In a linear algebraic equation, Cramer’s rule is the explicit formula to provide the answer to the structure of the linear algebraic equation with so many equations that are not known, are valid whenever there is a unique solution provided by the equation. It defines the answer in words of determinants of that (square)matrix which is coefficient and are of matrices which are obtained through it by substituting a single column with the vector of the column that is at the right side of the given equations. It is termed on Gabriel Cramer the person who published all the laws or the arbitrary digit of not knowns in 1750, though Colin Maclaurin too came out with a special collection of the laws in 1748.

## Cramer’s Rule

Cramer’s rule was applied in a very naïve process that is computationally not efficient for the structure of excess than two, max three equations.] Looking at the problem of n mathematical problem in n, but knowns, it needs to show that + 1 is determinant Gaussian elimination produced the output with the Amilcar difficulty to compute it saying o y one determinant. Cramer’s rule could also be digitally not stable even for that of a 2×2 structure. Though it has lately been displayed that the Cramer’s rule could be applied in also O(n3) times, which is being compared to the more usual way of the calculating structure of the linear equations, for example, Gaussian elimination (continuously requires 2.5x times the arithmetic equation for all the matrix structure).

## Condition for infinite and no solution

Condition for infinite and no solution using Cramer’s rule is explained below with the help of a sum:

Solving the given problem on condition for infinite and no solution as follows.

x – 2y + 3z = 0 ———–(i)

3x + y – 2z = 0————-(ii)

2x – 4y + 6z = 0 ———–(iii)

Let us first find out the determinant. Setting up the matrix which is augmented by the starting two columns

1 -2 3 1 -2

|3 1 -2 | | 3 1 |

2 -4 6 2 -4

Hence,

1 (1) (6) + (-2) (-2) (2) + 3 (3) (-4) -2 (1) (3) – (-4) (-2) (1) – 6 (3) (-2) = 0

As the determinant value is equal to zero, two possibilities are either having no answer or having an infinite number of answers. We need to do substitution to pick out.

Multiply problem (1) by -2 then add the answer to problem (3)

-2x + 4y – 6x = 0

2x – 4y + 6x = 0

______________

0 = 0

_______________

Now, to obtain a solution of 0 = 0 a conclusion that is every time true, states that the structure has answers which are infinite in numbers. Graphically representing the structure, we can see that the two planes are the same, and both meet the plane which is in the third, on a line.

X – 2y + 3z = 0

2x – 4y + 6z = 0

3x + y + 2z = 0

For no solution,

There are just three possible output cases for Cramer’s rule.

As per the Cramer’s rule, x = Dx / D, y = Dy / D and z = Dz / D.

Cramer’s rule is applied only at that time when the Determinant answer is unequal to zero. The reason is that in the above-given expressions for variables x, y, and z, if the common divisor = 0, then there would not be any solution.

If D is unequal to 0, and if the minimum one is the Dx, Dy and also Dz are unequal to 0, therefore the structure of the problem is Compatible and has a Unique answer.

Assuming D = 0 hence the Dx, Dy, and Dz = 0 but if minimum one of the components that Co-efficient matrices (adj) have become unequal to 0, after that structure of the mathematical problem is Consistent by having Infinite number of answers.

If D = 0 and minimum one of the Dx, Dy, and Dz is unequal to zero, after that the structure of the sum can be termed inconsistent (No solution).

### Conclusion

Finding matrix inversion by using Cramer’s rule

Let variable A be the n × n matrix having its entries in the field F. Hence

Aadj (A) = Adj (A) A = det(A)I

where adj(A) indicates towards adjugate matrix, det(A) is the determinant, I represent the Identity matrix. Consider det(A) to be a number that is not zero, therefore the matrix inversion of A will be

A-1 = 1/det(A) * adj(A)

This produces a method for the matrix inversion of A, given that det(A) is not equal to 0. This method functions when F acts as a commutative ring, given that the det(A) is a unit. Assuming that det(A) isn’t a unit, so A also fails to be invertible upon the ring (exception – it can be invertible, but upon a ring which is larger within which some of the non- unit’s variable of F can also be invertible).