Cramer's Rule: Formula, Steps and Examples

Cramer’s rule solves a system of linear equations with determinants. For each variable, replace that variable’s column of coefficients with the column of constants, find the determinant, and divide it by the determinant of the coefficient matrix. It works only when that main determinant is not 0.

Before this lesson: Determinant of a Matrix, Systems of Equations and Systems of Equations in Three Variables

What Cramer’s rule does

A system like $3x + 2y = 16$ and $x - 4y = -4$ has three columns of numbers: the x coefficients, the y coefficients and the constants on the right side. Cramer’s rule builds one determinant from the coefficients and one more for each variable.

The main determinant, D, uses only the coefficients. To get $D_x$ , swap the x column for the constants. To get $D_y$ , swap the y column for the constants. Then each variable equals its own determinant divided by D.

Three determinants for 3x + 2y = 16 and x − 4y = −4: D = −14 from the coefficients, Dx = −56 with the constants in the x column, and Dy = −28 with the constants in the y column. — Swap the constants into one column at a time. Then x = −56 / (−14) = 4 and y = −28 / (−14) = 2.

Why does this work? If you solve the system by elimination using letters instead of numbers, every answer ends up with the same denominator, $ad - bc$ . That denominator is D, and each numerator is another determinant.

Cramer’s rule formula

For the system $ax + by = e$ and $cx + dy = f$ :

D = \begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad - bc

D_x = \begin{vmatrix} e & b \\ f & d \end{vmatrix} = ed - bf, \quad D_y = \begin{vmatrix} a & e \\ c & f \end{vmatrix} = af - ec

x = \frac{D_x}{D}, \quad y = \frac{D_y}{D}, \quad D \ne 0

For three equations in x, y and z, use 3 × 3 determinants the same way: $x = \frac{D_x}{D}$ , $y = \frac{D_y}{D}$ and $z = \frac{D_z}{D}$ .

If $D = 0$ , Cramer’s rule can’t give one answer. The system has no solution or infinitely many. For two equations, if $D = 0$ and $D_x$ or $D_y$ is not 0, there is no solution. If all three are 0, check the equations: when one is a multiple of the other, there are infinitely many solutions.

How to use Cramer’s rule

Write each equation in standard form, $ax + by = c$ , with the variables lined up in the same order.
Find D from the coefficients. If $D = 0$ , stop and check the system another way.
Replace the x column with the constants and find $D_x$ . Do the same for each other variable.
Divide: $x = \frac{D_x}{D}$ , $y = \frac{D_y}{D}$ , and so on.
Check the answer in every original equation.

Cramer’s rule examples

Example 1: A 2 × 2 system

Use Cramer’s rule to solve $3x + 2y = 16$ and $x - 4y = -4$ .

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

Write D from the coefficients.

D = 3(-4) - 2(1)

Use $ad - bc$ (see determinant of a matrix).

D = -12 - 2

Multiply (see multiplying integers).

D = -14

Subtract.

D_x = \begin{vmatrix} 16 & 2 \\ -4 & -4 \end{vmatrix}

Replace the x column with the constants 16 and −4.

D_x = 16(-4) - 2(-4)

Use $ad - bc$ .

D_x = -64 + 8

Multiply. Subtracting −8 is the same as adding 8.

D_x = -56

Add (see adding integers).

D_y = \begin{vmatrix} 3 & 16 \\ 1 & -4 \end{vmatrix}

Replace the y column with the constants.

D_y = 3(-4) - 16(1)

Use $ad - bc$ .

D_y = -12 - 16

Multiply.

D_y = -28

Subtract.

x = \frac{D_x}{D} = \frac{-56}{-14}

Divide $D_x$ by D.

x = 4

Simplify. A negative divided by a negative is positive.

y = \frac{D_y}{D} = \frac{-28}{-14}

Divide $D_y$ by D.

y = 2

Simplify.

3(4) + 2(2) = 12 + 4 = 16

Check the first equation.

4 - 4(2) = 4 - 8 = -4

Check the second equation.

Answer: $(4, 2)$

Example 2: Answers that are fractions

Solve $4x - 3y = 2$ and $2x + 5y = 7$ .

D = \begin{vmatrix} 4 & -3 \\ 2 & 5 \end{vmatrix}

Write D from the coefficients.

D = 4(5) - (-3)(2)

Use $ad - bc$ .

D = 20 + 6

Multiply. Subtracting −6 is the same as adding 6.

D = 26

Add.

D_x = \begin{vmatrix} 2 & -3 \\ 7 & 5 \end{vmatrix}

Replace the x column with the constants 2 and 7.

D_x = 2(5) - (-3)(7)

Use $ad - bc$ .

D_x = 10 + 21

Multiply.

D_x = 31

Add.

D_y = \begin{vmatrix} 4 & 2 \\ 2 & 7 \end{vmatrix}

Replace the y column with the constants.

D_y = 4(7) - 2(2)

Use $ad - bc$ .

D_y = 28 - 4

Multiply.

D_y = 24

Subtract.

x = \frac{D_x}{D} = \frac{31}{26}

Divide. 31 and 26 share no common factor, so the fraction is already simplified.

y = \frac{D_y}{D} = \frac{24}{26}

Divide.

y = \frac{12}{13}

Divide the top and bottom by 2 (see simplifying fractions).

4\left(\frac{31}{26}\right) - 3\left(\frac{12}{13}\right) = \frac{62}{13} - \frac{36}{13} = \frac{26}{13} = 2

Check the first equation.

2\left(\frac{31}{26}\right) + 5\left(\frac{12}{13}\right) = \frac{31}{13} + \frac{60}{13} = \frac{91}{13} = 7

Check the second equation.

Answer: $\left(\frac{31}{26}, \frac{12}{13}\right)$

Example 3: When D = 0

Solve $2x - 6y = 4$ and $-x + 3y = 5$ .

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

Write D from the coefficients.

D = 2(3) - (-6)(-1)

Use $ad - bc$ .

D = 6 - 6

Multiply. Two negatives make a positive: $(-6)(-1) = 6$ .

D = 0

Subtract. Cramer’s rule can’t give one answer.

D_x = \begin{vmatrix} 4 & -6 \\ 5 & 3 \end{vmatrix}

Find $D_x$ to tell the two cases apart.

D_x = 4(3) - (-6)(5)

Use $ad - bc$ .

D_x = 12 + 30

Multiply.

D_x = 42

Add. So $D = 0$ but $D_x \ne 0$ .

Since $D = 0$ and $D_x \ne 0$ , the system has no solution. You can see why. Divide the first equation by 2 to get $x - 3y = 2$ , and multiply the second by −1 to get $x - 3y = -5$ . The same expression can’t equal both 2 and −5, so the two lines are parallel.

Example 4: A 3 × 3 system

Solve the system $x + y + z = 6$ , $2x - y + z = 3$ and $x + 2y - z = 2$ .

D = \begin{vmatrix} 1 & 1 & 1 \\ 2 & -1 & 1 \\ 1 & 2 & -1 \end{vmatrix}

Write D from the coefficients.

D = 1[(-1)(-1) - 1(2)] - 1[2(-1) - 1(1)] + 1[2(2) - (-1)(1)]

Expand along the first row with the signs +, −, +.

D = 1(1 - 2) - 1(-2 - 1) + 1(4 + 1)

Multiply inside the brackets.

D = -1 + 3 + 5

Simplify each term.

D = 7

Add.

D_x = \begin{vmatrix} 6 & 1 & 1 \\ 3 & -1 & 1 \\ 2 & 2 & -1 \end{vmatrix}

Replace the x column with the constants 6, 3 and 2.

D_x = 6[(-1)(-1) - 1(2)] - 1[3(-1) - 1(2)] + 1[3(2) - (-1)(2)]

Expand along the first row.

D_x = 6(1 - 2) - 1(-3 - 2) + 1(6 + 2)

Multiply inside the brackets.

D_x = -6 + 5 + 8

Simplify each term.

D_x = 7

Add.

D_y = \begin{vmatrix} 1 & 6 & 1 \\ 2 & 3 & 1 \\ 1 & 2 & -1 \end{vmatrix}

Replace the y column with the constants.

D_y = 1[3(-1) - 1(2)] - 6[2(-1) - 1(1)] + 1[2(2) - 3(1)]

Expand along the first row.

D_y = 1(-3 - 2) - 6(-2 - 1) + 1(4 - 3)

Multiply inside the brackets.

D_y = -5 + 18 + 1

Simplify each term.

D_y = 14

Add.

D_z = \begin{vmatrix} 1 & 1 & 6 \\ 2 & -1 & 3 \\ 1 & 2 & 2 \end{vmatrix}

Replace the z column with the constants.

D_z = 1[(-1)(2) - 3(2)] - 1[2(2) - 3(1)] + 6[2(2) - (-1)(1)]

Expand along the first row.

D_z = 1(-2 - 6) - 1(4 - 3) + 6(4 + 1)

Multiply inside the brackets.

D_z = -8 - 1 + 30

Simplify each term.

D_z = 21

Add.

x = \frac{D_x}{D} = \frac{7}{7} = 1

Divide $D_x$ by D.

y = \frac{D_y}{D} = \frac{14}{7} = 2

Divide $D_y$ by D.

z = \frac{D_z}{D} = \frac{21}{7} = 3

Divide $D_z$ by D.

1 + 2 + 3 = 6

Check the first equation.

2(1) - 2 + 3 = 3

Check the second equation.

1 + 2(2) - 3 = 2

Check the third equation.

Answer: $(1, 2, 3)$

Example 5: Word problem

A coffee shop sells two roasts by the pound. An order of 3 pounds of dark roast and 2 pounds of blonde roast costs $47. An order of 1 pound of dark roast and 4 pounds of blonde roast costs $39. Find the price per pound of each roast.

Let d be the price of dark roast and b the price of blonde roast, in dollars per pound. The two orders give $3d + 2b = 47$ and $d + 4b = 39$ .

D = \begin{vmatrix} 3 & 2 \\ 1 & 4 \end{vmatrix}

Write D from the coefficients.

D = 3(4) - 2(1)

Use $ad - bc$ .

D = 12 - 2

Multiply.

D = 10

Subtract.

D_d = \begin{vmatrix} 47 & 2 \\ 39 & 4 \end{vmatrix}

Replace the d column with the costs 47 and 39.

D_d = 47(4) - 2(39)

Use $ad - bc$ .

D_d = 188 - 78

Multiply.

D_d = 110

Subtract.

D_b = \begin{vmatrix} 3 & 47 \\ 1 & 39 \end{vmatrix}

Replace the b column with the costs.

D_b = 3(39) - 47(1)

Use $ad - bc$ .

D_b = 117 - 47

Multiply.

D_b = 70

Subtract.

d = \frac{D_d}{D} = \frac{110}{10} = 11

Divide $D_d$ by D.

b = \frac{D_b}{D} = \frac{70}{10} = 7

Divide $D_b$ by D.

3(11) + 2(7) = 33 + 14 = 47

Check the first order.

11 + 4(7) = 11 + 28 = 39

Check the second order.

Answer: Dark roast costs $11 per pound and blonde roast costs $7 per pound.

Common mistakes

Mistake	Example of the error	Fix
Replacing the wrong column	For $D_x$ , putting the constants in the y column	The constants replace the column of the variable you are solving for.
Using $D_x$ as the answer	Writing $x = -56$ in Example 1	Divide: $x = \frac{D_x}{D} = \frac{-56}{-14} = 4$ .
Sign errors in $ad - bc$	$4(5) - (-3)(2) = 20 - 6$	Subtracting a negative product adds: $20 + 6 = 26$ .
Equations not lined up	Reading $2y + x = 5$ as if 2 were the x coefficient	Rewrite it as $x + 2y = 5$ before you read the coefficients.
Dividing by zero	Writing $x = \frac{42}{0}$ when $D = 0$	When $D = 0$ , decide between no solution and infinitely many instead.
Wrong signs in a 3 × 3 expansion	Adding all three terms of the first row	Use the signs +, −, + across the first row.

Cramer’s rule vs. other methods

Cramer’s rule is quickest for 2 × 2 systems, or when you need only one variable. For two equations, the elimination method is often just as fast.

A 3 × 3 system needs four determinants, so augmented matrices with row operations are often quicker. Inverse matrices help when you solve many systems with the same coefficients.

Method	Best for	Main step
Cramer’s rule	2 × 2 systems, or finding one variable	Divide determinants
Elimination or substitution	Small systems by hand	Combine or substitute equations
Augmented matrix	Three or more variables	Row operations
Inverse matrix	Many systems with the same coefficients	$X = A^{-1}B$

Practice problems

Solve each system with Cramer’s rule. If $D = 0$ , say whether there is no solution or infinitely many.

$x + y = 7$ and $2x - y = 5$
$5x + 2y = 1$ and $3x + y = 0$
$2x + 3y = 4$ and $5x - 2y = 1$
$6x - 4y = 10$ and $-3x + 2y = 7$
$3x - y = 4$ and $-6x + 2y = -8$
$4x + y = -2$ and $x - 3y = 19$
Find only y: $7x + 2y = 3$ and $3x + 4y = 17$
$x + y + z = 2$ , $2x + y - z = 2$ and $x - y + 2z = 5$
Find only z: $2x + y - z = 3$ , $x - y + 2z = 2$ and $3x + 2y + z = 11$
A school play sells adult and student tickets. Four adult tickets and 6 student tickets cost $74. Three adult tickets and 2 student tickets cost $38. Find the price of each kind of ticket.

Show answers

D = 1(-1) - 1(2)

Find D from the coefficients with $ad - bc$ .

D = -1 - 2

Multiply.

D = -3

Subtract.

D_x = 7(-1) - 1(5)

Replace the x column with 7 and 5.

D_x = -7 - 5

Multiply.

D_x = -12

Subtract.

D_y = 1(5) - 7(2)

Replace the y column with 7 and 5.

D_y = 5 - 14

Multiply.

D_y = -9

Subtract.

x = \frac{-12}{-3} = 4

Divide $D_x$ by D.

y = \frac{-9}{-3} = 3

Divide $D_y$ by D.

Check: $4 + 3 = 7$ and $2(4) - 3 = 5$ . Answer: $(4, 3)$

D = 5(1) - 2(3)

Find D with $ad - bc$ .

D = 5 - 6

Multiply.

D = -1

Subtract.

D_x = 1(1) - 2(0)

Replace the x column with 1 and 0.

D_x = 1 - 0

Multiply.

D_x = 1

Subtract.

D_y = 5(0) - 1(3)

Replace the y column with 1 and 0.

D_y = 0 - 3

Multiply.

D_y = -3

Subtract.

x = \frac{1}{-1} = -1

Divide $D_x$ by D.

y = \frac{-3}{-1} = 3

Divide $D_y$ by D.

Check: $5(-1) + 2(3) = 1$ and $3(-1) + 3 = 0$ . Answer: $(-1, 3)$

D = 2(-2) - 3(5)

Find D with $ad - bc$ .

D = -4 - 15

Multiply.

D = -19

Subtract.

D_x = 4(-2) - 3(1)

Replace the x column with 4 and 1.

D_x = -8 - 3

Multiply.

D_x = -11

Subtract.

D_y = 2(1) - 4(5)

Replace the y column with 4 and 1.

D_y = 2 - 20

Multiply.

D_y = -18

Subtract.

x = \frac{-11}{-19} = \frac{11}{19}

Divide $D_x$ by D.

y = \frac{-18}{-19} = \frac{18}{19}

Divide $D_y$ by D.

Check: $2\left(\frac{11}{19}\right) + 3\left(\frac{18}{19}\right) = \frac{76}{19} = 4$ and $5\left(\frac{11}{19}\right) - 2\left(\frac{18}{19}\right) = \frac{19}{19} = 1$ . Answer: $\left(\frac{11}{19}, \frac{18}{19}\right)$

D = 6(2) - (-4)(-3)

Find D with $ad - bc$ .

D = 12 - 12

Multiply.

D = 0

Subtract. Cramer’s rule can’t give one answer.

D_x = 10(2) - (-4)(7)

Replace the x column with 10 and 7.

D_x = 20 + 28

Multiply.

D_x = 48

Add. So $D_x \ne 0$ .

Since $D = 0$ and $D_x \ne 0$ , there is no solution. Dividing the first equation by −2 gives $-3x + 2y = -5$ , which can’t also equal 7.

D = 3(2) - (-1)(-6)

Find D with $ad - bc$ .

D = 6 - 6

Multiply.

D = 0

Subtract.

D_x = 4(2) - (-1)(-8)

Replace the x column with 4 and −8.

D_x = 8 - 8

Multiply.

D_x = 0

Subtract.

D_y = 3(-8) - 4(-6)

Replace the y column with 4 and −8.

D_y = -24 + 24

Multiply.

D_y = 0

Add.

All three determinants are 0, and the second equation is −2 times the first. Both equations describe the same line, so there are infinitely many solutions: every point with $y = 3x - 4$ .

D = 4(-3) - 1(1)

Find D with $ad - bc$ .

D = -12 - 1

Multiply.

D = -13

Subtract.

D_x = (-2)(-3) - 1(19)

Replace the x column with −2 and 19.

D_x = 6 - 19

Multiply.

D_x = -13

Subtract.

D_y = 4(19) - (-2)(1)

Replace the y column with −2 and 19.

D_y = 76 + 2

Multiply.

D_y = 78

Add.

x = \frac{-13}{-13} = 1

Divide $D_x$ by D.

y = \frac{78}{-13} = -6

Divide $D_y$ by D.

Check: $4(1) + (-6) = -2$ and $1 - 3(-6) = 19$ . Answer: $(1, -6)$

D = 7(4) - 2(3)

Find D with $ad - bc$ .

D = 28 - 6

Multiply.

D = 22

Subtract.

D_y = 7(17) - 3(3)

Replace the y column with 3 and 17.

D_y = 119 - 9

Multiply.

D_y = 110

Subtract.

y = \frac{D_y}{D} = \frac{110}{22} = 5

Divide $D_y$ by D.

Answer: $y = 5$ . Check: then $7x + 10 = 3$ , so $x = -1$ , and $3(-1) + 4(5) = 17$ .

D = 1[1(2) - (-1)(-1)] - 1[2(2) - (-1)(1)] + 1[2(-1) - 1(1)]

Expand D along the first row.

D = 1(2 - 1) - 1(4 + 1) + 1(-2 - 1)

Multiply inside the brackets.

D = 1 - 5 - 3

Simplify each term.

D = -7

Add.

D_x = 2[1(2) - (-1)(-1)] - 1[2(2) - (-1)(5)] + 1[2(-1) - 1(5)]

Replace the x column with 2, 2 and 5, then expand.

D_x = 2(2 - 1) - 1(4 + 5) + 1(-2 - 5)

Multiply inside the brackets.

D_x = 2 - 9 - 7

Simplify each term.

D_x = -14

Add.

D_y = 1[2(2) - (-1)(5)] - 2[2(2) - (-1)(1)] + 1[2(5) - 2(1)]

Replace the y column with 2, 2 and 5, then expand.

D_y = 1(4 + 5) - 2(4 + 1) + 1(10 - 2)

Multiply inside the brackets.

D_y = 9 - 10 + 8

Simplify each term.

D_y = 7

Add.

D_z = 1[1(5) - 2(-1)] - 1[2(5) - 2(1)] + 2[2(-1) - 1(1)]

Replace the z column with 2, 2 and 5, then expand.

D_z = 1(5 + 2) - 1(10 - 2) + 2(-2 - 1)

Multiply inside the brackets.

D_z = 7 - 8 - 6

Simplify each term.

D_z = -7

Add.

x = \frac{-14}{-7} = 2

Divide $D_x$ by D.

y = \frac{7}{-7} = -1

Divide $D_y$ by D.

z = \frac{-7}{-7} = 1

Divide $D_z$ by D.

Check: $2 - 1 + 1 = 2$ , $2(2) - 1 - 1 = 2$ and $2 + 1 + 2(1) = 5$ . Answer: $(2, -1, 1)$

D = 2[(-1)(1) - 2(2)] - 1[1(1) - 2(3)] + (-1)[1(2) - (-1)(3)]

Expand D along the first row.

D = 2(-1 - 4) - 1(1 - 6) - 1(2 + 3)

Multiply inside the brackets.

D = -10 + 5 - 5

Simplify each term.

D = -10

Add.

D_z = 2[(-1)(11) - 2(2)] - 1[1(11) - 2(3)] + 3[1(2) - (-1)(3)]

Replace the z column with 3, 2 and 11, then expand.

D_z = 2(-11 - 4) - 1(11 - 6) + 3(2 + 3)

Multiply inside the brackets.

D_z = -30 - 5 + 15

Simplify each term.

D_z = -20

Add.

z = \frac{D_z}{D} = \frac{-20}{-10} = 2

Divide $D_z$ by D.

Answer: $z = 2$ . (The full solution is $(1, 3, 2)$ .)

10.

D = 4(2) - 6(3)

Let a and s be the adult and student prices: $4a + 6s = 74$ and $3a + 2s = 38$ . Find D.

D = 8 - 18

Multiply.

D = -10

Subtract.

D_a = 74(2) - 6(38)

Replace the a column with 74 and 38.

D_a = 148 - 228

Multiply.

D_a = -80

Subtract.

D_s = 4(38) - 74(3)

Replace the s column with 74 and 38.

D_s = 152 - 222

Multiply.

D_s = -70

Subtract.

a = \frac{-80}{-10} = 8

Divide $D_a$ by D.

s = \frac{-70}{-10} = 7

Divide $D_s$ by D.

Check: $4(8) + 6(7) = 32 + 42 = 74$ and $3(8) + 2(7) = 24 + 14 = 38$ . Answer: adult tickets cost $8 and student tickets cost $7.

Cramer’s rule FAQ

When can you use Cramer’s rule?

Use it when the system has as many equations as variables and the determinant of the coefficients is not 0. Then the system has exactly one solution, and Cramer’s rule finds it.

What happens if the determinant is 0?

The system has no solution or infinitely many solutions, and Cramer’s rule alone can’t find them. Solve it another way, such as with elimination or row operations.

Is Cramer’s rule faster than elimination?

For two equations, the two methods are about equally quick. For three or more equations, Cramer’s rule needs many determinants, so row operations are usually faster, unless you need only one variable.

Does Cramer’s rule work for 4 × 4 systems?

Yes. The rule works for any square system, but each 4 × 4 determinant takes a lot of arithmetic. Larger systems are usually solved with row operations or technology.

Summary

Cramer’s rule solves a square system with determinants: $x = \frac{D_x}{D}$ , $y = \frac{D_y}{D}$ , and so on.
D uses the coefficients. To get $D_x$ , replace the x column with the constants.
The rule gives one answer only when $D \ne 0$ .
If $D = 0$ , the system has no solution or infinitely many.
Always check the answer in every original equation.

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