Solving Linear Equations Determining A, B, And C Values

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Introduction to Linear Equation Systems

In the realm of mathematics, systems of linear equations hold a pivotal position, serving as a fundamental tool for modeling and solving real-world problems across diverse fields, including engineering, physics, economics, and computer science. These systems involve a set of equations, each of which is linear, meaning that the variables involved are raised to the power of one. The goal is to find the values of the variables that simultaneously satisfy all the equations in the system. This article delves into the intricacies of solving a specific system of linear equations, focusing on determining the values of unknown coefficients within the system's matrix representation.

Linear equation systems are indispensable in various applications. In engineering, they are used to analyze electrical circuits, structural frameworks, and fluid dynamics. Physicists employ them to model the motion of particles, the flow of heat, and the behavior of electromagnetic fields. Economists utilize linear systems to study market equilibrium, input-output analysis, and economic forecasting. Computer scientists leverage them in computer graphics, optimization problems, and machine learning algorithms. Understanding the principles and techniques for solving linear equations is thus crucial for students and professionals in these fields.

In this article, we will tackle a specific system of linear equations represented in matrix form:

[a1−35b12−1c][pqr]=[8−75]{ \begin{bmatrix} a & 1 & -3 \\ 5 & b & 1 \\ 2 & -1 & c \end{bmatrix} \begin{bmatrix} p \\ q \\ r \end{bmatrix} = \begin{bmatrix} 8 \\ -7 \\ 5 \end{bmatrix} }

Our primary objective is to determine suitable values for the unknown coefficients a, b, and c that render this system solvable. We will explore the concepts of system consistency, matrix invertibility, and potential solution approaches. By delving into this problem, we aim to provide a comprehensive understanding of how to approach and solve linear equation systems, equipping readers with the skills and knowledge necessary to tackle similar problems in their respective fields.

Understanding the System of Equations

To effectively solve the given system of linear equations, a thorough understanding of its structure is essential. The system is represented in matrix form as follows:

[a1−35b12−1c][pqr]=[8−75]{ \begin{bmatrix} a & 1 & -3 \\ 5 & b & 1 \\ 2 & -1 & c \end{bmatrix} \begin{bmatrix} p \\ q \\ r \end{bmatrix} = \begin{bmatrix} 8 \\ -7 \\ 5 \end{bmatrix} }

This equation can be broken down into a set of three linear equations:

  1. ap + q - 3r = 8
  2. 5p + bq + r = -7
  3. 2p - q + cr = 5

Here, a, b, and c are the unknown coefficients we aim to determine. p, q, and r are the variables, and the constants on the right-hand side (8, -7, and 5) define the system's solution space. The matrix on the left-hand side is the coefficient matrix, and the vector egin{bmatrix} p \ q \ r \end{bmatrix} is the variable vector.

The crucial question we need to address is: Can we find values for a, b, and c that make this system consistent? A system of linear equations is considered consistent if it has at least one solution. This means there exists at least one set of values for p, q, and r that satisfies all three equations simultaneously. Conversely, if no such set of values exists, the system is inconsistent, indicating that there is no solution.

To determine the consistency of the system, we need to delve into the properties of the coefficient matrix and the relationships between the equations. One approach is to analyze the determinant of the coefficient matrix. If the determinant is non-zero, the matrix is invertible, and the system has a unique solution. However, if the determinant is zero, the matrix is singular, and the system may have either infinitely many solutions or no solution at all. We must also look at the relationships between the equations. For instance, if one equation is a linear combination of the other two, then the system will have infinite solutions or no solutions.

Determining Values for a, b, and c

To determine suitable values for a, b, and c, we need to consider the consistency of the system. As previously discussed, a system of linear equations is consistent if it has at least one solution. To achieve this, we can strategically choose values for a, b, and c that do not lead to contradictions within the equations. We can utilize techniques from linear algebra to guide our choices.

One approach is to consider the determinant of the coefficient matrix:

∣a1−35b12−1c∣{ \begin{vmatrix} a & 1 & -3 \\ 5 & b & 1 \\ 2 & -1 & c \end{vmatrix} }

If the determinant is non-zero, the matrix is invertible, and the system has a unique solution for any values on the right-hand side. The determinant can be calculated as:

Determinant = a(bc + 1) - 1(5c - 2) - 3(-5 - 2b)

= abc + a - 5c + 2 + 15 + 6b

= abc + a + 6b - 5c + 17

To ensure the system has a unique solution, we need to choose a, b, and c such that this expression is not equal to zero.

Let's choose some simple values to illustrate:

If we set a = 1, b = 1, and c = 1, the determinant becomes:

Determinant = (1)(1)(1) + 1 + 6(1) - 5(1) + 17 = 1 + 1 + 6 - 5 + 17 = 20

Since the determinant is 20, which is non-zero, these values of a, b, and c will result in a consistent system with a unique solution.

Another approach is to analyze the equations directly. Let’s revisit our system of equations:

  1. ap + q - 3r = 8
  2. 5p + bq + r = -7
  3. 2p - q + cr = 5

We can choose values for a, b, and c that simplify these equations or make them independent. For example, we can choose a = 1, b = 1, and c = 1, as we did before. This gives us:

  1. p + q - 3r = 8
  2. 5p + q + r = -7
  3. 2p - q + r = 5

This system can be solved using various methods, such as substitution, elimination, or matrix inversion. The fact that we could find a non-zero determinant suggests that solving the system should give us unique values for p, q, and r.

Alternative scenarios: If we choose values that make the determinant zero, such as setting a = 0, b = 0, and c = -17/5, we might encounter scenarios where the system has either infinite solutions or no solution. For example, if we set a = 0, b = 0, and c = 0, our equations become:

  1. q - 3r = 8
  2. 5p + r = -7
  3. 2p - q = 5

This system might still be solvable, but the solution process and the number of possible solutions will be different from the case when the determinant is non-zero.

Solving the System with Chosen Values

Having determined suitable values for a, b, and c that ensure the consistency of the system, the next step involves solving the system of equations to find the values of p, q, and r. Previously, we chose a = 1, b = 1, and c = 1, which resulted in the following system of equations:

  1. p + q - 3r = 8
  2. 5p + q + r = -7
  3. 2p - q + r = 5

There are several methods to solve this system, including substitution, elimination, and matrix inversion. Here, we will use the elimination method, which is a systematic way of combining equations to eliminate variables one by one.

Step 1: Eliminate q from equations (1) and (2)

Subtract equation (1) from equation (2):

(5p + q + r) - (p + q - 3r) = -7 - 8

4p + 4r = -15 (Equation 4)

Step 2: Eliminate q from equations (1) and (3)

Add equation (1) to equation (3):

(p + q - 3r) + (2p - q + r) = 8 + 5

3p - 2r = 13 (Equation 5)

Step 3: Solve the system of two equations (4) and (5) for p and r

We now have a system of two equations with two variables, p and r:

  1. 4p + 4r = -15
  2. 3p - 2r = 13

Multiply equation (5) by 2 to make the coefficients of r opposites:

6p - 4r = 26 (Equation 6)

Add equation (4) to equation (6):

(4p + 4r) + (6p - 4r) = -15 + 26

10p = 11

p = 11/10

Now, substitute the value of p into equation (4) to find r:

4(11/10) + 4r = -15

44/10 + 4r = -15

4r = -15 - 44/10

4r = -194/10

r = -194/40

r = -97/20

Step 4: Substitute the values of p and r into equation (1) to find q

p + q - 3r = 8

(11/10) + q - 3(-97/20) = 8

q = 8 - 11/10 - 291/20

q = (160 - 22 - 291)/20

q = -153/20

Solution

Thus, the solution to the system of equations with a = 1, b = 1, and c = 1 is:

p = 11/10

q = -153/20

r = -97/20

This solution confirms that with our chosen values of a, b, and c, the system is consistent and has a unique solution.

Alternative Values and Scenarios

While we successfully found a solution for the system of equations with a = 1, b = 1, and c = 1, it is essential to explore how different choices of these coefficients can impact the system's behavior. As discussed earlier, the determinant of the coefficient matrix plays a crucial role in determining the consistency and uniqueness of the solution.

If the determinant is non-zero, the system has a unique solution, as we observed in our previous example. However, if the determinant is zero, the system may have either infinitely many solutions or no solution at all. Let's consider a scenario where the determinant is zero to illustrate this.

Recall that the determinant of the coefficient matrix is given by:

Determinant = abc + a + 6b - 5c + 17

To make the determinant zero, we need to find values of a, b, and c that satisfy the equation:

abc + a + 6b - 5c + 17 = 0

Let's choose a = -17, b = 0, and c = 1. The determinant becomes:

(-17)(0)(1) + (-17) + 6(0) - 5(1) + 17 = 0 - 17 + 0 - 5 + 17 = -5

This choice doesn't result in a zero determinant. Let's try a = 1, b = -3, and c = -2:

(1)(-3)(-2) + 1 + 6(-3) - 5(-2) + 17 = 6 + 1 - 18 + 10 + 17 = 16

That did not result in 0 either. Let's try to set a = 0, b = 1 and solve for c:

(0)(1)c + 0 + 6(1) - 5c + 17 = 0

6 - 5c + 17 = 0

23 - 5c = 0

5c = 23

c = 23/5

So we have a = 0, b = 1, and c = 23/5. With these values, our system of equations becomes:

  1. q - 3r = 8
  2. 5p + q + r = -7
  3. 2p - q + (23/5)r = 5

Now, let’s add equations 1 and 3:

q - 3r + 2p - q + (23/5)r = 8 + 5

2p + (8/5)r = 13

Now we have:

  1. q - 3r = 8
  2. 5p + q + r = -7
  3. 2p + (8/5)r = 13

Multiply the third equation by 5/2:

5p + 4r = 65/2

Subtract the second equation:

5p + 4r - (5p + q + r) = 65/2 - (-7)

3r - q = 79/2

We also know that q - 3r = 8, and multiply this equation by -1:

-q + 3r = -8

Now it is obvious the system of equations does not have any solution because 3r - q and -q + 3r cannot have different values. Therefore, with those values, the system does not have a solution, meaning it is inconsistent.

Infinite Solutions: To demonstrate infinite solutions, the equations would have to reduce in such a way that one or more equations are redundant, meaning they provide no unique information. For instance, one equation becomes a linear combination of the others. This requires more complex choices for a, b, and c.

Conclusion

In this comprehensive exploration, we delved into solving systems of linear equations, focusing on the determination of unknown coefficients within the matrix representation. We highlighted the fundamental importance of linear equation systems across various fields and illustrated a step-by-step approach to solving a specific system. We first chose values for a, b, and c to ensure that the system was consistent, meaning it had at least one solution. The determinant of the coefficient matrix was a crucial tool in making this determination.

When the determinant is non-zero, the system has a unique solution. We demonstrated this by setting a = 1, b = 1, and c = 1, solving the resulting system using the elimination method. This led us to specific values for p, q, and r that satisfied all three equations. This showcased the practical application of linear algebra techniques in finding unique solutions to linear systems.

Furthermore, we examined scenarios where the determinant could be zero. In these cases, the system could have either no solution or infinitely many solutions. We demonstrated a scenario where setting the coefficients to specific values (a = 0, b = 1, c = 23/5) led to an inconsistent system with no solution. This underscores the importance of the determinant in understanding the nature of the solutions.

Solving systems of linear equations is a fundamental skill in mathematics and is applied widely across various disciplines. Understanding the conditions for consistency and uniqueness of solutions, as well as the practical methods for finding those solutions, is invaluable. The techniques and insights provided here offer a solid foundation for tackling more complex problems in linear algebra and its applications. Whether you're a student, engineer, scientist, or anyone working with quantitative data, mastering these concepts will empower you to model and solve a wide range of real-world problems.