Application of Matrices and Determinants

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Application of Determinants

Now we shall discuss the use of determinants in finding the area of a triangle and in the solution of simultaneous equations.

Area of a Triangle

We have already learnt in the previous class that the area of triangle whose vertices are (x₁, y₁), (x₂, y₂), (x₃, y₃) is given by

Hence area of a triangle having vertices at (x₁, y₁), (x₂, y₂) and (x₃, y₃) is given by

Since the area has to be a positive quantity, we always take the absolute value of the above determinant.

Example:

Using determinants, find the area of triangle whose vertices are

(2, -7), (1, 3), (10, 8).

Solution:

(x₁, y₁) = (2, -7)

(x₂, y₂) = (1, 3)

(x₃, y₃) = (10, 8)

Area of the triangle

= -47.5

Since area has to be a positive quantity, it is given by 47.5 sq.units.

Cramer's rule for the solution of a system of equations in 2 variables

We recall from our earlier classes that a system of linear equation with two variables is given by

This system of linear equation may have either one solution or infinitely many solutions or no solution.

Consistency of a system of linear equation

If a system of linear equations has at least one solution, then the system is called consistent, otherwise it is called inconsistent.

Solve the system of linear equations (1) by using method of elimination as studied earlier

Multiplying the first equation by a₂ and the second equation by a₁, we get

Similarly a₁a₂ x + a₁a₂ y = a₁c₂ ...(3)

Subtracting (2) from (3), we get

(a₁b₂ - a₂b₁)y = a₁c₂ - a₂c₁

The above discussion leads to find the solution of a system of linear equations in two variables by using Cramer's rule.

Cramer's rule suggests the use of determinants to solve a system of linear equations.

Let us denote a₁b₂ - a₂b₁ (Denominators of x and y in (4) and (5))

as the determinant

Then the solution of the given system of linear equation (1) is given by

Now consider the system of linear equations with three variable

Let the determinant of the coefficients be denoted by D

i.e.,

Multiplying both sides by x, we have

Adding y times column 2 and z times column 3 to column (1), we have

Similarity it can be shown that,

and

This method of solving a system of linear equations with three unknowns, is known as Cramer's rule.

Note:

In the above method note that

To obtain D₁, replace a₁, a₂, a₃ by d₁, d₂, d₃ in D

To obtain D₂, replace b₁, b₂, b₃ by d₁, d₂, d₃ in D

To obtain D₃, replace c₁, c₂, c₃ by d₁, d₂, d₃ in D

Example:

Solve the system of linear equations.

x +2y + 3z = 6

2x + 4y + z = 7

3x + 2y + 9z = 14

using Cramer's rule.

Suggested answer:

The determinant of coefficients

Note that we can also evaluate the determinant D₁, D₂ and D₃ directly without using the properties of determinant.

The solution of the system is given by

It is important to mention here the consistency and inconsistency of a system of linear equations with three unknowns.

Case I:

Case II:

If D = 0 and D₁, D₂ and D₃ are not all zero, then the system is inconsistent, that is the system has no solution.

Case III:

If D = 0 and all D₁, D₂ and D₃ are zeros, this system has either infinite solution or no solution.

In this case, put x = k(y = k or z = k), in any two of the equations, find y and z in terms of k. Substitute these values of x, y and z in terms of k, in the third equation. If the third equation is satisfied by this solution, the system has infinitely many solutions.

If the third equation is not satisfied, the system has no solution.

Application of Matrices

Homogeneous Equations (Constant = 0)

Consider the homogeneous equations

a₁x + b₁y + c₁z = 0

a₂x + b₂y + c₂z = 0

a₃x + b₃y + c₃z = 0

The homogenous system of equations is always consistent because x = 0, y = 0, z = 0 satisfies all the equations in the system. This solution is called the trivial solution.

then the system has a non-trivial solution also. In fact, it has an infinite number of solutions and is said to be dependent.

then the system has only the trivial solution x = 0, y = 0, z = 0.

Non Homogenous Equations (Solution by the Matrix Method)

Consider the non-homogeneous equations

a₁x + b₁y + c₁z = d₁

a₂x + b₂y + c₂z = d₂

a₃x + b₃y + c₃z = d₃

This can be written as

|A| may or may not be zero.

Case I:

Pre-multiply by A^-1,

\ A^-1 (AX) = A^-1B

\ (A^-1A) X = A^-1B

\ I X = A^-1B

or X = A^-1B

This is the matrix method to solve the equations.

However,

Case II:

A^-1 does not exist

But if (adj A) B = 0, then the system is consistent with infinite number of solutions or has no solution.

the system is inconsistent i.e., it has no solution.

Example 1:

Using matrix method solve the following systems of linear equations

2x - y + z = -3

3x - z = - 8

2x + 6y = 2

Suggested answer:

The given equations are

2x - y + z = -3

3x - 0.y - z = - 8

2x + 6y + 0.z= 2

= 2(6) +1(2) + 1(18)

= 12 +2 + 18

= 32

The system has a unique solutions.

A₁₁ = (0 + 6) = 6, A₁₂ = -(0 + 2) = -2, A₁₃ = 18

A₂₁ = 6, A₂₂ = -2, A₂₃ = -14

A₃₁ = 1, A₃₂ = 5, A₃₃ = 3

The solution is

\ x = -2, y = 1, z = 2

Example 2:

Using matrix method, solve the following system of linear equations

x + y + z = 6 …(1)

x + 2y + 3z = 14 …(2)

x + 4y + 7z = 30 …(3)

Suggested answer:

= (14 - 12) - (7 - 3) + (4 - 2)

= 2 - 4 + 2 = 0

The system may have infinite number of solutions or no solution.

Put x = k in (1) and (2) and solve

y + z = 6 - k

2y + 3z = 14 - k.

Solving the above two equations, we have

z = k + 2 and y = 4 - 2k

When x = k, substituting these values of x, y and z in (3), we get

x + 4y + 7z = k + 16 - 8k + 7k +14

= 30

Since these values of x, y, z satisfy the 3^rd equation, the system has infinitely many solutions x = k, y = 4 - 2k and z = k + 2, where k is any number.