Cramer's Rule

Cramer's Rule is used to find the unknowns in the given system of linear equations. Cramer's Rule is the most commonly used formula for finding the solution to the given system of linear equations in matrix form.

• This rule is valid only if the given system of equations has a unique solution.
• It doesn't work with a system of equations with infinitely many solutions or no solution.
• This rule is used to find solutions for variables with the same number of equations.
• This rule uses determinants to find the solution of the given equations or the value of unknowns.

Cramer's Rule Formula

Cramer's Rule Formula is used to solve the system of equations in the form 

AX = B

where, 

• A is the coefficient matrix
• B is the column matrix of constants
• X is the column matrix of unknowns

Now the value of X is calculated using the formula given in the diagram below,

The individual values of x, y, and z are calculated with the help of the formula discussed in the above figure.

Cramer’s Rule Conditions

Cramer’s rule is applicable only when certain conditions are satisfied. The important condition of Cramer's rules are,

1. The system AX = B has a unique solution if D ≠ 0 i.e. determinant is non-zero.
2. If D = 0 we have two conditions and any of them can be true,

First Condition: Infinitely many solutions, this situation arises when D = 0 and at least one determinant of the numerator is zero.

Second Condition: No solution, this situation arises when D = 0 and at no determinant of the numerator is zero.

Cramer's Rule For 2 x 2

Now let us solve a system of 2 equations in 2 variables using Cramer's rule. Given equation,

a_₁x + b_₁y = c_₁\\[2pts]a_₂x + b_₂y = c_₂

Follow the steps to solve the system of 2 × 2 equations with two unknowns x and y using Cramer's rule.

Step 1: Write the given system of the equation in matrix form as AX = B

Step 2: Find the determinant (D) of A and find Dₓ and Dᵧ where

Dₓ = det (A) where B replaces the first column of A

Dᵧ = det (A) where B replaces the second column of A

Step 3: Find the values of the variables x and y as,

• x = D_x / D
• y = D_y / D

Cramer's Rule For 3 × 3

Now let us solve a system of 3 equations in 3 variables using Cramer's rule. Given equation,

• a₁x + b₁y  c₁z = d₁
• a₂x + b₂y  c₂z = d₂
• a₃x + b₃y  c₃z = d₃

Follow the steps to solve the system of 3 × 3 equations with two unknowns x and y using Cramer's rule.

Step 1: Write the given system of the equation in matrix form as AX = B

Step 2: Find the determinant (D) of A and find D_x, D_y, and D_z where

• D_x = det (A) where B replaces the first column of A
• D_y = det (A) where B replaces the second column of A
• D_z = det (A) where B replaces the third column of A

Step 3: Find the values of the variables x, y, and z as,

• x = D_x / D
• y = D_y / D
• z = D_z / D

Steps to Solve Using Cramer's Rule

Study the following steps to solve the linear equations using Creamer's Rule,

• Write the given system of equations in AX = B form.
• Find the value of determinant (D) of matrix A. (Note: If the determinant is zero, then a system of equations does not have a unique solution, which is invalid in Cramer's Rule).
• Now, find the value D_x which is the determinant of matrix A in which constants of the given linear equations replace the coefficient of x. 
• Now, find the value D_y which is the determinant of matrix A in which constants of the given linear equations replace the coefficient of y. 
• Now, find the value D_z which is the determinant of matrix A in which constants of the given linear equations replace the coefficient of z. (find this determinant only if 3 variables are present in the given equation).
• Similarly, find determinants for all the unknowns if more than three unknowns are present.
• Find the values of x = D_x/D, y = D_y/D, and z = D_z/D.

Related Articles

• Determinant of a Matrix
• Transpose of a Matrix
• Inverse of a Matrix

Solved Examples on Cramer's Rule

Example 1: Solve  \left\{ \begin{array}{c} 12x-10y= 46\\ 3x+20y=-11 \\ \end{array} \right\}

The given equations in the form of AX = B 

A =\begin{bmatrix}    12 & -10 \\    3 & 20   \end{bmatrix}          , 

B = \begin{bmatrix}    46  \\    -11       \end{bmatrix}         , 

X =  \begin{bmatrix}     x \\     y    \end{bmatrix}

Then, the determinant D of matrix 

A = \begin{vmatrix}    12 & -10 \\    3 & 20  \\ \end{vmatrix}        

   = 12 × 20 - 3 × (-10) = 240 + 30 = 270

Now, find D_x and D_y 

D_x = \begin{vmatrix}    46 & -10 \\    -11& 20  \\ \end{vmatrix}        

    = [46×20 - (-10)×(-11)] = 920 - 110 = 810

D_y = \begin{vmatrix}    12 & 46 \\    3 & -11 \\ \end{vmatrix}         

     =[12×(-11) - 3×46] = -132 -138 = -270

Now, find x = D_x/D, y = D_y/D 

x = 810/270 = 3, y = -270/270 = -1

x = 3, y = -1

Example 2: Solve \left\{ \begin{array}{c} 6.6x+0.95y= 5.2\\ 4.2x+8.6y=19.3 \\ \end{array}\right\}  

The given equations in the form of AX = B 

A = \begin{bmatrix}    6.6& 0.95 \\    4.2& 8.6  \\ \end{bmatrix}               

B = \begin{bmatrix}    5.2  \\    19.3       \end{bmatrix}     

X  =  \begin{bmatrix}     x \\     y    \end{bmatrix}

Then, the determinant D of matrix 

A = \begin{vmatrix}    6.6& 0.95 \\    4.2& 8.6  \\ \end{vmatrix}        = 6.6×8.6 - 4.2×0.95 = 56.76 - 3.99 =52.77

Now, find D_x and D_y 

D_x = \begin{vmatrix}    5.2& 0.95 \\    19.3& 8.6  \\ \end{vmatrix}        = 5.2 × 8.6 - 19.3 × 0.95 = 44.72 - 18.335 = 26.385

D_y = \begin{vmatrix}    6.6& 5.2 \\    4.2& 19.3  \\ \end{vmatrix}        = 6.6×19.3 - 4.2×5.2 = 127.38 - 21.84 = 105.54

Now, find x = D_x/D  , y = D_y/D

x = 26.385/52.77 = 0.5, y = 105.54/52.77 = 2

x = 0.5,  y = 2

Example 3: Solve \left\{ \begin{array}{c} 3x^2+4y^2= 91\\ 6x^2-y^2=38 \\ \end{array} \right\}

Let x² = a, y² = b ---(1)

Then, the equation can be written as,

\left\{ \begin{array}{c} 3a+4b= 91\\ 6a-b=38 \\ \end{array} \right.         

The given equations in the form of AX = B

A = \begin{bmatrix}    3& 4 \\    6&-1  \\ \end{bmatrix}      

B = \begin{bmatrix}    91  \\    38     \end{bmatrix}     

X = \begin{bmatrix}     a \\     b    \end{bmatrix}

Then, the determinant D of matrix A = \begin{vmatrix}    3& 4 \\    6& -1   \end{vmatrix}        = 3×(-1) - 6×4 = -3-24 = -27

Now, find D_a and D_b

D_a =\begin{vmatrix}    91& 4 \\    38& -1   \end{vmatrix}        

     = 91×(-1) - 38×4 = - 91 - 152 

     = -243

D_b = \begin{vmatrix}    3& 91 \\    6& 38   \end{vmatrix}     

     = 3×38 - 6×91 = 114 - 546 

     = -432

Now, find a = D_a/D, b = D_b/D

a = -243/-27 = 9, b = -432/-27 = 16

a = 9, b = 16

We have to find x and y, so we substitute the values into equation (1).

Now x² = a = 9, x = √9 = 3

y² = b = 16, y = √16 = +4 or -4           

Example 4: Solve \left\{ \begin{array}{c} 3x-4y+8z= 34\\ 4x+y-2z=1\\ -6x-13y+20z=61\\ \end{array} \right. 

The given equations in the form of AX = B

A = \begin{bmatrix}    3& -4& 8 \\    4&1&-2  \\-6&-13&20 \end{bmatrix}      

B =\begin{bmatrix}    34  \\    1 \\    61     \end{bmatrix}     

X =\begin{bmatrix}     x \\     y \\     z    \end{bmatrix}     

Then, the determinant D of matrix A = \begin{vmatrix}    3& -4& 8 \\    4&1&-2  \\-6&-13&20 \end{vmatrix}         = 3(20 - 26) - (-4)(80 - 12) + 8(-52-(-6)) = 3×(-6) + 4×68 - 46×8 = -18 + 272 - 368

= -114

Now, find D_x , D_y and D_z

D_x = \begin{vmatrix}    34& -4& 8 \\    1&1&-2  \\61&-13&20 \end{vmatrix}         

    = 34(20 - 26) - (-4)(20 + 122) + 8(-13 - 61) 

    = 34 × (-6) + 4 × 142 + 8 × (-74) = -204 + 568 - 592 = -228 

D_y = \begin{vmatrix}    3& 34& 8 \\    4&1&-2  \\-6&61&20 \end{vmatrix}         

    = 3(20 + 2 × 61) - 34(80 - 12) + 8(61 × 4 + 6) 

    = 3 × 142 - 34 × 68 + 8 × 250 = 426 - 2312 + 2000 = 114

D_z = \begin{vmatrix}    3& -4& 34 \\    4&1&1  \\-6&-13&61 \end{vmatrix}          

    = 3(61+13) - (-4)(61×4 + 6) + 34(-52+6) 

    = 3 × 74 + 4 × 250 + 34 × (-46) = 222 + 1000 -1564  = -342

Now, find x = D_x/D, y = D_y/D, z = D_z/D

x = -228/-114 = 2, y = 114/-114 = -1,  z = -342/-114 = 3   

x = 2,  y = -1,  z = 3

Example 5: Solve \left\{ \begin{array}{c} 3x-8y+10z= 8\\ -x+10y+9z=-15\\ 2x-6y+z=11\\ \end{array} \right. 

The given equations in the form of AX = B 

A = \begin{bmatrix}    3& -8& 10 \\    -1&10&9  \\2&-6&1 \end{bmatrix}      

B = \begin{bmatrix}    8  \\    -15 \\    11     \end{bmatrix}     

X = \begin{bmatrix}     x \\     y \\     z    \end{bmatrix}     

Then, the determinant D of matrix A = \begin{vmatrix}    3& -8& 10 \\    -1&10&9  \\2&-6&1 \end{vmatrix}         = 3(10+54) + 8(-1-18) +10(6-20)                                                                     

 = 3 × 64 - 8 × 19 + 10 × (-14) = 192 -152 - 140 = -100

Now, find D_x , D_y and D_z 

D_x = \begin{vmatrix}    8& -8& 10 \\    -15&10&9  \\11&-6&1 \end{vmatrix}         

     = 8(10+54) + 8(-15-99) + 10(90 -110) 

     = 8 × 64 + 8 × (-114) + 10 × (-20) = 512 - 912 - 200 = -600 

D_y = \begin{vmatrix}    3& 8& 10 \\    -1&-15&9  \\2&11&1 \end{vmatrix}          

    = 3(-15-99) - 8(-1-18) + 10(-11+30) 

    = 3 × (-114) + 8 × 19 + 10 × 19 = -342 + 152 +190 = 0

D_z = \begin{vmatrix}    3& -8& 8 \\    -1&10&-15  \\2&-6&11 \end{vmatrix}          

    = 3(110-90) + 8(-11+30) + 8(6-20) 

    = 3 × 20 + 8 × 19 + 8 × (-14) = 60 + 152 - 112 = 100 

Now, find x = D_x/D, y = D_y/D, z = D_z/D 

x = -600/-100 = 6, y = 0/-100 = 0, z = 100/-100 = -1

x = 6, y = 0, z = -1

Example 6: Solve \left\{ \begin{array}{c} 2x+4y-6z= 19\\ 3x+6y-9z=30\\ 4x-7y+z=15\\ \end{array} \right. 

The given equations in the form of AX = B  

A = \begin{bmatrix}    2& 4& -6 \\    3&6&-9  \\4&-7&1 \end{bmatrix}      

B = \begin{bmatrix}    19  \\    30 \\    15    \end{bmatrix}     

X =  \begin{bmatrix}     x \\     y \\     z    \end{bmatrix}     

Then, the determinant D of matrix A =  = 2(6 - 63) - 4(3 + 36) - 6(-21 - 24) = 2 × (-57) - 4 × 39 - 6 × (-45) = -114 - 156 + 270  = 0

Since |D| = 0, 

which means the given system of equations does not have a unique solution, which is invalid in Cramer's Rule as it is defined only for the system of equations that have a unique solution. This means that the given system of equations either has an infinite solution or no solution.

Example 7: Solve: \left\{ \begin{array}{c} x+y+z= 6\\ 5x-6y+8z=17\\ 2x+3y-z=5\\ \end{array} \right. 

The given equations in the form of AX = B  

A = \begin{bmatrix}    1& 1& 1 \\    5&-6&8  \\2&3&-1 \end{bmatrix}      

B = \begin{bmatrix}    6  \\    17 \\    5    \end{bmatrix}     

X = \begin{bmatrix}     x \\     y \\     z    \end{bmatrix}

Then, the determinant D of matrix A = \begin{vmatrix}    1& 1& 1 \\    5&-6&8  \\2&3&-1 \end{vmatrix}         = 1(6 - 24) - 1(-5 - 16) + 1(15 + 12)

= -18 + 21 + 27 = 30

Now, find D_x , D_y and D_z  

D_x = \begin{vmatrix}    6& 1& 1 \\    17&-6&8  \\5&3&-1 \end{vmatrix}         

     = 6(6-24) -1(-17-40) +1(51+30) 

     = 6(-18) + 57 + 81 = -108 + 138 = 30

D_y = \begin{vmatrix}    1& 6& 1 \\    5&17&8  \\2&5&-1 \end{vmatrix}         

     = 1(-17 - 40) - 6(-5 - 16) + 1(25 - 34) 

     = -57 + 126 - 9 = 60

D_z = \begin{vmatrix}    1& 1& 6 \\    5&-6&17  \\2&3&5 \end{vmatrix}         

     = 1(-30 - 51) - 1(25 - 34) + 6(15 + 12) 

     = -81 + 9 + 162 = 90  

Now, find x = D_x/D, y = D_y/D, z = D_z/D 

x = 30/30 = 1, y = 60/30 = 2, z = 90/30 = 3

x = 1,  y = 2, z = 3

Practice Problems on Cramer's Rule

Question 1: Solve the following system using Cramer's Rule:

• 2x+3y = 5
• 4x+y = 11

Question 2: Find the values of 𝑥 y, and z using Cramer's Rule:

• x−2y+3z = 7
• 2x+y−z = 4
• -x+y+z = 2

Question 3: Determine if the following system has a unique solution using Cramer's Rule:

• 3x+5y = 9
• 6x+10y = 18

Question 4: Apply Cramer's Rule to solve:

• 5x+2y−z = 4
• -3x+4y+2z = −1
• 7x−y+5z = 10

Question 5: Use Cramer's Rule to solve for x and 𝑦:

• 4x+y = 3
• x−2y = 1

Answers:

1. x = 14/5, y = -1/5
2. x = 32/13, y = 23/13, z= 35/13
3. The given system of equations does not have a unique solution, which is invalid in Cramer's Rule as it is defined only for the system of equations that have a unique solution. This means that the given system of equations either has an infinite solution or no solution.
4. x = 177/193, y = 14/193, z = 141/193
5. x = 7/9, y = -1/9