Solving a System of Linear Equations using a Matrix

Oct 13, 2025 by wbrum.org 44 views

Solving a System of Linear Equations using a Matrix

Introduction

In mathematics, a system of linear equations is a set of two or more equations in which the variables are of the first degree. These equations can be represented in a matrix form, which makes it easier to solve them. In this article, we will discuss how to solve a system of linear equations using a matrix.

The Matrix Representation

The given system of equations is:

${ \begin{array}{l} y = 700x + 200 \\ y = 5,000 - 75x \end{array} \}$

To represent this system in a matrix form, we need to rewrite the equations in the form of:

${ \begin{array}{l} ax + by = c \\ dx + ey = f \end{array} \}$

where a, b, c, d, e, and f are constants.

In this case, we have:

${ \begin{array}{l} 700x + 200y = c \\ -75x + 5,000y = f \end{array} \}$

Now, we can represent this system in a matrix form as:

\begin{bmatrix} c \ f \end{bmatrix} \end{array} }$

Solving the System using a Matrix

To solve this system, we can use the method of substitution or elimination. However, using a matrix makes it easier to solve the system.

We can represent the matrix as:

${ A = \begin{bmatrix} 700 & 200 \\ -75 & 5,000 \end{bmatrix} }$

and the vector of constants as:

${ b = \begin{bmatrix} c \\ f \end{bmatrix} }$

The solution to the system is given by:

${ x = A^{-1}b }$

where A^{-1} is the inverse of the matrix A.

Finding the Inverse of the Matrix

To find the inverse of the matrix A, we can use the formula:

${ A^{-1} = \frac{1}{\det(A)} \begin{bmatrix} e_{11} & e_{12} \\ e_{21} & e_{22} \end{bmatrix} }$

where det(A) is the determinant of the matrix A, and e_{ij} are the cofactors of the matrix A.

The determinant of the matrix A is given by:

${ \det(A) = 700(5,000) - 200(-75) = 3,500,000 + 15,000 = 3,515,000 }$

The cofactors of the matrix A are:

${ e_{11} = 5,000, e_{12} = 75, e_{21} = -200, e_{22} = 700 }$

Now, we can find the inverse of the matrix A as:

${ A^{-1} = \frac{1}{3,515,000} \begin{bmatrix} 5,000 & 75 \\ -200 & 700 \end{bmatrix} }$

Solving for x and y

Now that we have the inverse of the matrix A, we can solve for x and y as:

${ x = A^{-1}b = \frac{1}{3,515,000} \begin{bmatrix} 5,000 & 75 \\ -200 & 700 \end{bmatrix} \begin{bmatrix} c \\ f \end{bmatrix} }$

Substituting the values of c and f, we get:

${ x = \frac{1}{3,515,000} \begin{bmatrix} 5,000 & 75 \\ -200 & 700 \end{bmatrix} \begin{bmatrix} 700x + 200 \\ -75x + 5,000 \end{bmatrix} }$

Simplifying the expression, we get:

${ x = \frac{1}{3,515,000} \begin{bmatrix} 5,000(700x + 200) + 75(-75x + 5,000) \\ -200(700x + 200) + 700(-75x + 5,000) \end{bmatrix} }$

Simplifying further, we get:

${ x = \frac{1}{3,515,000} \begin{bmatrix} 3,500,000x + 1,000,000 - 5,625x + 375,000 \\ -140,000x - 40,000 + 49,500x - 3,500,000 \end{bmatrix} }$

Simplifying the expression, we get:

${ x = \frac{1}{3,515,000} \begin{bmatrix} -1,125x + 1,375,000 \\ -90,500x - 3,540,000 \end{bmatrix} }$

Simplifying further, we get:

${ x = \frac{1}{3,515,000} \begin{bmatrix} -1,125x \\ -90,500x \end{bmatrix} }$

Simplifying the expression, we get:

${ x = \frac{-1,125x}{3,515,000} }$

Simplifying further, we get:

${ x = \frac{-1,125}{3,515,000}x }$

Simplifying the expression, we get:

${ x = -\frac{1,125}{3,515,000}x }$

Simplifying further, we get:

${ x = -\frac{1}{3,115}x }$

Simplifying the expression, we get:

${ x = -\frac{1}{3,115}x }$

Simplifying further, we get:

${ x = 0 }$

Now that we have the value of x, we can substitute it into one of the original equations to solve for y.

Substituting x = 0 into the first equation, we get:

${ y = 700(0) + 200 }$

Simplifying the expression, we get:

${ y = 200 }$

Therefore, the solution to the system of equations is x = 0 and y = 200.

Conclusion

In this article, we discussed how to solve a system of linear equations using a matrix. We represented the system in a matrix form, found the inverse of the matrix, and solved for x and y. The solution to the system is x = 0 and y = 200.

Discussion

The system of equations we solved in this article is a simple example of a system of linear equations. However, in real-world applications, systems of linear equations can be much more complex and involve many variables. In such cases, using a matrix to solve the system can be a powerful tool.

References

[1] "Linear Algebra and Its Applications" by Gilbert Strang
[2] "Introduction to Linear Algebra" by Gilbert Strang
[3] "Linear Algebra and Its Applications" by David C. Lay

Final Answer

The final answer is:

x = 0 y = 200
Solving a System of Linear Equations using a Matrix: Q&A

Introduction

In our previous article, we discussed how to solve a system of linear equations using a matrix. We represented the system in a matrix form, found the inverse of the matrix, and solved for x and y. In this article, we will answer some common questions related to solving systems of linear equations using a matrix.

Q: What is a system of linear equations?

A: A system of linear equations is a set of two or more equations in which the variables are of the first degree. These equations can be represented in a matrix form, which makes it easier to solve them.

Q: How do I represent a system of linear equations in a matrix form?

A: To represent a system of linear equations in a matrix form, you need to rewrite the equations in the form of:

${ \begin{array}{l} ax + by = c \\ dx + ey = f \end{array} \}$

where a, b, c, d, e, and f are constants.

Q: What is the inverse of a matrix?

A: The inverse of a matrix is a matrix that, when multiplied by the original matrix, results in the identity matrix. The inverse of a matrix is denoted by A^{-1}.

Q: How do I find the inverse of a matrix?

A: To find the inverse of a matrix, you can use the formula:

${ A^{-1} = \frac{1}{\det(A)} \begin{bmatrix} e_{11} & e_{12} \\ e_{21} & e_{22} \end{bmatrix} }$

where det(A) is the determinant of the matrix A, and e_{ij} are the cofactors of the matrix A.

Q: What is the determinant of a matrix?

A: The determinant of a matrix is a scalar value that can be used to determine the solvability of a system of linear equations. The determinant of a 2x2 matrix is given by:

${ \det(A) = ad - bc }$

where a, b, c, and d are the elements of the matrix.

Q: How do I solve a system of linear equations using a matrix?

A: To solve a system of linear equations using a matrix, you need to follow these steps:

Represent the system of linear equations in a matrix form.
Find the inverse of the matrix.
Multiply the inverse of the matrix by the vector of constants to get the solution.

Q: What are the advantages of using a matrix to solve a system of linear equations?

A: The advantages of using a matrix to solve a system of linear equations include:

It makes it easier to solve the system.
It allows you to use the inverse of the matrix to solve the system.
It makes it easier to find the determinant of the matrix.

Q: What are the disadvantages of using a matrix to solve a system of linear equations?

A: The disadvantages of using a matrix to solve a system of linear equations include:

It can be difficult to find the inverse of the matrix.
It can be difficult to find the determinant of the matrix.
It can be difficult to solve the system if the matrix is not invertible.

Q: When should I use a matrix to solve a system of linear equations?

A: You should use a matrix to solve a system of linear equations when:

The system has two or more equations.
The variables are of the first degree.
The matrix is invertible.

Q: When should I not use a matrix to solve a system of linear equations?

A: You should not use a matrix to solve a system of linear equations when:

The system has only one equation.
The variables are not of the first degree.
The matrix is not invertible.

Conclusion

In this article, we answered some common questions related to solving systems of linear equations using a matrix. We discussed the advantages and disadvantages of using a matrix to solve a system of linear equations and provided some tips on when to use a matrix and when not to use a matrix.

Final Answer

The final answer is:

A system of linear equations is a set of two or more equations in which the variables are of the first degree.
To represent a system of linear equations in a matrix form, you need to rewrite the equations in the form of: ${ \begin{array}{l} ax + by = c \\ dx + ey = f \end{array} \}$
The inverse of a matrix is a matrix that, when multiplied by the original matrix, results in the identity matrix.
To find the inverse of a matrix, you can use the formula: ${ A^{-1} = \frac{1}{\det(A)} \begin{bmatrix} e_{11} & e_{12} \\ e_{21} & e_{22} \end{bmatrix} }$
The determinant of a matrix is a scalar value that can be used to determine the solvability of a system of linear equations.
To solve a system of linear equations using a matrix, you need to follow these steps:

Represent the system of linear equations in a matrix form.
Find the inverse of the matrix.
Multiply the inverse of the matrix by the vector of constants to get the solution.