Matrices and Determinants


   
 
Application of Matrices and Determinants
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 3rd equation, the system has infinitely many solutions x = k, y = 4 - 2k and z = k + 2, where k is any number.
 
  
     
   
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