Solving 3 Systems of Equations Calculator

3 Systems of Equations Solver

Enter the coefficients for each equation in the form:

aX + bY + cZ = d

Solution:

Understanding and Solving Systems of Three Linear Equations

A system of three linear equations involves three equations with three unknown variables, typically denoted as X, Y, and Z. Each equation represents a plane in a 3D coordinate system, and the solution to the system is the point (X, Y, Z) where all three planes intersect. Such systems are fundamental in various fields, including physics, engineering, economics, and computer graphics, for modeling complex relationships and finding specific points of equilibrium or intersection.

The General Form

A system of three linear equations can be written in the general form:

a₁X + b₁Y + c₁Z = d₁

a₂X + b₂Y + c₂Z = d₂

a₃X + b₃Y + c₃Z = d₃

Where a₁, b₁, c₁, d₁, a₂, b₂, c₂, d₂, a₃, b₃, c₃, and d₃ are known coefficients and constants, and X, Y, Z are the variables we aim to solve for.

Methods for Solving

There are several methods to solve systems of linear equations, each with its advantages:

1. Substitution Method: Involves solving one equation for one variable in terms of the others, then substituting that expression into the remaining equations to reduce the system's complexity.
2. Elimination Method (Gaussian Elimination): Involves adding or subtracting multiples of equations to eliminate one variable at a time, simplifying the system into a triangular form that is easy to solve.
3. Matrix Methods (Cramer's Rule, Inverse Matrix): These methods use matrices and determinants to find the solution. They are particularly powerful for larger systems and are often implemented in computational tools.

Cramer's Rule Explained

Our calculator utilizes Cramer's Rule, a method that uses determinants to find the solution. It's particularly elegant for systems with a unique solution.

Here's how it works:

1. Form the Coefficient Matrix (A):

   | a₁ b₁ c₁ |
   | a₂ b₂ c₂ |
   | a₃ b₃ c₃ |

2. Calculate the Determinant of A (D):

   D = a₁(b₂c₃ - b₃c₂) - b₁(a₂c₃ - a₃c₂) + c₁(a₂b₃ - a₃b₂)

   If D = 0, the system either has no solution or infinitely many solutions, and Cramer's Rule cannot provide a unique solution.

3. Form and Calculate Determinants for Each Variable (Dx, Dy, Dz):
   • Dx: Replace the X-coefficients column in matrix A with the constant terms (d₁, d₂, d₃).

     | d₁ b₁ c₁ |
     | d₂ b₂ c₂ |
     | d₃ b₃ c₃ |

     Dx = d₁(b₂c₃ - b₃c₂) - b₁(d₂c₃ - d₃c₂) + c₁(d₂b₃ - d₃b₂)

   • Dy: Replace the Y-coefficients column in matrix A with the constant terms.

     | a₁ d₁ c₁ |
     | a₂ d₂ c₂ |
     | a₃ d₃ c₃ |

     Dy = a₁(d₂c₃ - d₃c₂) - d₁(a₂c₃ - a₃c₂) + c₁(a₂d₃ - a₃d₂)

   • Dz: Replace the Z-coefficients column in matrix A with the constant terms.

     | a₁ b₁ d₁ |
     | a₂ b₂ d₂ |
     | a₃ b₃ d₃ |

     Dz = a₁(b₂d₃ - b₃d₂) - b₁(a₂d₃ - a₃d₂) + d₁(a₂b₃ - a₃b₂)

4. Calculate the Solutions:

   X = Dx / D

   Y = Dy / D

   Z = Dz / D

How to Use the Calculator

Our calculator simplifies this process for you. Follow these steps:

1. Identify Coefficients: For each of your three equations, identify the coefficients for X, Y, Z, and the constant term. Ensure your equations are in the standard form aX + bY + cZ = d.
2. Input Values: Enter the corresponding numerical values into the input fields for a1, b1, c1, d1 (for Equation 1), a2, b2, c2, d2 (for Equation 2), and a3, b3, c3, d3 (for Equation 3).
3. Click "Solve System": The calculator will instantly compute the values of X, Y, and Z.
4. Interpret Results: The result section will display the unique solution (X, Y, Z) if one exists. If the determinant D is zero, it will indicate that there is no unique solution, meaning the system either has no solution or infinitely many solutions.

Example Calculation

Let's solve the following system:

2X + Y - Z = 8

-3X - Y + 2Z = -11

-2X + Y + 2Z = -3

Inputs for the calculator:

• a1 = 2, b1 = 1, c1 = -1, d1 = 8
• a2 = -3, b2 = -1, c2 = 2, d2 = -11
• a3 = -2, b3 = 1, c3 = 2, d3 = -3

Using Cramer's Rule (as the calculator does):

First, calculate D:

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

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

D = 2(-4) - 1(-2) - 1(-5)

D = -8 + 2 + 5 = -1

Since D is not zero, a unique solution exists.

Next, calculate Dx:

Dx = 8((-1)(2) - (2)(1)) - 1((-11)(2) - (2)(-3)) + (-1)((-11)(1) - (-1)(-3))

Dx = 8(-2 - 2) - 1(-22 + 6) - 1(-11 - 3)

Dx = 8(-4) - 1(-16) - 1(-14)

Dx = -32 + 16 + 14 = -2

Then, calculate Dy:

Dy = 2((-11)(2) - (2)(-3)) - 8((-3)(2) - (2)(-2)) + (-1)((-3)(-3) - (-11)(-2))

Dy = 2(-22 + 6) - 8(-6 + 4) - 1(9 - 22)

Dy = 2(-16) - 8(-2) - 1(-13)

Dy = -32 + 16 + 13 = -3

Finally, calculate Dz:

Dz = 2((-1)( -3) - (1)(-11)) - 1((-3)(-3) - (-11)(-2)) + 8((-3)(1) - (-1)(-2))

Dz = 2(3 + 11) - 1(9 - 22) + 8(-3 - 2)

Dz = 2(14) - 1(-13) + 8(-5)

Dz = 28 + 13 - 40 = 1

Solutions:

X = Dx / D = -2 / -1 = 2

Y = Dy / D = -3 / -1 = 3

Z = Dz / D = 1 / -1 = -1

The calculator will output: X = 2.0000, Y = 3.0000, Z = -1.0000.

This calculator is a powerful tool for quickly and accurately solving systems of three linear equations, saving time and reducing the potential for manual calculation errors.