linear Equations:A solution of the system (1) is any set of values of which satisfy simultaneously the equations.
Consistent and Inconsistent System of Linear Equations: When the system has a solution, it is said to be consistent; otherwise, the system is said to be inconsistent. A consistent system has either a unique solution or infinitely many solutions.
Equivalent System of Equations: Two systems of linear equations having the same number of unknowns are called equivalent if every solution of either system is a solution of the other.
Solution to the matrix equation: A column matrix X is a solution to the matrix equation AX = B if it satisfies the matrix equation.
Augmented Matrix The matrix [AB] which is obtained by placing the column matrix B to the right of the matrix A is called the augmented matrix. Thus the augmented matrix of the system AX = B is
i.e., is the matrix where first n columns are columns of A and the th column is the column matrix B.
REFERENCES
Books
- I.N. Herstein, Topics in Algebra, Wiley Easten Ltd., New Delhi, 1975.
- K. B. Datta, Matrix and Linear Algebra, Prentice Hall of India Pvt. Ltd., New Delhi, 2000.
- P.B. Bhattacharya, S.K. Jain and S.R. Nagpaul, First Course in Linear Algebra, Wiley Eastern, New Delhi, 1983.
- P.B. Bhattacharya, S.K. Jain and S.R. Nagpaul, Basic Abstract Algebra (2nd Edition), CambridgeUniversity Press, Indian Edition, 1997., New Delhi, 1983.
- S.K. Jain, A. Gunawardena and P.B. Bhattacharya, Basic Linear Algebra with MATLAB,KeyCollege Publishing (Springer-Verlag), 2001.
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