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:

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

*a*_{11}*x*_{1} +* a*_{12}*x*_{2} +* a*_{13}*x*_{3} + ……… +* a*_{1n}*x*_{n} + = 0

*a*_{21}*x*_{1} +* a*_{22}*x*_{2} +* a*_{23}*x*_{3} + ……… +* a*_{2n}*x*_{n} + = 0

*a*_{31}*x*_{1} +* a*_{32}*x*_{2} +* a*_{3}*x*_{3} + ……… +* a*_{3n}*x*_{n} + = 0

….. …. …..
….. ….. ...

*a*_{m1}*x*_{1} +* a*_{m2}*x*_{2} +* a*_{m3}*x*_{3} + ……… +* a*_{mn}*x*_{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 *x*_{1} = 0, *x*_{2} = 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*_{m}_{x1}, 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:

*a*_{11}*x*_{1} +* a*_{12}*x*_{2} +* a*_{13}*x*_{3} + ……… +* a*_{1n}*x*_{n} + = 0

*a*_{21}*x*_{1} +* a*_{22}*x*_{2} +* a*_{23}*x*_{3} + ……… +* a*_{2n}*x*_{n} + = 0

*a*_{31}*x*_{1} +* a*_{32}*x*_{2} +* a*_{3}*x*_{3} + ……… +* a*_{3n}*x*_{n} + = 0

….. …. …..
….. ….. ...

*a*_{n1}*x*_{1} +* a*_{n2}*x*_{2} +* a*_{n3}*x*_{3} + ……… +* a*_{nn}*x*_{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*.

Solve the following
system:

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

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.

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

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.

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

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:

*C*_{5}*H*_{8} + *O*_{2} → *CO*_{2} + *H*_{2}*O*.

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

**Solution**

We are searching for
positive integers *x*_{1}, *x*_{2}, *x*_{3} and *x*_{4} such that

x_{1}C_{5}H_{8} + x_{2}O_{2} = x_{3}CO_{2} + x_{4}H_{2}O
....
(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

5x_{1} = x_{3} ⇒ 5x_{1} – x_{3} =0
……..(2)

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

8x_{1} = 2x_{4} ⇒ 4x_{1} – x_{4} = 0, ……. (3)

2x_{2} = 2x_{3} + x_{4} ⇒ 2x_{2} – 2x_{3} – x_{4} =
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

4x_{1} – x_{4} = 0, 2x_{2 }- 2x_{3 }- x_{4 }= 0, -4x_{3} + 5x_{4} = 0.

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

Let us choose *x*_{4} = t. t ≠ 0. Then, by
back substitution, we get

Since x_{1}, x_{2}, x_{3}, and x_{4} are positive integers,
let us choose t = 4.

Then, we get *x*_{1} =1, *x*_{2} =7, *x*_{3} =5, and *x*_{4} = 4.

So, the balanced
equation is C_{5}H_{8} + 7O_{2} → 5CO_{2} + H_{2}O.

**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 R_{2} → R_{2} − R_{1} and R_{3} → R_{3} − R_{1} 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

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