Cramer's Rule part 2

Cramer's Rule is a method that uses determinants for solving systems of linear equations. To explain the method and how these determinants are generated, this lecture will first illustrate the solution to a system of linear equations by the addition/elimination method without simplifying the indicated products and sums of the coefficients. The intent is to demonstrate that these products and sums can be represented by determinants.

 

Cramer's Rule Formula

Consider a system of (2 x 2) matrix:

a₁x + b₁y = C₁

a₂x + b₂y = C₂

Cramer's Rule formula for 2x2 matrix is

Cramer's Rule Proof

Consider a system of n linear equations in n unknowns x1, x2,........xn whose coefficient matrix is A (square matrix), with det(A) assumed to be nonzero.

a11x1+a12x2+..................+a1nxn=b1

a21x1+a22x2+..................+a2nxn=b2

.......................... so on ...........................

an1x1+an2x2+..................+annxn=bn

The system Ax = b has a unique solution provided the determinant of coefficient, det(A) is nonzero.

∴ The solution to the system Ax = b is given by

x1 = D1D, x2 = D2D,.........., xn = DnD
The determinant Dj equals det Bj where matrix Bj is matrix with column j replaced by b = (b1,b2,.......,bn). The result is called Cramer's Rule for nxn systems.

Cramer's Rule 2x2

Consider the 2-equation system below with variables x and y:
Cramer’s rule for (2 x 2) matrix:

a₁x + b₁y = C₁

a₂x + b₂y = C₂

then x = D1D and y = D2D

Where, D = ∣∣∣a1a2b1b2∣∣∣, D₁ = ∣∣∣c1c2b1b2∣∣∣ and D₂ = ∣∣∣a1a2c1c2∣∣∣

Let us solve system of equations by Cramer rule, 2x + y = 1 and x + y = 2

Given 2x + y = 1 and x + y = 2

D = ∣∣∣2111∣∣∣ = 2 - 1 = 1

D₁ = ∣∣∣1211∣∣∣ = 1 - 2 = -1

D₂ = ∣∣∣2112∣∣∣ = 4 - 1 = 3

Using Cramer Rule, solution of the given system is

x = D1D = −11 = -1

y = D2D = 31 = 3

The solution of the system is (-1, 3).

Cramer's Rule 3x3

Consider the three-equations below with variables x, y and z:
Cramer’s rule for (3 x 3) matrix:

a₁x + b₁y + c₁z = d₁

a₂x + b₂y + c₂z = d₂

a₃x + b₃y + c₃z = d₃

Here,
D = ∣∣∣∣a1a2a3b1b2b3c1c2c3∣∣∣∣

D₁ = ∣∣∣∣d1d2d3b1b2b3c1c2c3∣∣∣∣

D₂ = ∣∣∣∣a1a2a3d1d2d3c1c2c3∣∣∣∣

D₃ = ∣∣∣∣a1a2a3b1b2b3d1d2d3∣∣∣∣

The solution of the system is

x = D1D, y = D2D and z = D3D.

Cramer's Rule 4x4

Solution of 4x4 matrix can be find as same as we find for 2x2 and 3x3 matrices.

Find the values of x, y, z and t using Cramer's rule. 2x - y + 4z + t = -2
3x + 2y - t = -3
x + 2y + 2t = 10
x + y + 2z = 2
D = ∣∣∣∣∣∣2311−122140021−120∣∣∣∣∣∣ = 2∣∣∣∣221002−120∣∣∣∣ + 1∣∣∣∣311002−120∣∣∣∣ + 4∣∣∣∣311221−120∣∣∣∣ - 1∣∣∣∣311221002∣∣∣∣
= -50 (Using 3x3 determinant method)
D₁ = ∣∣∣∣∣∣−2−3102−122140021−120∣∣∣∣∣∣ = -2∣∣∣∣221002−120∣∣∣∣ + 1∣∣∣∣−3102002−120∣∣∣∣ + 4∣∣∣∣−3102221−120∣∣∣∣ - 1∣∣∣∣−3102221002∣∣∣∣
= 100
D₂ = ∣∣∣∣∣∣2311−2−310240021−120∣∣∣∣∣∣ = 2∣∣∣∣−3102002−120∣∣∣∣ + 2∣∣∣∣311002−120∣∣∣∣ + 4∣∣∣∣311−3102−120∣∣∣∣ - 1∣∣∣∣311−3102002∣∣∣∣
= -150
D₃ = ∣∣∣∣∣∣2311−1221−2−31021−120∣∣∣∣∣∣ = 2∣∣∣∣221−3102−120∣∣∣∣ + 1∣∣∣∣311−3102−120∣∣∣∣ - 2∣∣∣∣311221−120∣∣∣∣ - 1∣∣∣∣311221−3102∣∣∣∣
= -25
D₄ = ∣∣∣∣∣∣2311−12214002−2−3102∣∣∣∣∣∣ = 2∣∣∣∣221002−3102∣∣∣∣ + 1∣∣∣∣311002−3102∣∣∣∣ + 4∣∣∣∣311221−3102∣∣∣∣ + 2∣∣∣∣311221002∣∣∣∣
= -150
Now
x = D1D = 100−50 = -2

y = D2D = −150−50 = 3

z = D3D = −25−50 = 12

t = D3D = −150−50 = 3
The solution of the system is (-2, 3, 12, 3).

Cramer's Rule Example

Given below are some of the examples based on Cramer's rule.

Solved Examples

Question 1: Solve the system:
3x - 2y =10
4x + y = 6

Solution:

D = ∣∣∣34−21∣∣∣ = 3 + 8 = 11

D₁ = ∣∣∣106−21∣∣∣ = 10 + 12 = 22

D₂ = ∣∣∣34106∣∣∣ = 18 - 40 = -22

Using Cramer Rule, solution of the given system is

x = D1D = 2211 = 2

y = D2D = −2211 = -2

The solution is (2, -2).

Question 2: Solve the system
3x - 2y + z = 9
x +2y - 2z = -5
x + y - 4z = -2

Solution:
Given system is 3x - 2y + z = 9
x + 2y - 2z = -5
x + y - 4z = -2

D = ∣∣∣∣311−2211−2−4∣∣∣∣ = 3(-8 + 2) + 2(-4 + 2) + 1(1 - 2) = -18 - 4 - 1 = -23

D₁ = ∣∣∣∣9−5−2−2211−2−4∣∣∣∣ = 9(-8 + 2) + 2(20 - 4) + 1(-5 + 4) = -54 + 32 - 1 = -23

D₂ = ∣∣∣∣3119−5−21−2−4∣∣∣∣ = 3(20 - 4) - 9(-4 + 2) + 1(-2 + 5) = 48 + 18 + 3 = 69

D₃ = ∣∣∣∣311−2219−5−2∣∣∣∣ = 3(-4 + 5) + 2(-2 + 5) + 9(1 - 2) = 3 + 6 - 9 = 0

Now

x = D1D = −23−23 = 1

y = D2D = 69−23 = -3

z = D3D = 0−23 = 0

Hence, the solution for the system is (1, -3, 0).