Type The Correct Answer In The Box. Use Numerals Instead Of Words.This System Of Equations Has Been Placed In A Matrix:$\[ \begin{array}{l} y = 700x + 200 \\ y = 5000 - 75x \end{array} \\]Complete The Matrix By Filling In The Missing Numbers.

Feb 27, 2025 by ADMIN 243 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 that involve variables raised to the power of 1. These equations can be represented in a matrix form, which makes it easier to solve them. In this article, we will discuss how to complete a matrix of a system of linear equations and provide a step-by-step solution.

The Matrix Representation

The given system of equations is:

{ \begin{array}{l} y = 700x + 200 \\ y = 5000 - 75x \end{array} \}

To represent this system in a matrix form, we need to rewrite the equations in the form of $Ax = b$ , where $A$ is the coefficient matrix, $x$ is the variable matrix, and $b$ is the constant matrix.

Step 1: Rewrite the Equations

Let's rewrite the given equations in the form of $Ax = b$ :

{ \begin{array}{l} 700x + 200 = y \\ -75x + 5000 = y \end{array} \}

Step 2: Create the Coefficient Matrix

The coefficient matrix $A$ is a 2x2 matrix that contains the coefficients of the variables $x$ and $y$ . In this case, the coefficient matrix is:

{ \begin{array}{l} A = \begin{bmatrix} 700 & 1 \\ -75 & 1 \end{bmatrix} \end{array} \}

Step 3: Create the Variable Matrix

The variable matrix $x$ is a 2x1 matrix that contains the variables $x$ and $y$ . In this case, the variable matrix is:

{ \begin{array}{l} x = \begin{bmatrix} x \\ y \end{bmatrix} \end{array} \}

Step 4: Create the Constant Matrix

The constant matrix $b$ is a 2x1 matrix that contains the constants on the right-hand side of the equations. In this case, the constant matrix is:

{ \begin{array}{l} b = \begin{bmatrix} 200 \\ 5000 \end{bmatrix} \end{array} \}

Step 5: Solve the System of Equations

To solve the system of equations, we need to find the values of $x$ and $y$ that satisfy both equations. We can do this by using the inverse of the coefficient matrix $A$ to multiply the constant matrix $b$ .

The Solution

To find the solution, we need to multiply the inverse of the coefficient matrix $A$ by the constant matrix $b$ :

{ \begin{array}{l} x = A^{-1}b \end{array} \}

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

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

where $\det(A)$ is the determinant of the coefficient matrix $A$ .

Calculating the Determinant

The determinant of the coefficient matrix $A$ is:

{ \begin{array}{l} \det(A) = 700(1) - 1(-75) = 775 \end{array} \}

Calculating the Inverse

The inverse of the coefficient matrix $A$ is:

{ \begin{array}{l} A^{-1} = \frac{1}{775} \begin{bmatrix} 1 & -1 \\ 75 & 700 \end{bmatrix} \end{array} \}

Multiplying the Inverse by the Constant Matrix

Now, we can multiply the inverse of the coefficient matrix $A$ by the constant matrix $b$ :

{ \begin{array}{l} x = A^{-1}b = \frac{1}{775} \begin{bmatrix} 1 & -1 \\ 75 & 700 \end{bmatrix} \begin{bmatrix} 200 \\ 5000 \end{bmatrix} \end{array} \}

Simplifying the Expression

Simplifying the expression, we get:

{ \begin{array}{l} x = \begin{bmatrix} \frac{200}{775} - \frac{5000}{775} \\ \frac{15000}{775} + \frac{140000}{775} \end{bmatrix} \end{array} \}

Simplifying Further

Simplifying further, we get:

{ \begin{array}{l} x = \begin{bmatrix} -\frac{4800}{775} \\ \frac{160000}{775} \end{bmatrix} \end{array} \}

Conclusion

In this article, we discussed how to complete a matrix of a system of linear equations and provided a step-by-step solution. We rewrote the given equations in the form of $Ax = b$ , created the coefficient matrix $A$ , variable matrix $x$ , and constant matrix $b$ . We then solved the system of equations by multiplying the inverse of the coefficient matrix $A$ by the constant matrix $b$ . The solution is:

{ \begin{array}{l} x = \begin{bmatrix} -\frac{4800}{775} \\ \frac{160000}{775} \end{bmatrix} \end{array} \}

Q: What is a system of linear equations?

A: A system of linear equations is a set of two or more equations that involve variables raised to the power of 1. These equations can be represented in a matrix form, which makes it easier to solve them.

Q: What is a matrix?

A: A matrix is a rectangular array of numbers, symbols, or expressions, arranged in rows and columns. In the context of solving a system of linear equations, a matrix is used to represent the coefficients of the variables and the constants on the right-hand side of the equations.

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 $Ax = b$ , where $A$ is the coefficient matrix, $x$ is the variable matrix, and $b$ is the constant matrix.

Q: What is the coefficient matrix?

A: The coefficient matrix is a 2x2 matrix that contains the coefficients of the variables $x$ and $y$ . In the example given in the article, the coefficient matrix is:

{ \begin{array}{l} A = \begin{bmatrix} 700 & 1 \\ -75 & 1 \end{bmatrix} \end{array} \}

Q: What is the variable matrix?

A: The variable matrix is a 2x1 matrix that contains the variables $x$ and $y$ . In the example given in the article, the variable matrix is:

{ \begin{array}{l} x = \begin{bmatrix} x \\ y \end{bmatrix} \end{array} \}

Q: What is the constant matrix?

A: The constant matrix is a 2x1 matrix that contains the constants on the right-hand side of the equations. In the example given in the article, the constant matrix is:

{ \begin{array}{l} b = \begin{bmatrix} 200 \\ 5000 \end{bmatrix} \end{array} \}

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 find the inverse of the coefficient matrix $A$ and multiply it by the constant matrix $b$ . The solution is then given by:

{ \begin{array}{l} x = A^{-1}b \end{array} \}

Q: What is the inverse of a matrix?

A: The inverse of a matrix is a matrix that, when multiplied by the original matrix, gives the identity matrix. In the context of solving a system of linear equations, the inverse of the coefficient matrix $A$ is used to multiply the constant matrix $b$ and find the solution.

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

A: To calculate the inverse of a matrix, you need to use the following formula:

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

where $\det(A)$ is the determinant 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 invertibility of the matrix. In the context of solving a system of linear equations, the determinant of the coefficient matrix $A$ is used to calculate the inverse of the matrix.

Q: How do I calculate the determinant of a matrix?

A: To calculate the determinant of a matrix, you need to use the following formula:

{ \begin{array}{l} \det(A) = a_{11}a_{22} - a_{12}a_{21} \end{array} \}

where $a_{11}$ , $a_{12}$ , $a_{21}$ , and $a_{22}$ are the elements of the matrix $A$ .

Q: What is the solution to a system of linear equations?

A: The solution to a system of linear equations is the set of values of the variables that satisfy all the equations in the system. In the context of solving a system of linear equations using a matrix, the solution is given by:

{ \begin{array}{l} x = A^{-1}b \end{array} \}

This solution represents the values of the variables that satisfy all the equations in the system.