Solution of simultaneous linear equations

Consider a system of simultaneous linear equations given by

A set of values of the variables x, y, z which simultaneously satisfy these three equations is called a solution.

A system of linear equations has a solution (whether unique or not) the system is said to be consistent. If it has no solution it is called an inconsistent system.

If d₁ = d₂ = d₃ = 0 in (1), then the system of equations is said to be a homogenous system. Otherwise, it is called a non-homogenous system of equations.

Solution of a system of linear equations (Cramer’s rule):

The solution of the system of simultaneous linear equations.

a₁x + b₁y = c₁ … (1)

a₂x + b₂y = c₂ … (2)

Is given by x = D₁/D, y = D₂/D,

Where  provided that D ≠ 0.

The solution of the system of linear equations

a₁x + b₁y + c₁z = d₁ … (1)

a₂x + b₂y + c₂z = d₂ … (2)

a₃x + b₃y + c₃z = d₃ … (3)

Is given by x = D₁/D, y = D₂/D and z = D₃/D where 

Conditions for consistency:

For a system of simultaneous linear equations.

i) If D ≠ 0 then the given system of equation is consistent and has a unique solution.

ii) If D = 0 and D₁ = D₂ = D₃ = … = D_n = 0, then the system is consistent and has infinitely many solutions.

iii) If D = 0 and at least one of the determinant D₁, D₂, …, D_n is non-zero, then the given system of equations is inconsistent.

Solution of Homogenous System of Equations:

Let a₁x + b₁y + c₁z = d₁

a₂x + b₂y + c₂z = d₂

a₃x + b₃y + c₃z = d₃

Be a homogenous system of linear equations. This system of equations has a unique trivial solution. x, y, z = 0. If 

It has infinitely many solutions, if D = 0. These solutions are also known as non-trivial solution.

Ex:

If  are not all zero such that

Ax + y + z = 0

X + by + z =0

X + y + cz = 0

Then prove that (1/1-a) + (1/1- b) + (1/1-c) = 1

Sol:

It is given that x, y, z are not all zero. This means that the system of equations has non-trivial solution. Therefore,

 

=> abc – a – c – b + 2 = 0

=> abc = a + b + c - 2

Now,

=(1/1-a) + (1/1- b) + (1/1-c)

= [(1 - b)(1 - c) + (1 - c)(1 - a) + (1 - a)(1 - b)] / [(1 - a)(1 - b)(1 - c)]

= [3 – 2 (a + b + c) + ab + bc + ca] / [1 – (a + b + c) + (ab + bc + ca) - abc] 

(·.· abc = a + b + c – 2)

=> [3 – 2 (a + b + c) + ab + bc + ca] / [3 – 2 (a + b + c) + ab + bc + ca] = 1