4  SLE

A System of Linear Equations (SLE) consists of two or more linear equations involving the same set of variables. The goal of SLE is to find the values of these variables that satisfy all the equations in the system. SLE is widely used in various fields, including computer science, engineering, economics, and social sciences, as many real-life problems can be modeled using linear equations.

4.1 Introduction

Before we delve into the System of Linear Equations (SLE), it’s crucial to first understand the basic concepts of functions and equations. These concepts form the foundation for understanding and solving systems of equations.

4.1.1 Function

A function is a mathematical relationship where each input (often represented by the variable \(x\)) has exactly one output (represented by \(y\)). A function describes how the output depends on the input and is typically written as:

\[ \begin{array}{rcl} y &=& f(x) \quad ,\text{or}\\ ay &=& bx+c \end{array} \]

where:

  • \(a, b,\) and \(c\) are constants.
  • \(x\) and \(y\) are variables.

Example of a linear function:

\[ y = 2x + 3 \]

Notes: \(y\) is determined by the value of \(x\). For each input value of \(x\), there is one unique output value of \(y\). This is a key property of a function: every input corresponds to exactly one output.

4.1.2 Equation

An equation, on the other hand, is a mathematical statement that asserts the equality of two expressions. It contains an equal sign “=” and is often used to find unknown values that satisfy the equality between the two sides of the equation.

Example of a Linear Equation:

\[ 2x + 3y = 6 \]

Notes: While functions represent how one variable depends on another, equations represent a balance or equality between expressions.

4.1.3 Functions vs Equations

Key Differences Between Functions and Equations

Aspect Function Equation
Definition Describes a relationship where one variable depends on another. States the equality between two expressions.
Symbol Uses \(y = f(x)\) to express the relationship. Uses the “=” sign to indicate equality.
Goal To define how one variable depends on another (input/output). To find values that balance both sides of the equation.
Example \(y = 2x + 3\) \(2x + 3y = 6\)
Graph A graph of a function shows how the output changes based on the input. A graph of an equation shows the relationship between two variables.

4.2 SLE in 2D

A System of Linear Equations in 2D consists of two linear equations with two variables, typically represented as \(x\) and \(y\). The solutions to this system represent the points at which the lines defined by the equations intersect.

The standard form of a system of linear equations in two dimensions is: \[ \begin{aligned} a_1x + b_1y &= c_1 \quad \text{(Equation 1)} \\ a_2x + b_2y &= c_2 \quad \text{(Equation 2)} \end{aligned} \] where \(a_1, b_1, c_1, a_2, b_2, c_2\) are constants.

Let consider the following system:

\[ \begin{aligned} 2x + 3y &= 6 \quad \text{(Equation 1)} \\ 4x - y &= 5 \quad \text{(Equation 2)} \end{aligned} \]

There are several methods to solve a system of linear equations in two dimensions. Below are the most common methods:

4.2.1 Substitution

The substitution method involves solving one equation for one variable and substituting it into the other equation:

Step 1: Solve one equation for one variable

We can start with Equation 1 and solve for \(y\):

\[ 3y = 6 - 2x \]

Dividing both sides by 3 gives:

\[ y = \frac{6 - 2x}{3} \quad \text{(Equation 3)} \]

Step 2: Substitute into the other equation

Now, we will substitute Equation 3 into Equation 2:

\[ 4x - y = 5 \]

Substituting \(y\) from Equation 3:

\[ 4x - \frac{6 - 2x}{3} = 5 \]

Step 3: Eliminate the fraction

To eliminate the fraction, multiply every term by 3:

\[ 3(4x) - (6 - 2x) = 3(5) \]

This simplifies to:

\[ 12x - 6 + 2x = 15 \]

Step 4: Combine like terms

Combine the \(x\) terms:

\[ 14x - 6 = 15 \]

Step 5: Solve for \(x\)

Add 6 to both sides:

\[ 14x = 21 \]

Now divide by 14:

\[ x = \frac{21}{14} = \frac{3}{2} \]

Step 6: Substitute back to find \(y\)

Now substitute \(x = \frac{3}{2}\) back into Equation 3 to find \(y\):

\[ y = \frac{6 - 2\left(\frac{3}{2}\right)}{3} \]

Calculating inside the parentheses:

\[ y = \frac{6 - 3}{3} = \frac{3}{3} = 1 \]

Thus, the solution to the system of equations is:

\[ \left( x, y \right) = \left( \frac{3}{2}, 1 \right) \]

This means that \(x = \frac{3}{2}\) and \(y = 1\) satisfy both equations in the system.

4.2.2 Elimination

The elimination method involves adding or subtracting equations to eliminate one variable:

Step 1: Align the equations

We can keep the equations as they are for now:

\[ \begin{aligned} 2x + 3y &= 6 \quad \text{(Equation 1)} \\ 4x - y &= 5 \quad \text{(Equation 2)} \end{aligned} \]

Step 2: Multiply the equations if necessary

To eliminate \(y\), we can multiply Equation 2 by 3 so that the coefficients of \(y\) will match:

\[ 3(4x - y) = 3(5) \]

This gives us:

\[ 12x - 3y = 15 \quad \text{(Equation 3)} \]

Now we have:

\[ \begin{aligned} 2x + 3y &= 6 \quad \text{(Equation 1)} \\ 12x - 3y &= 15 \quad \text{(Equation 3)} \end{aligned} \]

Step 3: Add the equations

Now, we can add Equation 1 and Equation 3 together to eliminate \(y\):

\[ (2x + 3y) + (12x - 3y) = 6 + 15 \]

This simplifies to:

\[ 14x = 21 \]

Step 4: Solve for \(x\)

Divide both sides by 14:

\[ x = \frac{21}{14} = \frac{3}{2} \]

Step 5: Substitute back to find \(y\)

Now substitute \(x = \frac{3}{2}\) back into Equation 1 to find \(y\):

\[ 2\left(\frac{3}{2}\right) + 3y = 6 \]

Calculating:

\[ 3 + 3y = 6 \]

Subtract 3 from both sides:

\[ 3y = 3 \]

Now divide by 3:

\[ y = 1 \]

Thus, the solution to the system of equations is:

\[ \left( x, y \right) = \left( \frac{3}{2}, 1 \right) \]

This means that \(x = \frac{3}{2}\) and \(y = 1\) satisfy both equations in the system.

4.2.3 Augmented Matrix

Step 1: Set up the Augmented Matrix

We can represent the system of equations as an augmented matrix:

\[ \begin{bmatrix} 2 & 3 & | & 6 \\ 4 & -1 & | & 5 \end{bmatrix} \]

Step 2: Perform Row Operations

Our goal is to convert this matrix into row-echelon form using row operations.

  1. Scale the First Row

First, we can scale the first row to make the leading coefficient (the coefficient of \(x\) equal to 1. However, we can also keep it as is for now:

\[ R_1: \begin{bmatrix} 2 & 3 & | & 6 \end{bmatrix} \]

  1. Eliminate \(x\) from the Second Row

To eliminate \(x\) from the second row, we can replace \(R_2\) with \(R_2 - 2R_1\):

\[ R_2: \begin{bmatrix} 4 & -1 & | & 5 \end{bmatrix} - 2 \times \begin{bmatrix} 2 & 3 & | & 6 \end{bmatrix} \]

Calculating this gives us:

\[ R_2: \begin{bmatrix} 4 - 4 & -1 - 6 & | & 5 - 12 \end{bmatrix}=\begin{bmatrix} 0 & -7 & | & -7 \end{bmatrix} \]

So now our augmented matrix is:

\[ \begin{bmatrix} 2 & 3 & | & 6 \\ 0 & -7 & | & -7 \end{bmatrix} \]

Step 3: Solve for the Variables

  1. Back Substitute

From the second row, we can solve for \(y\):

\[ -7y = -7 \implies y = 1 \]

  1. Substitute \(y\) back into the first row

Now substitute \(y = 1\) back into the first equation (from the first row of the augmented matrix):

\[ 2x + 3(1) = 6 \]

This simplifies to:

\[ 2x + 3 = 6 \]

Subtract 3 from both sides:

\[ 2x = 3 \]

Now divide by 2:

\[ x = \frac{3}{2} \]

Thus, the solution to the system of equations is:

\[ \left( x, y \right) = \left( \frac{3}{2}, 1 \right) \]

This means that \(x = \frac{3}{2}\) and \(y = 1\) satisfy both equations in the system.

4.2.4 Invers Matrix

The matrix method uses matrix operations to solve the system of equations:

Step 1: Write the system in matrix form

We can express the system in the form \(AX = B\), where:

  • \(A\) is the coefficient matrix,
  • \(X\) is the column matrix of variables, and
  • \(B\) is the column matrix of constants.

For our system, this looks like:

\[ \begin{bmatrix} 2 & 3 \\ 4 & -1 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 6 \\ 5 \end{bmatrix} \]

Thus, we have:

  • \(A = \begin{bmatrix} 2 & 3 \\ 4 & -1 \end{bmatrix}\)
  • \(X = \begin{bmatrix} x \\ y \end{bmatrix}\)
  • \(B = \begin{bmatrix} 6 \\ 5 \end{bmatrix}\)

Step 2: Find the Inverse of Matrix \(A\)

To solve for \(X\), we need to calculate \(A^{-1}\) (the inverse of matrix \(A\)). The formula for the inverse of a 2x2 matrix

Step 2.1: Calculate the determinant

The determinant \(D\) is calculated as:

\[ D = ad - bc = (2)(-1) - (3)(4) = -2 - 12 = -14 \]

Step 2.2: Apply the formula for the inverse

Now we apply the formula:

\[ A^{-1} = \frac{1}{-14} \begin{bmatrix} -1 & -3 \\ -4 & 2 \end{bmatrix} = \begin{bmatrix} \frac{1}{14} & \frac{3}{14} \\ \frac{4}{14} & -\frac{2}{14} \end{bmatrix} = \begin{bmatrix} \frac{1}{14} & \frac{3}{14} \\ \frac{2}{7} & -\frac{1}{7} \end{bmatrix} \]

Step 3: Multiply \(A^{-1}\) by \(B\)

Now we can find \(X\) by multiplying the inverse of \(A\) by \(B\):

\[ X = A^{-1}B \]

Calculating this gives:

\[ X = \begin{bmatrix} \frac{1}{14} & \frac{3}{14} \\ \frac{2}{7} & -\frac{1}{7} \end{bmatrix} \begin{bmatrix} 6 \\ 5 \end{bmatrix} \]

Step 3.1: Perform the matrix multiplication

Calculating each element:

\[ X = \begin{bmatrix} \frac{1}{14}(6) + \frac{3}{14}(5) \\ \frac{2}{7}(6) + -\frac{1}{7}(5) \end{bmatrix} = \begin{bmatrix} \frac{6}{14} + \frac{15}{14} \\ \frac{12}{7} - \frac{5}{7} \end{bmatrix} = \begin{bmatrix} \frac{21}{14} \\ \frac{7}{7} \end{bmatrix} = \begin{bmatrix} \frac{3}{2} \\ 1 \end{bmatrix} \]

Thus, the solution to the system of equations is:

\[ \left( x, y \right) = \left( \frac{3}{2}, 1 \right) \]

This means that \(x = \frac{3}{2}\) and \(y = 1\) satisfy both equations in the system.

4.2.5 Graphical

Step 1: Convert Each Equation to Slope-Intercept Form

First, we convert each equation to the slope-intercept form \(y = mx + b\):

  • For Equation 1:

\[ 2x + 3y = 6 \]

Rearranging gives:

\[ 3y = 6 - 2x \quad \Rightarrow \quad y = -\frac{2}{3}x + 2 \]

  • For Equation 2:

\[ 4x - y = 5 \] Rearranging gives: \[ y = 4x - 5 \]

Step 2: Plot Each Equation

  1. Plotting Equation 1: \(y = -\frac{2}{3}x + 2\)
  • The y-intercept is \(2\) (where the line crosses the y-axis).
  • The slope is \(-\frac{2}{3}\), which means for every 3 units you move to the right (increasing \(x\)), you move 2 units down (decreasing \(y\)).

Plot points for this line:

  • When \(x = 0: y = 2\) → Point: \((0, 2)\)
  • When \(x = 3: y = 0\) → Point: \((3, 0)\)
  1. Plotting Equation 2: \(y = 4x - 5\)
  • The y-intercept is -5.
  • The slope is 4, meaning for every 1 unit you move to the right, you move 4 units up.

Plot points for this line:

  • When \(x = 0: y = -5\) → Point: \((0, -5)\)
  • When \(x = 2: y = 3\) → Point: \((2, 3)\)

Step 3: Identify the Intersection Point

After plotting both lines on the same Cartesian plane, identify the point where the two lines intersect. This intersection point represents the solution to the system.

Example of Intersection:

Upon plotting, you may find the intersection at the point \(\left(\frac{3}{2}, 1\right)\).

Step 4: Write the Solution

Thus, the solution to the system of equations is:

\[ \left( x, y \right) = \left( \frac{3}{2}, 1 \right) \]

This means that \(x = \frac{3}{2}\) and \(y = 1\) satisfy both equations in the system.

4.3 SLE in 3D

To solve a system of linear equations (SLE) in three dimensions (3D), you can follow similar methods as in 2D (like substitution, elimination, and matrix methods), but the visualization and interpretation will be different due to the third variable.

4.3.1 Invers Matrix Method

Given the system of linear equations:

  1. \(x + y + z = 6\) (Equation 1)
  2. \(2x + 2y + z = 10\) (Equation 2)
  3. \(x - y + 2z = 3\) (Equation 3)

Step 1: Write in Matrix Form

We can represent the system as a matrix equation:

\[ AX = B \]

Where:

  • Coefficient matrix \(A\):

    \[ A = \begin{bmatrix} 1 & 1 & 1 \\ 2 & 2 & 1 \\ 1 & -1 & 2 \end{bmatrix} \]

  • Variable matrix \(X\):

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

  • Constant matrix \(B\):

    \[ B = \begin{bmatrix} 6 \\ 10 \\ 3 \end{bmatrix} \]

Step 2: Find the Inverse of Matrix \(A\)

To solve for \(X\), we need to calculate \(A^{-1}\) (the inverse of matrix \(A\)). First, calculate the determinant of \(A\):

\[ \det(A) = (1)(5) - (1)(3) + (1)(-4) = -2 \]

Since \(\det(A) \neq 0\), matrix \(A\) is invertible. Next, find the cofactor matrix:

\[ \text{Cofactor}(A) = \begin{bmatrix} 5 & -3 & -4 \\ -3 & 1 & -2 \\ -1 & 1 & 0 \end{bmatrix} \]

The adjugate matrix is the transpose of the cofactor matrix:

\[ \text{adj}(A) = \begin{bmatrix} 5 & -3 & -1 \\ -3 & 1 & 1 \\ -4 & -2 & 0 \end{bmatrix} \]

Now, the inverse of \(A\) is:

\[ A^{-1} = \frac{1}{-2} \text{adj}(A) = \begin{bmatrix} -\frac{5}{2} & \frac{3}{2} & \frac{1}{2} \\ \frac{3}{2} & -\frac{1}{2} & -\frac{1}{2} \\ 2 & 1 & 0 \end{bmatrix} \]

Step 3: Solve for \(X\)

Now, multiply \(A^{-1}\) by \(B\) to find \(X\):

\[ X = A^{-1} B = \begin{bmatrix} -\frac{5}{2} & \frac{3}{2} & \frac{1}{2} \\ \frac{3}{2} & -\frac{1}{2} & -\frac{1}{2} \\ 2 & 1 & 0 \end{bmatrix} \begin{bmatrix} 6 \\ 10 \\ 3 \end{bmatrix} \]

Carrying out the multiplication:

  • For \(x\): \(-\frac{5}{2}(6) + \frac{3}{2}(10) + \frac{1}{2}(3) = 1.5\)
  • For \(y\): \(\frac{3}{2}(6) - \frac{1}{2}(10) - \frac{1}{2}(3) = 2.5\)
  • For \(z\): \(2(6) + 1(10) + 0(3) = 22\)

Thus, the solution is:

\[ X = \begin{bmatrix} 1.5 \\ 2.5 \\ 22 \end{bmatrix} \]

The solution to the system of equations is:

  • \(x = 1.5\)
  • \(y = 2.5\)
  • \(z = 22\)

4.3.2 Graphical Method

To solve the system of equations using the graphical method, we will visualize each equation as a plane in a 3D space. The solution to the system corresponds to the intersection of these planes.

Step 1: Rearrange Each Equation

To visualize each equation as a plane in 3D space, we need to express them in terms of \(z\), i.e., \(z = f(x, y)\). This allows us to plot \(z\) as a function of \(x\) and \(y\).

  • Equation 1: \(x + y + z = 6\)

    Rearranging for \(z\): \[z_1 = 6 - x - y\]

  • Equation 2: \(2x + 2y + z = 10\)

    Rearranging for \(z\): \[z_2 = 10 - 2x - 2y\]

  • Equation 3: \(x - y + 2z = 3\)

    Rearranging for \(z\): \[z_3 = \frac{3 - x + y}{2}\]

Step 2: Interpret the Geometry

Each equation represents a plane in 3D space:

  • The first plane is \(z_1 = 6 - x - y\).
  • The second plane is \(z_2 = 10 - 2x - 2y\).
  • The third plane is \(z_3 = \frac{3 - x + y}{2}\).

Step 3: Graphical Visualization

We can plot these three planes and identify their intersection:

  • Plane 1: \(z_1 = 6 - x - y\)

This plane intercepts at \(z = 6\) when \(x = 0\) and \(y = 0\), and slopes downward as \(x\) and \(y\) increase (negative coefficients of \(x\) and \(y\)).

  • Plane 2: \(z_2 = 10 - 2x - 2y\)

This plane has an intercept at \(z = 10\) when \(x = 0\) and \(y = 0\), and slopes down more steeply since the coefficients of \(x\) and \(y\) are larger (\(-2\)).

  • Plane 3: \(z_3 = \frac{3 - x + y}{2}\)

This plane behaves differently as it involves a fraction. It intercepts at \(z = 1.5\) (when \(x = 0\) and \(y = 0\)) and has different slopes along the \(x\) and \(y\) axes.

After plotting the planes, we can look for the intersection point visually. The solution to the system corresponds to the intersection of these three planes. Identify the Intersection Points:

  • Unique Solution: If all hyperplanes intersect at a single point, that point represents the unique solution.
  • No Solution: If the hyperplanes do not intersect (are parallel), the system has no solution.
  • Infinite Solutions: If the hyperplanes coincide, there are infinitely many solutions.

However, to find the exact coordinates, we can solve the system algebraically (or using numerical methods) as discussed earlier.

4.4 SLE in n-Dimensions

A System of Linear Equations (SLE) in n dimensions involves multiple linear equations with \(n\) variables. The general form of such a system can be expressed as:

\[ \begin{aligned} a_{11}x_1 + a_{12}x_2 + \ldots + a_{1n}x_n &= b_1 \quad \text{(Equation 1)} \\ a_{21}x_1 + a_{22}x_2 + \ldots + a_{2n}x_n &= b_2 \quad \text{(Equation 2)} \\ &\vdots \\ a_{m1}x_1 + a_{m2}x_2 + \ldots + a_{mn}x_n &= b_m \quad \text{(Equation m)} \end{aligned} \]

Where:

  • \(x_1, x_2, \ldots, x_n\) are the variables.
  • \(a_{ij}\) are the coefficients of the variables.
  • \(b_i\) are the constants.

4.4.1 Write in Matrix Form

You can represent the system of equations in matrix form \(AX = B\):

  • Coefficient matrix \(A\):

\[ A = \begin{bmatrix} a_{11} & a_{12} & \ldots & a_{1n} \\ a_{21} & a_{22} & \ldots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{m1} & a_{m2} & \ldots & a_{mn} \end{bmatrix} \]

  • Variable matrix \(X\):

\[ X = \begin{bmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{bmatrix} \]

  • Constant matrix \(B\):

\[ B = \begin{bmatrix} b_1 \\ b_2 \\ \vdots \\ b_m \end{bmatrix} \]

The equation can then be expressed as:

\[ AX = B \]

4.4.2 Ensure A is Invertible

The matrix \(A\) must be invertible, meaning its determinant must be non-zero:

\[ \text{det}(A) \neq 0 \]

If the determinant is zero, the system either has no solution or an infinite number of solutions.

4.4.3 Find the Inverse of Matrix A

If \(A\) is invertible, calculate \(A^{-1}\), the inverse of matrix \(A\). This can be done using various methods:

  • Gaussian Elimination,
  • Cofactor Method,
  • Adjugate and Determinant Method.

4.4.4 Multiply \(A^{-1}\) by B

Once \(A^{-1}\) is computed, the solution to the system can be found by multiplying \(A^{-1}\) by the constant matrix \(B\):

\[ X = A^{-1}B \]

\[ \begin{bmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{bmatrix} = \begin{bmatrix} a_{11} & a_{12} & \ldots & a_{1n} \\ a_{21} & a_{22} & \ldots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{m1} & a_{m2} & \ldots & a_{mn} \end{bmatrix}^{-1} \times \begin{bmatrix} b_1 \\ b_2 \\ \vdots \\ b_m \end{bmatrix} \]

This matrix multiplication will give the values of the variables \(x_1, x_2, \dots, x_n\)

4.5 Case Study of SLE

4.5.1 Overview: XYZ Manufacturing Co.

XYZ Manufacturing Co. is a well-known company in the consumer goods industry. Our mission is to make everyday life easier and better through our innovative products. Since we started, we’ve focused on delivering quality and keeping our customers happy. We offer a variety of products that cater to the needs of our consumers.

4.5.2 Industry: Consumer Goods

We operate in the fast-moving consumer goods sector, where competition is strong, and customer preferences change quickly. To succeed, we prioritize efficient production and stay up-to-date with market trends.

4.5.3 Products

We have a diverse range of products that are designed to improve the lives of our customers:

  • Product A: This kitchen appliance has become a favorite in many homes. It’s easy to use and can perform multiple tasks, making cooking simpler and faster.

  • Product B: A small electronic device that has changed how people use technology in their daily lives. It’s compact and packed with features, making it a must-have for anyone who loves gadgets.

  • Product C: This home cleaning product is effective and environmentally friendly. With more people caring about the planet, we’ve created a solution that keeps homes clean without harming the environment.

  • Product D: A personal care item that shows our commitment to quality and health. Made with natural ingredients, it appeals to customers who want to take care of themselves without using harsh chemicals.

4.5.4 Objective

XYZ Manufacturing Co. aims to optimize its production plan to meet customer demand while efficiently utilizing its resources. The company has received orders for various quantities of its products and needs to determine how many units of each product to produce to maximize profits and maintain customer satisfaction.

4.5.5 Constraints

The company faces certain resource constraints represented by the following factors:

  • Material Resources: Each product requires different amounts of raw materials, which are limited due to supplier agreements.
  • Labor Hours: The production of each product requires a specific number of labor hours, which is also limited by the workforce available.
  • Production Capacity: The manufacturing facility can only produce a certain total number of units across all products due to machinery and operational limits.

4.5.6 System of Equations

To formalize the production planning, the company develops a system of equations that represents the relationships between the products and the resources available. The equations account for the material requirements, labor hours, and production capacity:

  1. \(2x_1 + 3x_2 + 4x_3 + 2x_4 = 20\)
    • Represents the total material resources available.
  2. \(x_1 + 2x_2 + 2x_3 + 3x_4 = 15\)
    • Represents the total labor hours available.
  3. \(2x_1 + 2x_2 + 3x_3 + x_4 = 20\)
    • Represents the total production capacity for the facility.
  4. \(2x_1 + x_2 + 2x_3 + 3x_4 = 25\)
    • Represents additional constraints based on customer demand and supply chain considerations.

4.5.7 Decision Variables

  • \(x_1\): Units of Product A to produce
  • \(x_2\): Units of Product B to produce
  • \(x_3\): Units of Product C to produce
  • \(x_4\): Units of Product D to produce

4.5.8 Goals

The primary goals for the production planning include:

  • Maximizing Output: Produce enough units to meet customer demand while adhering to resource constraints.
  • Cost Efficiency: Minimize production costs by optimizing resource allocation across products.
  • Customer Satisfaction: Ensure that production levels align with customer orders to avoid stockouts.

4.5.9 Coefficient Matrix \(A\)

The coefficient matrix \(A\) can be represented as:

\[ A = \begin{bmatrix} 2 & 3 & 4 & 2 \\ 1 & 2 & 2 & 3 \\ 2 & 2 & 3 & 1 \\ 2 & 1 & 2 & 3 \end{bmatrix} \]

4.5.10 Constant Matrix \(B\)

The constant matrix \(B\) is given by:

\[ B = \begin{bmatrix} 20 \\ 15 \\ 20 \\ 25 \end{bmatrix} \]

4.5.11 Python Code to Solve the System

You can use the following Python code to solve the system of equations:

import numpy as np

# Define the coefficient matrix A
A = np.array([[2, 3, 4, 2],
              [1, 2, 2, 3],
              [2, 2, 3, 1],
              [2, 1, 2, 3]])

# Define the constant matrix B
B = np.array([20, 15, 20, 25])

# Calculate the solution
X = np.linalg.solve(A, B)

# Print the results
print(f'Units of Product A: {X[0]:.2f}')
print(f'Units of Product B: {X[1]:.2f}')
print(f'Units of Product C: {X[2]:.2f}')
print(f'Units of Product D: {X[3]:.2f}')

Expected Results: Upon running the code, the results for the number of units to produce for each product will be:

Units of Product A: 3.00
Units of Product B: 1.00
Units of Product C: 3.00
Units of Product D: 4.00

4.6 SLE in Python

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