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Cramer’s Rule

By Vighnesh Hemnani

There are many ways to solve a system of equations and Cramer’s Rule is a simple way to solve them. Learn more about what it is, steps in using Cramer’s Rule, solved examples and sample problems.

Why is this concept useful?

Cramer’s Rule is especially useful because it is an efficient method. The method only requires the knowledge of finding the determinant of a matrix, unlike Gaussian Elimination, which requires long and tedious row operations. This makes solving the system of equations faster and easier and can be extrapolated to a matrix of any size (n by n).

Where does this concept fit into the curriculum?

High School

What is the Cramer’s Rule?


The Cramer’s Rule is a method to solve a system of equations that is simple and straight-forward. Developed by Gabriel Cramer in the 18th century, it is a method that uses only determinants to be able to find the solution to a system of linear equations and is valid of any n number system of linear equations as long as they have a unique solution.

How to use this concept?

To be able to use the Cramer’s Rule, there are several steps to follow. To start with, there is a condition where Cramer’s Rule breaks down, so the first step is to find the determinant of your coefficient matrix (from your system of equations) and check if the determinant is equal to 0. (You can use any method to find the determinant, a simple formula for 2x2 or even Cofactor Expansion).

system of equations: {a1x + b1y = c1 a2x + b2y = c2   →   D = |[ a1 b1 a2 b2 ]|

if D=0,   do not use Cramer's Rule

if D≠0,   use Cramer's Rule

If the determinant is equal to 0, then Cramer’s Rule does not work and you will need to try another method. This is because if it is 0, then there are either no solutions or infinitely many solutions. If it is non-zero, then you may proceed to the next step as it would imply that there is a unique solution.


To find the solution to an arbitrary system of two linear equations, we can simply say that x and y as below. D is the determinant of the coefficient matrix, Dx is the determinant of the numerator in solution for x, and Dy is the determinant of the numerator in solution for y.

{a1x + b1y = c1 a2x + b2y = c2

Solutions via Cramer’s Rule (for 2 by 2 matrix):

x=DxDx = \frac{Dx}{D}
,
y=DyDy = \frac{Dy}{D}

where:

D = |[ a1 b1 a2 b2]|

Dx = |[c1 b1 c2 b2 ]|

Dy= |[ a1 c1 a2 c2 ]|

The above is only for a system of two linear equations, however, the same can be extrapolated to a system of three linear equations like the following:

{a1x + b1y + c1z = d1 a2x + b2y + c2z = d2a3x + b3y + c3z = d3

x=DxDx = \frac{Dx}{D}
,
y=DyDy = \frac{Dy}{D}
,
z=DzDz = \frac{Dz}{D}

D= |[ a1 b1 c1 a2 b2 c2 a3 b3 c3 ]|

Dx = |[ d1 b1 c1 d2 b2 c2 d3 b3 c3]|

Dy= |[ a1 d1 c1 a2 d2 c2 a3 d3 c3 ]|

Sample Math Problems

Question

Attempt to solve the following system of linear equations using Cramer’s Rule.

{x + 2y = 8 - x - 2y = -8

Answer

Simply from inspection, you may be able to notice that the above two equations are essentially the same. To show this, we can find the determinant of the coefficient matrix and check it against 0:

D = |[ 1 2 - 1 -2 = ad - bc = -2 - (2( -1 )=0

Question

Solve the following system of linear equations using Cramer’s Rule. If it is not possible, explain why.

{ 5x + y = 6 - 4x + 12y = 8

Answer

First, we attempt to find D:

D= |[ 5 1 - 4 12 ]| = ad - bc = 5∙12 - ( 1 ∙ -4) = 64≠0


Continuing, we will try to find x and y by first finding Dx and Dy:

Dx = |[ c1 b1 c2 b2 ]| = |[ 6 1 8 12]| = ad-bc = 6∙12 - 8∙1 = 64

Dy = |[ a1 c1 a2 c2 ]| = |[ 5 6 -4 8]| = ad - bc = 5∙8 - 6∙ (-4) = 64

So:

x=DxD=6464=1x = \frac{Dx}{D} = \frac{64}{64} = 1
y=DyD=6464=1y = \frac{Dy}{D} = \frac{64}{64} = 1

Thus, the solution is (1,1).

Question

Using Cramer’s Rule, try to solve the following system of linear equations.

{ -4x + 10y = 18 3x + 4y = -2

Answer

First, we will try to find determinant D:

D= |[ -4 10 3 4 ]| = ad - bc = -4 ∙ 4 - (3∙ 10) = -46 ≠ 0

Continuing, we will try to find x and y by first finding Dx and Dy:

Dx= |[ c1 b1 c2 b2]| = |[ 18 10 -2 4 ]| = ad - bc = 18 ∙ 4 - 10 ∙ -2 = 92

Dy = |[ a1 c1 a2 c2]| = |[-4 18 3 -2]| = ad - bc =-4 ∙ -2 -3 ∙ 18 = -46

So:

x=DxD=9246=2x = \frac{Dx}{D} = \frac{92}{-46} = -2
y=DyD=4646=1y = \frac{Dy}{D} = \frac{-46}{-46} = 1

Thus, the solution is (-2,1).

Question

Attempt to solve the following system of linear equations using Cramer’s Rule.

{ 15x - 9y = -3 6x - 8y = 12

Answer

First, we attempt to find D:

D = |[ 15 - 9 6 -8]| = ad - bc = -8 ∙ 15 -( 6∙ -9) = -66 ≠ 0

Continuing, we will try to find x and y by first finding Dx and Dy:

Dx= |[ c1 b1 c2 b2 ]| = |[-3 -9 12 -8 ]| = ad - bc = -3∙ -8 -12 ∙ -9 =132

Dy= |[ a1 c1 a2 c2 ]| = |[ 15 -3 6 12 ]| = ad - bc = 15 ∙ 12 -6∙(-3)=198

So:

x=DxD=13266=2x = \frac{Dx}{D} = \frac{132}{-66} = -2
y=DyD=19866=3y = \frac{Dy}{D} = \frac{198}{-66} = -3

Thus, the solution is (-2,-3).

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Practice Math Problems

1. { 2x - 2y = -21 6x + 6y = 9

2. { 8x - 2y = 0 2x + 3y = 14

3. { 7x - 8y -9z = 39 -4x - 5y + 6z = -13 -2x + 4y + 6z = -14

4. { -2x - 2y - 5z = 1 6x + 2y + 2z = 6 2x - 3y -4z = 4

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