# Solution of simultaneous linear equations

Consider a system of simultaneous linear equations given by \left. \begin{align}& {{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}z={{d}_{1}} \\ & {{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}z={{d}_{2}} \\& {{a}_{3}}x+{{b}_{3}}y+{{c}_{3}}z={{d}_{3}} \\ \end{align} \right\} … (1)

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 $$D=\left| \begin{matrix}{{a}_{1}} & {{b}_{1}} \\{{a}_{2}} & {{b}_{2}} \\\end{matrix} \right|$$, $${{D}_{1}}=\left| \begin{matrix}{{c}_{1}} & {{b}_{1}} \\{{c}_{2}} & {{b}_{2}} \\\end{matrix}\right|$$ and $${{D}_{2}}=\left|\begin{matrix}{{a}_{1}} & {{c}_{1}} \\{{a}_{2}} & {{c}_{2}} \\\end{matrix}\right|$$ 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 $$D=\left| \begin{matrix}{{a}_{1}} & {{b}_{1}} & {{c}_{1}} \\{{a}_{2}} & {{b}_{2}} & {{c}_{2}} \\{{a}_{3}} & {{b}_{3}} & {{c}_{3}} \\\end{matrix} \right|$$, $${{D}_{1}}=\left| \begin{matrix}{{d}_{1}} & {{b}_{1}} & {{c}_{1}} \\{{d}_{2}} & {{b}_{2}} & {{c}_{2}} \\{{d}_{3}} & {{b}_{3}} & {{c}_{3}} \\\end{matrix} \right|$$, $${{D}_{2}}=\left| \begin{matrix}{{a}_{1}} & {{d}_{1}} & {{c}_{1}} \\{{a}_{2}} & {{d}_{2}} & {{c}_{2}} \\{{a}_{3}} & {{d}_{3}} & {{c}_{3}} \\\end{matrix} \right|$$ and $${{D}_{3}}=\left| \begin{matrix}{{a}_{1}} & {{b}_{1}} & {{d}_{1}} \\{{a}_{2}} & {{b}_{2}} & {{d}_{2}} \\{{a}_{3}} & {{b}_{3}} & {{d}_{3}} \\\end{matrix} \right|$$

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 = … = Dn = 0, then the system is consistent and has infinitely many solutions.

iii) If D = 0 and at least one of the determinant D1, D2, …, Dn 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 $$D=\left| \begin{matrix}{{a}_{1}} & {{b}_{1}} & {{c}_{1}} \\{{a}_{2}} & {{b}_{2}} & {{c}_{2}} \\{{a}_{3}} & {{b}_{3}} & {{c}_{3}} \\\end{matrix} \right|\ne 0$$

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

Ex: If x, y, z 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, $$\left| \begin{matrix}a & 1 & 1 \\1 & b & 1 \\1 & 1 & c \\\end{matrix} \right|=0$$

abc – a – c – b + 2 = 0

abc = a + b + c – 2

Now,

= $$\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}$$

= $$\frac{(1-b)(1-c)\,+\,(1-c)(1-a)+(1-a)(1-b)}{(1-a)(1-b)(1-c)}$$

= $$\frac{3-2\,(a+b+c)+ab+bc+ca}{1-(a+b+c)+\left( ab+bc+ca \right)-abc}$$

= $$\frac{3-2\,(a+b+c)+ab+bc+ca}{3-2\,(a+b+c)+ab+bc+ca}=1$$               (∵ abc = a + b + c – 2)