Let us consider a system of linear equations be
`a_1x + b_1y+ c_1z= d_1`
`a_2 x + b_2y + c_2z = d_2`
`a_3x + b_3y + c_3z = d_3`

`Delta = |(a_1,b_1,c_1),(a_2,b_2,c_2),(a_3,b_3,c_3)| \ \ \ \ Delta_1 = |(d_1,b_1,d_1),(d_2,b_2,d_2),(d_3,b_3,d_3)|`


`Delta_2 = |(a_1,d_1,c_1),(a_2,d_2,c_2),(a_3,d_3,c_3)| \ \ \ \ Delta_3 = |(a_1,b_1,d_1),(a_2,b_2,d_2),(a_3,b_3,d_3)|`

Now, there are two cases arise:

`text(Case I :)` If `Delta ne 0`

In this case , `x = Delta_1/Delta , y = Delta_2/Delta , z = Delta_3/ Delta`

Then, system will have unique finite solutions and so equations are consistent.

`text(Case II :)` If `Delta = 0`

`(a)` `text(When atleast one of)` `Delta_1 ,Delta_2,Delta_3` `text( be non-zero)`

`(i)` Let `Delta_1 ne 0` then from `Delta_1 = x Delta` will not be satisfied for any value of x
because `Delta = 0` and `Delta_1 ne 0` and hence no value of xis possible.

`(ii)` Let `Delta_2= 0`, then from `Delta_2 = yDelta` will not be satisfied for any value of y
because `Delta= 0` and `Delta_2 ne 0` and hence no value of y is possible.

`(iii)` Let `Delta_3 ne 0`, then from `Delta_3 = z Delta` will not be satisfied for any value of z
because `Delta = 0` and `Delta_3 ne 0` and hence no value of z is possible.
Thus, if `Delta = 0` and any of`delta_1,Delta_2,Delta_3` is non-zero. Then, the system has
no solution i.e., equations are inconsistent.

`(b)` `text(When)` `Delta_1 = Delta_2 = Delta_3=0`

In this case , `tt[(Delta_1, = , xDelta),(Delta_2, = , yDelta),(Delta_3, = , zDelta)]}` will be true for all values of x, y and z .

But, since `a_1x + b_1y + c_1z = d_1`, therefore only two of `x, y` and `z` will be
independent and third will be dependent on the other two.
Thus, the system will have infinite number of solutions i.e., equaLons are
`text(consistent)` .

`text(Fundas)`

`1.` If `Delta ne 0 ,` then system will have unique finite solution and so equation
are consistent.

`2.` If `Delta=0` and atleast on `Delta_1,Delta_2,Delta_3` be non-zero, then the system has
solution i.e., equations are inconsistent.

`3.` If `Delta = Delta_1 - Delta_2 = Delta_3=0` then equations will have infinite number
solutions i.e., equations are consistent.

Let us consider a system of linear equations in x and y
`a_1x + b_1y + c_1 = 0 ... (i)`
`a_2x + b_2y + c_2 = 0 ... (ii)`
`a_3x + b_3y + c_3 = 0 ... (iii)`

will be consistent, then values of x and y obtained from any two equations satisfy
the third equation.
On solving Eqs. (ii) and (iii) by Cramer's rule, we have

`x/|(b_2,c_2),(b_3,c_3)| = y/|(c_2,a_2),(c_3,a_3)| = 1/|(a_2,b_2),(a_3,b_3)|`

These values of x andy will satisfy Eq. (i), then

`a_1|(b_2,c_2),(b_3,c_3)| + b_1|(c_2,a_2),(c_3,a_3)| + c_1|(a_2,b_2),(a_3,b_3)| = 0`

` => a_1|(b_2,c_2),(b_3,c_3)| - b_1|(a_2,c_2),(c_3,a_3)| + c_1|(a_2,b_2),(a_3,b_3)| = 0`

or `|(a_1,b_1,c_1),(a_2,b_2,c_2),(a_3,b_3,c_3)| =0`

which is the required condition.

`text(Note)`

For consistency of three linear equations in two knowns, the number of
solution is one.

Let us consider a system of homogeneous linear equations in three unknown `x, y` and `z` be

`a_1x + b_1y+ c_1z= 0`
`a_2 x + b_2y + c_2z = 0`
`a_3x + b_3y + c_3z = 0`


`Delta = |(a_1,b_1,c_1),(a_2,b_2,c_2),(a_3,b_3,c_3)|`

`text(Case I : )` If `Delta != 0`, then `x = 0, y = 0, z = 0` is the only solution of above system, then given system of equations has only zero solution for all its variables and the given equations are said to have Trivial solution.

`text(Case II : )` If `Delta = 0`, atleast one of x, y and z is non-zero, then given system of equations has no solution or infinite
solutions for all its variables and the given equations are said to have Non-trivial solution.