Cramer's rule are used to solve a systems of n linear equations with n variables using explicit formulas.
We first start with a proof of Cramer's rule to solve a 2 by 2 systems of linear equations. Rules for 3 by 3 systems of equations are also presented. Then examples and questions with detailed solutions are presented.
To check answers when solving 2 by 2 and 3 by 3 systems of equations, you may want to use these online Systems of Equations Calculator and Solver.
Cramer's Rules for a 2 by 2 System of Equations
To find rules (or formulas) that may be used solve any 2 by 2 system of linear equations, we need to solve the general system of the form
We multiply equation (1) by b 2 and equation (2) by - b 1.
\(
\left\{
\begin{array}{lcl}
a_1 b_ 2 x + b_1 b_2 y & = & c_1 b_2\\
-a_2 b_1 x - b_2 b_1 y & = & - c_2 b_1
\end{array}
\right.
\)
Add the left sides and the right sides of the above equations and simplify to obtain an equation in one variable.
\(
a_1 b_2 x - a_2 b_1 x = c_1 b_2 - c_2 b_1
\)
Factor x out on the left side
\(
x(a_1 b_2 - a_2 b_1) = b_2 c_1 - b_1 c_2
\)
Solve the above equation for x
\(
x = \dfrac{c_1 b_2 - c_2 b_1}{a_1 b_2 - a_2 b_1}
\)
We may use similar steps to eliminate x and solve for y to obtain.
\(
y = \dfrac{a_1 c_2 - a_2 c_1}{a_1 b_2 - a_2 b_1}
\)
The solution to the given 2 by 2 system of linear equations is given by Cramer's rules as follows
\[ x = \dfrac{D_x}{D} , y = \dfrac{D_y}{D} \]
Using the determinant of a 2 by 2 Matrix notation, \( D \), \( D_x \) and \( D_y \) are defined by
Example 1
Use Cramer's rule to solve the system
\[
\left\{
\begin{array}{lcl}
2 x - 3 y & = & 5 \\
-4 x + 7 y & = & -11
\end{array}
\right.
\]
Solution to Example 1
Calculate the determinants \( D, D_x \text{and} D_y \)
\( D = \begin{vmatrix}2 & -3\\ -4 & 7\end{vmatrix} = (2)(7) - (-3)(-4) = 2 \)
We now use Cramer's rule to calculate the unknowns \(x\) and \(y\).
\( x = \dfrac{D_x}{D} = \dfrac{2}{2} = 1 \)
\( y = \dfrac{D_y}{D} = \dfrac{-2}{2} = -1 \)
As an exercise, substitute \(x\) and \(y\) by 1 and -1 respectively in the given system and check that the solution found satisfies the system of equation.
Note: If the determinant D is equal to zero, Cramer's rule cannot be used to solve a system of linear equations.
Cramer's Rules for a 3 by 3 System of Equations
A general 3 by 3 systems of linear equations may be written as follows:
\[
\left\{
\begin{array}{lcl}
a_1 x + b_1 y + c_1 z = & \color{red}{d_1} & (1)\\
a_2 x + b_2 y + c_2 = & \color{red}{d_2} & (2) \\
a_3 x + b_3 y + c_3 = & \color{red}{d_3} & (2) \\
\end{array}
\right.
\]
For a 3 by 3 system of linear equations, Cramer's rule gives the solution as follows
\[ x = \dfrac{D_x}{D} , y = \dfrac{D_y}{D} , z = \dfrac{D_z}{D} \]
where \( D, D_x, D_y \text{and} D_z \) are determinants of 3 by 3 matrices defined by
\( D = \begin{vmatrix}a_1&b_1&c_1\\ a_2&b_2&c_2\\a_3 & b_3 & c_3 \end{vmatrix}\)
\( D_z = \begin{vmatrix} -1 & - 3 & 6 \\ 2 & -5 & 14\\ 1&1&-1\end{vmatrix} = -1\cdot \begin{vmatrix}-5&14\\ 1&-1\end{vmatrix}-\left(-3\right) \begin{vmatrix}2&14\\ 1&-1\end{vmatrix}+6\cdot \begin{vmatrix}2&-5\\ 1&1\end{vmatrix} = 3\)
Use Cramer's rule to calculate the unknowns \(x\), \(y\) and \(z\).
\( x = \dfrac{D_x}{D} = \dfrac{-3}{-3} = 1 \)
\( y = \dfrac{D_y}{D} = \dfrac{6}{-3} = -2 \)
\( z = \dfrac{D_z}{D} = \dfrac{3}{-3} = -1 \)
As an exercise, substitute \(x\), \(y\) and \(y\) by 1, -2 and 1 respectively in the given system to check the solution found.
Questions with Solution
Part 1
Use Cramer's rule to solve the following systems of linear equations.
a)
\(
\left\{
\begin{array}{lcl}
5 x - \dfrac{2}{3} y & = & \dfrac{1}{3} \\
- x + \dfrac{1}{2} y & = & - \dfrac{1}{2}
\end{array}
\right.
\) b)
\(
\left\{
\begin{array}{lcl}
0.1 x - 0.3 y & = & 1.1 \\
- 0.2 x + 1.3 y & = & - 1.5
\end{array}
\right.
\)
c) \(
\left\{
\begin{array}{lcl}
- 3x + 5 y - z & = & \dfrac{1}{2} \\
\dfrac{1}{5} x - 5 y - \dfrac{3}{5} z & = & 0 \\
-4 x + \dfrac{4}{5} y - z & = & - 7
\end{array}
\right.
\)
Part 2
Use Cramer's rule to find the solution of the systems of linear equations in terms of the parameter k.
a)
\(
\left\{
\begin{array}{lcl}
5 x - k y & = & 6 \\
- 2 x + 2 k y & = & - 3
\end{array}
\right.
\) b)
\(
\left\{
\begin{array}{lcl}
2 x - 3 y & = & k \\
x + 2 y & = & -2k
\end{array}
\right.
\)
Part 3
Use Cramer's rule to find the solution of the systems of linear equations in terms of the parameters p and q.
a) \(
\left\{
\begin{array}{lcl}
- x - 3 y & = & 5 p \\
- 2 x - 5 y & = & - 2 q
\end{array}
\right.
\)
b) Use your answer to a) to solve the systems
1) \(
\left\{
\begin{array}{lcl}
- x - 3 y & = & 10 \\
- 2 x - 5 y & = & - 4
\end{array}
\right. \) 2) \(
\left\{
\begin{array}{lcl}
- x - 3 y & = & 5/2 \\
- 2 x - 5 y & = & - 2
\end{array}
\right. \) 3) \(
\left\{
\begin{array}{lcl}
- x - 3 y & = & 50 \\
- 2 x - 5 y & = & 6
\end{array}
\right. \)
Part 4
Solve the system and show that the solution does not depends on the parameter p.
\(
\left\{
\begin{array}{lcl}
5 x - 6 y + 6 z & = & -14 \\
9 x - p y - z & = & 22 \\
-2 x - 6 y & = & -4
\end{array}
\right.
\)
Solutions to the Above Questions
Part 1
a) Calculate the determinants
\( D = \begin{vmatrix} 5 & -2/3\\ -1 & 1/2 \end{vmatrix} = (5)(1/2)-(-2/3)(-1) = 11/6\)
Solution to the system
\( x = \dfrac{D_x }{D} = \dfrac{9k}{8k} = \dfrac{9}{8} \)
, \( y = \dfrac{D_y }{D} = \dfrac{-3}{8k} = -\dfrac{3}{8k} \)
b) The determinants are given by
\( D = \begin{vmatrix} 2 & - 3\\ 1 & 2 \end{vmatrix} = (2)(2) - (-3)(1) = 7 \)
\( D_x = \begin{vmatrix} k & - 3\\ -2 k & 2 \end{vmatrix} = (k)(2) - (-3)(-2k) = -4k \)
\( D_y = \begin{vmatrix} 2 & k\\ 1 & -2k \end{vmatrix} = (2)(-2k) - (k)(1) = -5k \)
Solution to the system
\( x = \dfrac{D_x }{D} = \dfrac{-4k}{7} = -\dfrac{4}{7}k \)
, \( y = \dfrac{D_y }{D} = \dfrac{-5k}{7} = -\dfrac{5}{7}k \)
Part 3
a) Using Cramer's rule, the solutions are given by
\( x = 6 q + 25 p \) \( y = -2 q - 10 p \)
b)
1) If we substitute the parameters p and q in the system of equations in a) by 2 and 2 respectively, we obtain the system of equation
\(
\left\{
\begin{array}{lcl}
- x - 3 y & = & 10 \\
- 2 x - 5 y & = & - 4
\end{array}
\right. \)
that we have to solve. But the system in a) has been solved for all values of p and q. Hence to solve the system in b) 1), we substitute p and q by their values (2 and 2) in the solutions obtained in a) which gives.
\( x = 6 q + 25 p = 6(2) + 25(2) = 62 \) \( y = -2 q - 10 p = -2 (2) - 10 (2) = -24 \)
2) For this system p = 1/2 and q = 1; hence the solution
\( x = 6 q + 25 p = 6(1) + 25(1/2) = 37/2 \) \( y = -2 q - 10 p = -2 (1) - 10 (1/2) = - 7 \)
3) For this system p = 10 and q = -3; hence the solution
\( x = 6 q + 25 p = 6(-3) + 25(10) = 232 \) \( y = -2 q - 10 p = -2(-3) - 10(10) = - 94 \)
Part 4
The determinants used in Cramer's rule are given by
