# Cramer’s Rule – Explanation & Examples

To solve a system of equations, we primarily use the substitution method, elimination method, or graphing method. We can also use matrix algebra to solve a system of equations. Processes such as Gaussian Elimination (also known as Gauss-Jordan Elimination) can help solve a system of equations with $3$ or more unknowns. We can also use Cramer’s Rule to solve a system.

What is the Cramer’s Rule?

The Cramer’s Rule is a method of solving a system of equations using determinants.

In this lesson, we will look at what the Cramer’s Rule is and how to solve a system of equations. Some examples and practice problems will follow.

## What is Cramer’s Rule?

The Cramer’s Rule is a method to solve a system of equations using determinants. That’s that beauty of the Cramer’s Rule. We can find a single variable’s value without solving the whole system (or other variables).

Remember determinants?

Consider the $2 \times 2$ matrix shown below:

$A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$

The determinant of this matrix is given by:

$det( A ) = | A | = \begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad – bc$

Note: We have used $3$ notations in denoting a determinant.

Now, consider the $3 \times 3$ matrix shown below:

$B = \begin{bmatrix} { a } & { b } & c \\ { d } & { e } & f \\ g & h & i \end {bmatrix}$

The determinant of this matrix is given by:

$det( B ) = | B | = \begin{vmatrix} { a } & { b } & c \\ { d } & { e } & f \\ g & h & i \end {vmatrix} = a \begin{vmatrix} { e } & f \\ h & i \end {vmatrix} – b \begin{vmatrix} d & f \\ g & i \end {vmatrix} + c \begin{vmatrix} d & e \\ g & h \end {vmatrix}$

Note that we have broken down the $3\times 3$ matrix into smaller $2\times 2$ matrices. The vertical bars outside the $2 \times 2$ matrices indicate that we have to take the determinant. From knowledge of determinant of $2 \times 2$ matrices, we can further simplify the formula to be:

$det(B)=| B | = a(ei-fh) – b(di – fg) + c(dh-eg)$

Let’s consider a system of equations shown below:

\begin{align*} { 2x } + 3y &= \, { 7 } \\ { – 3x } + 4y &= { 15 } \end{align*}

We will now name some matrices to help us out with using the Cramer’s Rule to solve this system later on.

• Following the $2 \times 2$ determinant formula, we can write the determinant matrix as:

$D = \begin{vmatrix} 2 & 3 \\ { – 3 } & 4 \end{vmatrix}$

We labeled it “D.”

• Putting the constant coefficients from the system in the first column (in place of $x$s), we can write another matrix:

$D_{ x } = \begin{vmatrix} 7 & 3 \\ { 15 } & 4 \end{vmatrix}$

We labeled it “$D_{ x }$” and call it the x-matrix.

• Similarly, putting the constants from the system in the second column (in place of $y$s), we can write another matrix:

$D_{ y } = \begin{vmatrix} 2 & 7 \\ { – 3 } & 15 \end{vmatrix}$

We labeled it “$D_{ y }$” and call it the y-matrix.

Now, the formula of Cramer’s Rule to solve for the variables $x$ and $y$ are shown below:

$x = \frac{ D_{ x } }{ D } = \frac{ \begin{vmatrix} 7 & 3 \\ { 15 } & 4 \end{vmatrix} }{ \begin{vmatrix} 2 & 3 \\ { – 3 } & 4 \end{vmatrix} }$

$y = \frac{ D_{ y } }{ D } = \frac{ \begin{vmatrix} 2 & 7 \\ { – 3 } & 15 \end{vmatrix} }{ \begin{vmatrix} 2 & 3 \\ { – 3 } & 4 \end{vmatrix} }$

The next section will show us how to actually use the rule and solve the system! Note that we cannot use the Cramer’s Rule when a matrix’s determinant is $0$! Zero determinant can mean:

• The system is inconsistent (it doesn’t have a solution)
• The system is dependent (it has infinite solutions)

In this case, we have to rely upon other methods in solving a system, such as the substitution/elimination method or Gaussian Elimination method.

## How to use Cramer’s Rule?

Let’s solve a system of equations ( $2$ variables ) using the Cramer’s Rule to see the concept live in action!

Solve the system of equations shown below using the Cramer’s Rule:

\begin{align*} { 2x } + y &= \, { 7 } \\ { 3x } – 2y &= { – 7 } \end{align*}

The first step is to write the determinants of this system of equations, determinant ( $D$ ), $x –$ determinant ( $D_{ x } ), and the$ y – $determinant ($ D_{ y } ). Let’s use the formula we learned and write them up:

$D = \begin{vmatrix} 2 & 1 \\ { 3 } & { – 2 } \end{vmatrix}$

$D_{ x } = \begin{vmatrix} 7 & 1 \\ { – 7 } & { – 2 } \end{vmatrix}$

$D_{ y } = \begin{vmatrix} 2 & { 7 } \\ { 3 } & { – 7 } \end{vmatrix}$

Recall the formula to evaluate a $2 \times 2$ determinant:

For a $2 \timess 2$ matrix —

$A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$

The determinant is calculated as —

$det( A ) = | A | = \begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad – bc$

Let’s calculate the determinants:

$D = \begin{vmatrix} 2 & 1 \\ { 3 } & { – 2 } \end{vmatrix} = ( 2 )( – 2 ) – ( 1 )( 3 ) = – 4 – 3 = – 7$

$D_{ x } = \begin{vmatrix} 7 & 1 \\ { – 7 } & { – 2 } \end{vmatrix} = ( 7 )( – 2 ) – ( 1 )( – 7 ) = – 14 – ( – 7 ) = – 14 + 7 = – 7$

$D_{ y } = \begin{vmatrix} 2 & { 7 } \\ { 3 } & { – 7 } \end{vmatrix} = ( 2 )( – 7 ) – ( 7 )( 3 ) = -14 – 21 = – 35$

Now, we can use the formulas and thus, the Cramer’s Rule, to solve for the variables $x$ and $y$. Shown below:

$x = \frac{ D_{ x } }{ D } = \frac{ – 7 }{ – 7 } = 1$

$y = \frac{D_{ y } }{ D } = \frac{ – 35 }{ – 7 } = 5$

The solution set of the system is (1, 5).

You can notice that if we wanted to solve for only $1$ variables without solving the whole system, we could have easily used the formula for a single variable to find it. Cramer’s Rule is a pretty nifty tool to use to find solutions to a system of equations. We will see some examples as well as one with $3$ variables.

#### Example 1

Solve the system of equations shown below using Cramer’s Rule:

\begin{align*} { – x } – y &= \, { 5 } \\ { 2x } + y &= { 4 } \end{align*}

Solution

The first step is to write the determinants of this system of equations, determinant ( $D$ ), $x –$ determinant ( $D_{ x } ), and the$ y – $determinant ($ D_{ y } ). Let’s use the formula we learned and write them up:

$D = \begin{vmatrix} – 1 & – 1 \\ { 2 } & { 1 } \end{vmatrix}$

$D_{ x } = \begin{vmatrix} 5 & – 1 \\ { 4 } & { 1 } \end{vmatrix}$

$D_{ y } = \begin{vmatrix} – 1 & { 5 } \\ { 2 } & { 4 } \end{vmatrix}$

We will use the formula to calculate the determinants of $2 \times 2$ matrices to calculate the matrices $D$, $D_{ x }$, and $D_{ y }$.

$D = \begin{vmatrix} – 1 & – 1 \\ { 2 } & { 1 } \end{vmatrix} = ( – 1 )( 1 ) – ( – 1 )( 2 ) = – 1 + 2 = 1$

$D_{ x } = \begin{vmatrix} 5 & – 1 \\ { 4 } & { 1 } \end{vmatrix} = ( 5 )( 1 ) – ( – 1 )( 4 ) = 5 + 4 = 9$

$D_{ y } = \begin{vmatrix} – 1 & { 5 } \\ { 2 } & { 4 } \end{vmatrix} = ( – 1 )( 4 ) – ( 5 )( 2 ) = – 4 – 10 = – 14$

Now, we use the formulas learned in Cramer’s Rule to find the values of the variables:

$x = \frac{ D_{ x } }{ D } = \frac{ 9 }{ 1 } = 9$

$y = \frac{D_{ y } }{ D } = \frac{ – 14 }{ 1 } = – 14$

The solution set of the system is $(9, – 14)$.

Let us look at an example with $3$ variables.

#### Example 2

Solve the system of equations shown below using Cramer’s Rule:

\begin{align*} 2a + b – 2c &= \, – 1 \\ 3a – 3b – c &= 5 \\ a – 2b + 3c = 6 \end{align*}

Solution

The first step is to write the determinants of this system of equations, determinant ( $D$ ), $a –$ determinant ( $D_{ a } ),$ b – $determinant ($ D_{ b } ),  and the $c –$ determinant ( $D_{ c } ). Let’s use the formula we learned and write them up:$  D = \begin{vmatrix} 2 & 1 & -2 \\ 3 & -3 & -1 \\ 1 & -2 & 3 \end{vmatrix}    D_{ a } = \begin{vmatrix} -1 & 1 & -2 \\ 5 & -3 & -1 \\ 6 & -2 & 3 \end{vmatrix}    D_{ b } = \begin{vmatrix} 2 & -1 & -2 \\ 3 & 5 & -1 \\ 1 & 6 & 3 \end{vmatrix}    D_{ c } = \begin{vmatrix} 2 & 1 & -1 \\ 3 & -3 & 5 \\ 1 & -2 & 6 \end{vmatrix}  $For a matrix in the form:$ B = \begin{bmatrix} { a }  & { b } & c \\ { d } & { e } & f \\ g & h & i  \end {bmatrix} $The determinant is calculated as:$ | B | = a(ei-fh) – b(di – fg) + c(dh-eg) $Now, we use the Cramer’s Rule and calculate the values of the variables$ a $,$b$, and$c$. The steps are shown below (we didn’t show the detailed steps of finding determinants of$ 3 \times 3 $matrices):$ a = \frac{ D_{ a } }{ D } = \frac{ \begin{vmatrix} -1 & 1 & -2 \\ 5 & -3 & -1 \\ 6 & -2 & 3 \end{vmatrix} }{ \begin{vmatrix} 2 & 1 & -2 \\ 3 & -3 & -1 \\ 1 & -2 & 3 \end{vmatrix} } = \frac{-26}{-26} =1  b = \frac{D_{ b } }{ D } = \frac{ \begin{vmatrix} 2 & -1 & -2 \\ 3 & 5 & -1 \\ 1 & 6 & 3 \end{vmatrix} }{ \begin{vmatrix} 2 & 1 & -2 \\ 3 & -3 & -1 \\ 1 & -2 & 3 \end{vmatrix} } = \frac{26}{-26} = -1  c = \frac{D_{ c } }{ D } = \frac{ \begin{vmatrix} 2 & 1 & -1 \\ 3 & -3 & 5 \\ 1 & -2 & 6 \end{vmatrix} }{ \begin{vmatrix} 2 & 1 & -2 \\ 3 & -3 & -1 \\ 1 & -2 & 3 \end{vmatrix} } = \frac{-26}{-26} = 1$The solution set of the system is$ (1, – 1,1) $. Practice Questions 1. Solve the system of equations shown below using Cramer’s Rule:$ \begin{align*} { 5x } + 2y &= \, { 10 } \\ { – x } + 4y  &= { 20 }  \end{align*} $2. Solve the system of equations shown below using Cramer’s Rule:$ \begin{align*} 3x – 4y + z &= \, -5 \\ x – y – z  &= – 10  \\ 6x – 8y + 2z = 10 \end{align*} $### Answers 1. The first step is to write the determinants of this system of equations, determinant ($ D $),$ x – $determinant ($ D_{ x } $), and the$ y – $determinant ($ D_{ y } $). Let’s use the formula we learned and write them up:$  D = \begin{vmatrix} 5 & 2 \\ {  – 1 } &  { 4 } \end{vmatrix}    D_{ x } = \begin{vmatrix} 10 & 2 \\ { 20 } & { 4 } \end{vmatrix}    D_{ y } = \begin{vmatrix} 5 & { 10 } \\ { – 1 } & { 20 } \end{vmatrix}  $We will use the formula to calculate the determinants of$ 2 \times 2 $matrices to calculate the matrices$ D $,$ D_{ x } $, and$ D_{ y } $.$  D = \begin{vmatrix} 5 & 2 \\ {  – 1 } &  { 4 } \end{vmatrix}  = ( 5 )( 4 ) – ( 2 )( -1 ) = 20 + 2 = 22     D_{ x } = \begin{vmatrix} 10 & 2 \\ { 20 } & { 4 } \end{vmatrix} = ( 10 )( 4 ) – ( 2 )( 20 ) = 40 – 40 = 0    D_{ y } = \begin{vmatrix} 5 & { 10 } \\ { – 1 } & { 20 } \end{vmatrix}  = ( 5 )( 20 ) – ( 10 )( -1 ) = 100 + 10 = 110    $Now, we use the formulas learned in Cramer’s Rule to find the values of the variables:$ x = \frac{ D_{ x } }{ D } = \frac{ 0 }{ 22 } = 0  y = \frac{D_{ y } }{ D } = \frac{ 110 }{ 22 } = 5 $The solution set of the system is$ (0, 5) $. 2. The first step is to write the determinants of this system of equations, determinant ($ D $),$ x – $determinant ($ D_{ x }  ), $y –$ determinant ( $D_{ y } )$,  and the $z –$ determinant ( $D_{ z } )$. Let’s use the formula we learned and write them up:

$D = \begin{vmatrix} 3 & -4 & 1 \\ 1 & -1 & -1 \\ 6 & -8 & 2 \end{vmatrix}$

$D_{ x } = \begin{vmatrix} 5 & -4 & 1 \\ -10 & -1 & -1 \\ 10 & -8 & 2 \end{vmatrix}$

$D_{ y } = \begin{vmatrix} 3 & 5 & 1 \\ 1 & -10 & -1 \\ 6 & 10 & 2 \end{vmatrix}$

$D_{ z } = \begin{vmatrix} 3 & -4 & 5 \\ 1 & -1 & -10 \\ 6 & -8 & 10 \end{vmatrix}$

Recall that a $3 \times 3$ matrix of the form:

$B = \begin{bmatrix} { a } & { b } & c \\ { d } & { e } & f \\ g & h & i \end {bmatrix}$

Has a determinant equals to:

$| B | = a(ei-fh) – b(di – fg) + c(dh-eg)$

First, let’s find the value of the determinant, $D$,

$D = \begin{vmatrix} 3 & -4 & 1 \\ 1 & -1 & -1 \\ 6 & -8 & 2 \end{vmatrix} = 3(-2-8) +4(2+6) +1(-8+6) = 3(-10) + 4(8) +1(-2) = 0$

This matrix’s determinant is $0$; thus, we cannot solve the system using Cramer’s Rule!!