Using Cramer’s Rule to Solve a System of Linear Equations

Using Cramer's Rule to solve a system of linear equations left image.Using Cramer's Rule to solve a system of linear equations right image.
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3×3 Cramer’s Rule
Cramer’s Rule, named for the Swiss mathematician Gabriel Cramer

Gabriel Cramer (1704-1752)

Cramer’s Rule, named for the Swiss mathematician Gabriel Cramer, requires the understanding of determinants to solve a system of linear equations.

If the determinate equals zero Cramer’s Rule does not apply and the system of linear equations has either; 1) No solution, represented geometrically as parallel lines, or 2)  Has infinite solutions, where at every point the equations represent the same line.

Cramer’s Rule for Solving a 2×2 System:

To determine a mechanical procedure for solving a system of two linear equations in two variables observe:

Matrix A

a1

b1

a2

b2

det(A) = a1b2 − a2b1

Matrix B

k1

b1

k2

b2

det(B) = k1b2 − k2b1

Now consider two linear equations in two variables.

a1x + b1y = k1

a2x + b2y = k2

To eliminate y, multiply both sides of A by b2 and both sides of B by −b1.

a1b2x + b1b2y = k1b2

−a2b1x + −b2b1y = −k2b1

Add the two equations.

x(a1b2 − a2b1) = k1b2 − k2b1

The result is

x det(A) = det(B)

If the left-side equation determinate given by (a1b2 − a2b1) is not zero then

x = det(B) / det(A)

The determinate of the denominator has the coefficients of x and y as they are in the original equations. Call this the Determinate D:

a1x + b1y = k1

a2x + b2y = k2

=

a1

b1

a2

b2

=

 det(A) = D

The determinate of the numerator, Dx, is obtained from D by replacing the a’s (coefficients of x) by the corresponding k’s:

Dx =

k1

b1

k2

b2

So that x = Dx / D  (if D ≠ 0)

Similarly it can be shown that Dy, the 2×2 determinant obtained from D by replacing the b’s (coefficients of y) by the corresponding k’s is:

Dy =

a1

k1

a2

k2

And if D ≠ 0 then y = Dy / D

Thus, to solve by Cramer’s Rule two linear equations having two variables:

Linear Equations System

10x − 7y = 12

3x − 2y = 5

D =

10

−7

3

−2

=

 10(−2) − 3(−7) = 1

Dx =

12

−7

5

−2

=

 12(−2) − 5(−7) = 11

Dy =

10

12

3

5

=

 10(5) − 3(12) = 14

If D ≠ 0; then by Cramer’s Rule…

x = Dx / D = 11 / 1 = 11

y = Dy / D = 14 / 1 = 14

The unique solution to the linear equation system is (11, 14).

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