## System of Linear Equations

System of Linear EquationsWe have already discussed the linear equations under the topic Quadratic Equations. The set of n (> 2) linear equations is called the system of linear equations and this system is said to be consistent if it has at least one solution.

Illustration:Check the consistency of the following system of equation.

(i) x + 2x = 4

2x + 4y = 9

This is inconsistent because it has no solution i.e. there is no value of x and y which satisfies both the equations.

(ii) x + y = 4

X - y = 0

This is consistent because it has a solution x = 2 and y = 2

Solving System of Linear Equations by using DeterminantsThere are several methods to solve the system of linear equations but determinant is one of the best mathematical tool from which we can solve the system of linear equations very easily.

CRAMER'S RULE

Case I:System of linear equations in two variables.Let, us have the system of equations

a

_{1}x + b_{1}y + c_{1}= 0a

_{2}x + b_{2}y + c_{2}= 0, where a_{1}/a_{2}≠ b_{1}/b_{2}Solving by cross multiplication, we get,

Case II:System of linear equations in three variables:Let, us have the system of equations.

a

_{1}x + b_{1}y + c_{1}= d_{1}a

_{2}x + b_{2}y + c_{2}= d_{2}a

_{3}x + b_{3}y + c_{3}= d_{3}

.·. x = Δ

_{1}/ΔSimilarly

y = Δ

_{2}/Δ and z = Δ_{3}/Δ, where Δ ≠ 0.

Illustration:Solve the following system of equations.

2x - y + 3z = 9

x + y + z = 6

x - y + z = 2

Solution:Here Δ = = -2 (≠0)

Δ

_{1}= = -2Δ

_{2}= = -4Δ

_{3}= = -6By Cramer's Rule,

x = Δ

_{1}/Δ = 1y = Δ

_{2}/Δ = 2z = Δ

_{3}/Δ = 3

Consistency of the System of the EquationsLet the given system of equation be

a

_{1}x + b_{1}y + c_{1}= d_{1}a

_{2}x + b_{2}y + c_{2}= d_{2}a

_{3}x + b_{3}y + c_{3}= d_{3}From the last enquiry,

Now, there are two possibilities.

Case I.when Δ ≠ 0In this case we have,

x = Δ

_{1}/Δ, y = Δ_{2}/Δ, z = Δ_{3}/Δ = 3Hence unique value of x, y, z will be obtained

Case II:when Δ = 0(a) When at least one of Δ

_{1}, Δ_{2}and Δ_{3}is non zero.Let Δ

_{1}¹ 0, then from case I, Δ_{1}= x Δ will not be satisfied for any value of x because Δ = 0 and Δ_{1}¹ 0 and hence no value of x is possible in this case. Similarly when Δ_{2}¹ 0, and Δ_{2}= y Δ and similarly for Δ_{3}¹ 0.Therefore, if Δ = 0 and any of Δ

_{1}, Δ_{2}and Δ_{3}is non zero, then no solution is possible and hence system of equations will be inconsistent.(b) When Δ = 0 and Δ

_{1}= Δ_{2}= Δ_{3}= 0.and we have,

Δ

_{1}= x Δ, Δ_{2}= y Δ and Δ_{3}= zΔ will be true for all values of x, y and z. But since a x + b_{1}y + c_{1}z = d_{1}, so, only two of x, y, z will be independent and third will be dependent on other two, therefore if Δ = Δ_{1}= Δ_{2}= Δ_{3}= 0, then the system of equations will be consistent and it will have infinitely many solutions.

System of homogeneous linear equationsA system of linear equations is said to be homogenous if sum of the powers of the variables in each term is same. In other words we can say that if constant term is a zero in a system of linear equations.

Let's consider the system of linear homogeneous equations to be

a

_{1}x + b_{1}y + c_{1}z = 0a

_{2}x + b_{2}y + c_{2}z = 0a

_{3}x + b_{3}y + c_{3}z = 0By clean observation, x = 0, y = 0, z = 0 is a solution of above system of equations. This solution is known as trivial solution. For non-trivial solution, consider first two equations from above system.

x/(b

_{1}c_{2}-b_{2}c_{1}) = y/(c_{1}a_{2}-c_{2}a_{2}) = z/(a_{1}b_{2}-b_{1}a_{2}) = k.·. x = k (b

_{1}c_{2}- b_{2}c_{1}), y = k (c_{1}a_{2}- c_{2}a_{1}), z = k (a_{1}b_{2}- b_{2}a_{2})Now putting these values in the third equation, we get,

k[a

_{3}(b_{1}c_{2}- c_{1}b_{2}) -b_{3}(a_{1}c_{2}- c_{1}a_{2}) -c_{3}(a_{1}b_{2}- b_{1}a_{2})] = 0or, a

_{3}(b_{1}c_{2}- c_{1}b_{2}) -b_{3}(a_{1}c_{2}- c_{1}a_{2}) -c_{3}(a_{1}b_{2}- b_{1}a_{2}) = 0 [k ≠ 0]or, = 0

This is required condition for the above system of above homogeneous linear equations to have non-trivial solution.

Illustration:For what value of k doe the following homogeneous system of equations posses a non-trivial solution: x + ky + 3z = 0, kx + ky - 2z = 0, 2x + 3y - 4z = 0.

Solution:For non-trivial solution i.e. D = 0.

D = = 0.

Apply R

_{2}- 3R_{1}and R_{3}- 2R_{1}=> = 0.Expanding by 1

^{st}column, we get k = 33/2.

Illustration:For what values of p and q, the system of equation 2x + py + 6z = 8, x + 2y + qz = 5, x + y + 3z = 4 has (i) no solution (ii) a unique solution (iii) infinitely many solutions.

Solution:Here Δ = = 2(6-q) - p(3-q) + 6(1-2) = 12 - 2q - 3p + pq - 6

=pq-2q-3p+6 = (p-2)(q-3). Also Δ

_{1 }6(6-p)-p(15-49)+6(5-8)= 48 - 8q - 15p + 4pq - 18 = 4pq - 18 = 4pq - 8q - 15p + 30 = 4q(p-2)

- 15(p - 2) = (4q - 15)(p - 2), Δ_{2}= 2(15-4q)-8(3-q)+6(4-5)=0.and Δ

_{3}= = 2(8 - 5) - p(4 - 5) + 8(1 - 2) = p - 2.

Case-1:When Δ ≠ 0, i.e. p ≠ 2, q ≠ 3, given system of equation has a unique solution.

Case-II:When Δ = 0, i.e. p = 2, or q = 3.When p = 2, Δ = 0, Δ

_{1}= 0, Δ_{2}= 0, Δ_{3}= 0.=> given system of equation has infinitely many solutions.

When q = 3, p ≠ 2, D = 0, D

_{1}≠ 0. The system has no solution.

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