Matrix: Homogeneous system of linear equations

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 (Rouché - 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, ρ ( A) = ρ ([ A | 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 + ……… + a_1nx_n + = 0 

a₂₁x₁ + a₂₂x₂ + a₂₃x₃ + ……… + a_2nx_n + = 0 

a₃₁x₁ + a₃₂x₂ + a₃x₃ + ……… + a_3nx_n + = 0 

…..    ….   …..   …..   …..    ...

a_m1x₁ + a_m2x₂ + a_m3x₃ + ……… + a_mnx_n + = 0               ………(1)

where          the  coefficients  a_ij , i = 1, 2,…., m; j = 1, 2,…., n are constants. The above system is always satisfied by x₁ = 0, x₂ = 0,….,, x_n = 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 = O_mx1, where

We will denote O_{m × 1} simply by the capital letter O. Since O is the zero column matrix, it is  always true that ρ ( A) = ρ ([ A | O]) ≤ m. So, by Rouché - 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 ρ ( A) = ρ ([ 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 + ……… + a_1nx_n + = 0 

a₂₁x₁ + a₂₂x₂ + a₂₃x₃ + ……… + a_2nx_n + = 0 

a₃₁x₁ + a₃₂x₂ + a₃x₃ + ……… + a_3nx_n + = 0 

…..    ….   …..   …..   …..    ...

a_n1x₁ + a_n2x₂ + a_n3x₃ + ……… + a_nnx_n + = 0               ………(1)

Two cases arise.

Case (i)

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

Since ρ ( A) = n, |A| ≠ 0. So for trivial solution | A | ≠ 0 . 

Case (ii)

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

Example 1.36

Solve the system: x + 3y − 2z = 0, 2x − y + 4z = 0, x −11y +14z = 0.

Solution

Here the number of unknowns is 3. 

Transforming into echelon form (Gaussian elimination method), the augmented matrix becomes

So, ρ(A) =  ρ ([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 – 2z =0, 7y – 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 − 2z = 0, 2x − 3y + z = 0, 3x − 7 y +10z = 0, 6x − 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, ρ(A) = ρ([A|O]) = 3 = Number of unknowns

Hence the system has trivial solution only.

Example 1.38

Determine the values of λ for which the following system of equations 

(3λ – 8 )x + 3y + 3z =0, 3x+(3λ-8)y + 3z = 0, 3x + 3y + (3λ -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 → 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 ⇒ 5x₁ – x₃ =0                   ……..(2)

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

8x₁ = 2x₄ ⇒ 4x₁ – x₄ = 0,                 ……. (3) 

2x₂ = 2x₃ + x₄ ⇒ 2x₂ – 2x₃ – 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, ρ(A) = ρ([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 – 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 ≠ 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 → 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 ≠ a, q ≠ b, r ≠ 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 → R₂ − R₁ and R₃ → R₃ − R₁ and in the above equation, 

we get