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# Matrix: Homogeneous system of linear equations

Applications of Matrices: Consistency of System of Linear Equations by Rank Method

## Applications of Matrices: Consistency of System of Linear Equations by Rank Method

In second previous section, we have already defined consistency of a system of linear equation. In this section, we investigate it by using rank method. We state the following theorem without proof:

## Theorem 1.14 (Rouch├ę - Capelli Theorem)

A system of linear equations, written in the matrix form as AX = B, is consistent if and only if the rank of the coefficient matrix is equal to the rank of the augmented matrix; that is, ¤ü A= ¤ü ([ B]).

We apply the theorem in the following examples.

Homogeneous system of linear equations

We recall that a homogeneous system of linear equations is given by

a11x1 + a12x2 + a13x3 + ÔÇŽÔÇŽÔÇŽ + a1nxn + = 0

a21x1 + a22x2 + a23x3 + ÔÇŽÔÇŽÔÇŽ + a2nxn + = 0

a31x1 + a32x2 + a3x3 + ÔÇŽÔÇŽÔÇŽ + a3nxn + = 0

ÔÇŽ..    ÔÇŽ.   ÔÇŽ..   ÔÇŽ..   ÔÇŽ..    ...

am1x1 + am2x2 + am3x3 + ÔÇŽÔÇŽÔÇŽ + amnxn + = 0               ÔÇŽÔÇŽÔÇŽ(1)

where          the  coefficients  aij , i = 1, 2,ÔÇŽ., m; j = 1, 2,ÔÇŽ., n are constants. The above system is always satisfied by x1 = 0, x2 = 0,ÔÇŽ.,, xn = 0.This solution is called the trivial solution of (1). In other words, the system (1) always possesses a solution.

The above system (1) can be put in the matrix form AX = Omx1, where We will denote Om ├Ś 1 simply by the capital letter O. Since O is the zero column matrix, it is  always true that ¤ü ( A) = ¤ü ([ A | O]) ÔëĄ m. So, by Rouch├ę - Capelli Theorem, any system of homogeneous linear equations is always consistent.

Suppose that m < n, then there are more number of unknowns than the number of equations. So ¤ü ( A) = ¤ü ([ A | O]) < n. Hence, system (1) possesses a non-trivial solution.

Suppose that m = n, then there are equal number of equations and unknowns:

a11x1 + a12x2 + a13x3 + ÔÇŽÔÇŽÔÇŽ + a1nxn + = 0

a21x1 + a22x2 + a23x3 + ÔÇŽÔÇŽÔÇŽ + a2nxn + = 0

a31x1 + a32x2 + a3x3 + ÔÇŽÔÇŽÔÇŽ + a3nxn + = 0

ÔÇŽ..    ÔÇŽ.   ÔÇŽ..   ÔÇŽ..   ÔÇŽ..    ...

an1x1 + an2x2 + an3x3 + ÔÇŽÔÇŽÔÇŽ + annxn + = 0               ÔÇŽÔÇŽÔÇŽ(1)

Two cases arise.

Case (i)

If ¤ü ( A) = ¤ü ([ A | O]) = n, then system (2) has a unique solution and it is the trivial solution.

Since ¤ü ( A) = n, |A| Ôëá 0. So for trivial solution | A | Ôëá 0 .

Case (ii)

If ¤ü ( A) = ¤ü ([ A | O]) < n, then system (2) has a non-trivial solution. Since ¤ü ( A) < n, |A| =0.

In other words, the homogeneous system (2) has a non-trivial solution if and only if the determinant of the coefficient matrix is zero.

Suppose that m > n, then there are more number of equations than the number of unknowns.

Reducing the system by elementary transformations, we get ¤ü (A) = ¤ü ([ A | O]) ÔëĄ n.

### Example 1.35

Solve the following system:

x + 2y + 3z = 0, 3x + 4y + 4z = 0, 7x + 10y + 12z =0.

### Solution

Here the number of equations is equal to the number of unknowns.

Transforming into echelon form (Gaussian elimination method), the augmented matrix becomes So, ¤ü(A) = ¤ü([A| O]) = 3 = Number of unknowns

Hence, the system has a unique solution. Since x = 0, y = 0, z = 0 is always a solution of the  homogeneous system, the only solution is the trivial solution x = 0, y = 0, z = 0.

Note

In the above example, we find that

|A|= = 1(48-40) - 2(36-28) + 3(30-28) = 8-16+6 = -2 Ôëá 0.

### Example 1.36

Solve the system: x + 3y Ôłĺ 2z = 0, 2x Ôłĺ y + 4z = 0, x Ôłĺ11y +14z = 0.

### Solution

Here the number of unknowns is 3.

Transforming into echelon form (Gaussian elimination method), the augmented matrix becomes So, ¤ü(A) =  ¤ü ([A | O ] ) = 2 < 3  = Number of unknowns

Hence, the system has a one parameter family of solutions.

Writing the equations using the echelon form, we get

x + 3y ÔÇô 2z =0, 7y ÔÇô 8z = 0, 0 = 0.

Taking z = t, where t is an arbitrary real number, we get by back substitution, , where is any real number.

### Example 1.37

Solve the system: x + y Ôłĺ 2z = 0, 2x Ôłĺ 3y + z = 0, 3x Ôłĺ 7 y +10z = 0, 6x Ôłĺ 9 y +10z = 0.

### Solution

Here the number of equations is 4 and the number of unknowns is 3. Reducing the augmented matrix to echelon-form, we get So, ¤ü(A) = ¤ü([A|O]) = 3 = Number of unknowns

Hence the system has trivial solution only.

Example 1.38

Determine the values of ╬╗ for which the following system of equations

(3╬╗ ÔÇô 8 )x + 3y + 3z =0, 3x+(3╬╗-8)y + 3z = 0, 3x + 3y + (3╬╗ -8)z = 0

has a non-trivial solution.

Solution

Here the number of unknowns is 3. So, if the system is consistent and has a non-trivial solution,  then the rank of the coefficient matrix is equal to the rank of the augmented matrix and is less than 3.  So the determinant of the coefficient matrix should be 0.

Hence we get We now give an application of system of linear homogeneous equations to chemistry. You are already aware of balancing chemical reaction equations by inspecting the number of atoms present on both sides. A direct method is explained as given below.

Example 1.39

By using Gaussian elimination method, balance the chemical reaction equation:

C5H8 + O2 Ôćĺ CO2 + H2O.

(The above is the reaction that is taking place in the burning of organic compound called isoprene.)

Solution

We are searching for positive integers x1, x2, x3 and x4 such that

x1C5H8 + x2O2 = x3CO2 + x4H2O          .... (1)

The number of carbon atoms on the left-hand side of (1) should be equal to the number of carbon atoms on the right-hand side of (1). So we get a linear homogenous equation

5x1 = x3 Ôçĺ 5x1 ÔÇô x3 =0                   ÔÇŽÔÇŽ..(2)

Similarly, considering hydrogen and oxygen atoms, we get respectively,

8x1 = 2x4 Ôçĺ 4x1 ÔÇô x4 = 0,                 ÔÇŽÔÇŽ. (3)

2x2 = 2x3 + x4 Ôçĺ 2x2 ÔÇô 2x3 ÔÇô x4  = 0. ... (4)

Equations (2), (3), and (4) constitute a homogeneous system of linear equations in four unknowns.

The augmented matrix is [ A|B]  = By Gaussian elimination method, we get Therefore, ¤ü(A) = ¤ü([A|B]) =  3 < 4  = umber of unknowns

The system is consistent and has infinite number of solutions.

Writing the equations using the echelon form, we get

4x1 ÔÇô x4 = 0, 2x2 - 2x3 - x4 = 0, -4x3 + 5x4 = 0.

So, one of the unknowns should be chosen arbitrarily as a non-zero real number.

Let us choose x4 = t. t Ôëá 0. Then, by back substitution, we get Since x1, x2, x3, and x4 are positive integers, let us choose t = 4.

Then, we get x1 =1, x2 =7, x3 =5, and x4 = 4.

So, the balanced equation is C5H8 + 7O2 Ôćĺ 5CO2 +  H2O.

Example 1.40

If the system of equations px + by + cz = 0, ax + qy + cz =0, ax + by + rz = 0 has a non-trivial solution and p Ôëá a, q Ôëá b, r Ôëá c, prove that Solution

Assume that the system px + by + cz = 0, ax + qy + cz = 0, ax + by + rz = 0 has a non-trivial solution

So, we have = 0.

Applying R2 Ôćĺ R2 Ôłĺ R1 and R3 Ôćĺ R3 Ôłĺ R1 and in the above equation,

we get Tags : Definition, Theorem, Formulas, Solved Example Problems | Applications of Matrices: Consistency of System of Linear Equations by Rank Method , 12th Mathematics : UNIT 1 : Applications of Matrices and Determinants
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12th Mathematics : UNIT 1 : Applications of Matrices and Determinants : Matrix: Homogeneous system of linear equations | Definition, Theorem, Formulas, Solved Example Problems | Applications of Matrices: Consistency of System of Linear Equations by Rank Method