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 for the given system of linear equations in matrix form. Cramer’s Rule uses the concept of the determinant to find its solution.
Let’s know How to Apply Cramer’s Rule and its explanation. It requires some prior knowledge of matrices, determinants, and the system of linear equations.
Cramer’s Rule Definition
Cramer’s rule is a rule which is used to find the unknowns from a given set of linear equations. 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,
From the above diagram, it is clear that if determinant (D) is not equal to zero then it gives a unique solution. In contrast, if D is equal to zero then no solution or infinitely many solutions are given.
The AX = B has a unique solution if D ≠ 0 i.e. determinant is non-zero.
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,
a₁x + b₁y = c₁
a₂x + b₂y = c₂
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,
Cramer’s Rule For 3 × 3
Now let us solve a system of 2 equations in 2 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 Dx, Dy, and Dz where
Dx = det (A) where B replaces the first column of A
Dy = det (A) where B replaces the second column of A
Dz = det (A) where B replaces the third column of A
Step 3: Find the values of the variables x, y, and z as,
- x = Dx / D
- y = Dy / D
- z = Dz / D
How to Use 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 Dx which is the determinant of matrix A in which constants of the given linear equations replace the coefficient of x.
- Now, find the value Dy which is the determinant of matrix A in which constants of the given linear equations replace the coefficient of y.
- Now, find the value Dz 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 = Dx/D, y = Dy/D, and z = Dz/D.
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Solved Examples on Cramer’s Rule
Example 1: Solve [Tex]\left\{ \begin{array}{c} 12x-10y= 46\\ 3x+20y=-11 \\ \end{array} \right. [/Tex]
Solution:
The given equations in the form of AX = B
A =[Tex]\begin{bmatrix} 12 & -10 \\ 3 & 20 \end{bmatrix} [/Tex] ,
B = [Tex]\begin{bmatrix} 46 \\ -11 \end{bmatrix} [/Tex] ,
X = [Tex] \begin{bmatrix} x \\ y \end{bmatrix}[/Tex]
Then, the determinant D of matrix
A = [Tex]\begin{vmatrix} 12 & -10 \\ 3 & 20 \\ \end{vmatrix} [/Tex]
= 12 × 20 – 3 × (-10) = 240 + 30 = 270
Now, find Dx and Dy
Dx = [Tex]\begin{vmatrix} 46 & -10 \\ -11& 20 \\ \end{vmatrix} [/Tex]
= [46×20 – (-10)×(-11)] = 920 – 110 = 810
Dy = [Tex]\begin{vmatrix} 12 & 46 \\ 3 & -11 \\ \end{vmatrix} [/Tex]
=[12×(-11) – 3×46] = -132 -138 = -270
Now, find x = Dx/D, y = Dy/D
x = 810/270 = 3, y = -270/270 = -1
x = 3, y = -1
Example 2: Solve [Tex]\left\{ \begin{array}{c} 6.6x+0.95y= 5.2\\ 4.2x+8.6y=19.3 \\ \end{array} \right. [/Tex]
Solution:
The given equations in the form of AX = B
A = [Tex]\begin{bmatrix} 6.6& 0.95 \\ 4.2& 8.6 \\ \end{bmatrix} [/Tex]
B = [Tex]\begin{bmatrix} 5.2 \\ 19.3 \end{bmatrix} [/Tex]
X = [Tex]\begin{bmatrix} x \\ y \end{bmatrix}[/Tex]
Then, the determinant D of matrix
A = [Tex]\begin{vmatrix} 6.6& 0.95 \\ 4.2& 8.6 \\ \end{vmatrix} [/Tex] = 6.6×8.6 – 4.2×0.95 = 56.76 – 3.99 =52.77
Now, find Dx and Dy
Dx = [Tex]\begin{vmatrix} 5.2& 0.95 \\ 19.3& 8.6 \\ \end{vmatrix} [/Tex] = 5.2 × 8.6 – 19.3 × 0.95 = 44.72 – 18.335 = 26.385
Dy = [Tex]\begin{vmatrix} 6.6& 5.2 \\ 4.2& 19.3 \\ \end{vmatrix} [/Tex] = 6.6×19.3 – 4.2×5.2 = 127.38 – 21.84 = 105.54
Now, find x = Dx/D , y = Dy/D
x = 26.385/52.77 = 0.5, y = 105.54/52.77 = 2
x = 0.5, y = 2
Example 3: Solve [Tex]\left\{ \begin{array}{c} 3x^2+4y^2= 91\\ 6x^2-y^2=38 \\ \end{array} \right. [/Tex]
Solution:
Let x2 = a, y2 = b
Then, the equation can be written as,
[Tex]\left\{ \begin{array}{c} 3a+4b= 91\\ 6a-b=38 \\ \end{array} \right. [/Tex]
The given equations in the form of AX = B
A = [Tex]\begin{bmatrix} 3& 4 \\ 6&-1 \\ \end{bmatrix} [/Tex]
B = [Tex]\begin{bmatrix} 91 \\ 38 \end{bmatrix} [/Tex]
X = [Tex]\begin{bmatrix} a \\ b \end{bmatrix}[/Tex]
Then, the determinant D of matrix A = [Tex]\begin{vmatrix} 3& 4 \\ 6& -1 \end{vmatrix} [/Tex] = 3×(-1) – 6×4 = -3-24 = -27
Now, find Da and Db
Da =[Tex]\begin{vmatrix} 91& 4 \\ 38& -1 \end{vmatrix} [/Tex]
= 91×(-1) – 38×4 = – 91 – 152
= -243
Db = [Tex]\begin{vmatrix} 3& 91 \\ 6& 38 \end{vmatrix} [/Tex]
= 3×38 – 6×91 = 114 – 546
= -432
Now, find a = Da/D, b = Db/D
a = -243/-27 = 9, b = -432/-27 = 16
a = 9, b = 16
Now x2 = a = 9, x = √9 = 3
y2 = b = 16, y = √16 = 4
Example 4: Solve [Tex]\left\{ \begin{array}{c} 3x-4y+8z= 34\\ 4x+y-2z=1\\ -6x-13y+20z=61\\ \end{array} \right. [/Tex]
Solution:
The given equations in the form of AX = B
A = [Tex]\begin{bmatrix} 3& -4& 8 \\ 4&1&-2 \\-6&-13&20 \end{bmatrix} [/Tex]
B =[Tex]\begin{bmatrix} 34 \\ 1 \\ 61 \end{bmatrix} [/Tex]
X =[Tex]\begin{bmatrix} x \\ y \\ z \end{bmatrix} [/Tex]
Then, the determinant D of matrix A = [Tex]\begin{vmatrix} 3& -4& 8 \\ 4&1&-2 \\-6&-13&20 \end{vmatrix} [/Tex] = 3(20 – 26) – (-4)(80 – 12) + 8(-52-(-6)) = 3×(-6) + 4×68 – 46×8 = -18 + 272 – 368
= -114
Now, find Dx , Dy and Dz
Dx = [Tex]\begin{vmatrix} 34& -4& 8 \\ 1&1&-2 \\61&-13&20 \end{vmatrix} [/Tex]
= 34(20 – 26) – (-4)(20 + 122) + 8(-13 – 61)
= 34 × (-6) + 4 × 142 + 8 × (-74) = -204 + 568 – 592 = -228
Dy = [Tex]\begin{vmatrix} 3& 34& 8 \\ 4&1&-2 \\-6&61&20 \end{vmatrix} [/Tex]
= 3(20 + 2 × 61) – 34(80 – 12) + 8(61 × 4 + 6)
= 3 × 142 – 34 × 68 + 8 × 250 = 426 – 2312 + 2000 = 114
Dz = [Tex]\begin{vmatrix} 3& -4& 34 \\ 4&1&1 \\-6&-13&61 \end{vmatrix} [/Tex]
= 3(61+13) – (-4)(61×4 + 6) + 34(-52+6)
= 3 × 74 + 4 × 250 + 34 × (-46) = 222 + 1000 -1564 = -342
Now, find x = Dx/D, y = Dy/D, z = Dz/D
x = -228/-114 = 2, y = 114/-114 = -1, z = -342/-114 = 3
x = 2, y = -1, z = 3
Example 5: Solve [Tex]\left\{ \begin{array}{c} 3x-8y+10z= 8\\ -x+10y+9z=-15\\ 2x-6y+z=11\\ \end{array} \right. [/Tex]
Solution:
The given equations in the form of AX = B
A = [Tex]\begin{bmatrix} 3& -8& 10 \\ -1&10&9 \\2&-6&1 \end{bmatrix} [/Tex]
B = [Tex]\begin{bmatrix} 8 \\ -15 \\ 11 \end{bmatrix} [/Tex]
X = [Tex]\begin{bmatrix} x \\ y \\ z \end{bmatrix} [/Tex]
Then, the determinant D of matrix A = [Tex]\begin{vmatrix} 3& -8& 10 \\ -1&10&9 \\2&-6&1 \end{vmatrix} [/Tex] = 3(10+54) + 8(-1-18) +10(6-20)
= 3 × 64 – 8 × 19 + 10 × (-14) = 192 -152 – 140 = -100
Now, find Dx , Dy and Dz
Dx = [Tex]\begin{vmatrix} 8& -8& 10 \\ -15&10&9 \\11&-6&1 \end{vmatrix} [/Tex]
= 8(10+54) + 8(-15-99) + 10(90 -110)
= 8 × 64 + 8 × (-114) + 10 × (-20) = 512 – 912 – 200 = -600
Dy = [Tex]\begin{vmatrix} 3& 8& 10 \\ -1&-15&9 \\2&11&1 \end{vmatrix} [/Tex]
= 3(-15-99) – 8(-1-18) + 10(-11+30)
= 3 × (-114) + 8 × 19 + 10 × 19 = -342 + 152 +190 = 0
Dz = [Tex]\begin{vmatrix} 3& -8& 8 \\ -1&10&-15 \\2&-6&11 \end{vmatrix} [/Tex]
= 3(110-90) + 8(-11+30) + 8(6-20)
= 3 × 20 + 8 × 19 + 8 × (-14) = 60 + 152 – 112 = 100
Now, find x = Dx/D, y = Dy/D, z = Dz/D
x = -600/-100 = 6, y = 0/-100 = 0, z = 100/-100 = -1
x = 6, y = 0, z = -1
Example 6: Solve [Tex]\left\{ \begin{array}{c} 2x+4y-6z= 19\\ 3x+6y-9z=30\\ 4x-7y+z=15\\ \end{array} \right. [/Tex]
Solution:
The given equations in the form of AX = B
A = [Tex]\begin{bmatrix} 2& 4& -6 \\ 3&6&-9 \\4&-7&1 \end{bmatrix} [/Tex]
B = [Tex]\begin{bmatrix} 19 \\ 30 \\ 15 \end{bmatrix} [/Tex]
X = [Tex]\begin{bmatrix} x \\ y \\ z \end{bmatrix} [/Tex]
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: [Tex]\left\{ \begin{array}{c} x+y+z= 6\\ 5x-6y+8z=17\\ 2x+3y-z=5\\ \end{array} \right. [/Tex]
Solution:
The given equations in the form of AX = B
A = [Tex]\begin{bmatrix} 1& 1& 1 \\ 5&-6&8 \\2&3&-1 \end{bmatrix} [/Tex]
B = [Tex]\begin{bmatrix} 6 \\ 17 \\ 5 \end{bmatrix} [/Tex]
X = [Tex]\begin{bmatrix} x \\ y \\ z \end{bmatrix}[/Tex]
Then, the determinant D of matrix A = [Tex]\begin{vmatrix} 1& 1& 1 \\ 5&-6&8 \\2&3&-1 \end{vmatrix} [/Tex] = 1(6 – 24) – 1(-5 – 16) + 1(15 + 12)
= -18 + 21 + 27 = 30
Now, find Dx , Dy and Dz
Dx = [Tex]\begin{vmatrix} 6& 1& 1 \\ 17&-6&8 \\5&3&-1 \end{vmatrix} [/Tex]
= 6(6-24) -1(-17-40) +1(51+30)
= 6(-18) + 57 + 81 = -108 + 138 = 30
Dy = [Tex]\begin{vmatrix} 1& 6& 1 \\ 5&17&8 \\2&5&-1 \end{vmatrix} [/Tex]
= 1(-17 – 40) – 6(-5 – 16) + 1(25 – 34)
= -57 + 126 – 9 = 60
Dz = [Tex]\begin{vmatrix} 1& 1& 6 \\ 5&-6&17 \\2&3&5 \end{vmatrix} [/Tex]
= 1(-30 – 51) – 1(25 – 34) + 6(15 + 12)
= -81 + 9 + 162 = 90
Now, find x = Dx/D, y = Dy/D, z = Dz/D
x = 30/30 = 1, y = 60/30 = 2, z = 90/30 = 3
x = 1, y = 2, z = 3
FAQs on Cramer’s Rule
Question 1: Why is Cramer’s Rule used?
Answer:
Cramer’s rule is used for finding solutions to linear equations which are represented in matrix form.
Question 2: Who Invented Cramer’s Rule of Matrices?
Answer:
A swiss mathematician named Gabriel Cramer was created to formulate Cramer’s rule. He wrote this formula in 1750 in his book.
Question 3: What are the Advantages of Cramer’s Rule?
Answer:
Cramer’s Rule helps us to find the solution to the linear equations where the number of the variable is equal to the number of equations. It is a shortcut method to solve the equation.
Question 4: What is the other name of Cramer’s rule?
Answer:
The other name of Cramer’s rule is the determinant method as it uses determinants to solve the linear equation.
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