/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 40 Use matrices to solve each syste... [FREE SOLUTION] | 91Ó°ÊÓ

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Use matrices to solve each system of equations. If the equations of a system are dependent or if a system is inconsistent, state this. $$ \left\\{\begin{array}{l} 2 x-y=0 \\ x+y=3 \end{array}\right. $$

Short Answer

Expert verified
The system has a unique solution: \( x = 1 \), \( y = 2 \).

Step by step solution

01

Write the System in Matrix Form

To solve the system of equations using matrices, we first write it in matrix form. The system is: \( \begin{align*} 2x - y &= 0 \ x + y &= 3 \end{align*} \). We can express this as a matrix equation \( A\mathbf{x} = \mathbf{b} \), where \( A \) is the coefficient matrix, \( \mathbf{x} \) is the column matrix of the variables, and \( \mathbf{b} \) is the column matrix of the constants. Thus, we have: \[ A = \begin{bmatrix} 2 & -1 \ 1 & 1 \end{bmatrix}, \quad \mathbf{x} = \begin{bmatrix} x \ y \end{bmatrix}, \quad \mathbf{b} = \begin{bmatrix} 0 \ 3 \end{bmatrix} \].
02

Calculate the Determinant of the Coefficient Matrix

Next, calculate the determinant of matrix \( A \) to determine if the system has a unique solution. The determinant of \( A = \begin{bmatrix} 2 & -1 \ 1 & 1 \end{bmatrix} \) is given by \( \det(A) = (2)(1) - (-1)(1) = 2 + 1 = 3 \). Since the determinant is non-zero, the system has a unique solution.
03

Find the Inverse of the Coefficient Matrix

Since the determinant is non-zero, matrix \( A \) is invertible. The inverse \( A^{-1} \) is calculated using the formula \[ A^{-1} = \frac{1}{\det(A)} \begin{bmatrix} d & -b \ -c & a \end{bmatrix} \]. Substituting the values from \( A \), we get \( A^{-1} = \frac{1}{3} \begin{bmatrix} 1 & 1 \ -1 & 2 \end{bmatrix} \).
04

Calculate the Solution Using the Inverse Matrix

Now, solve for \( \mathbf{x} \) using the formula \( \mathbf{x} = A^{-1}\mathbf{b} \). This gives \[ \begin{bmatrix} x \ y \end{bmatrix} = \frac{1}{3} \begin{bmatrix} 1 & 1 \ -1 & 2 \end{bmatrix} \begin{bmatrix} 0 \ 3 \end{bmatrix} = \frac{1}{3} \begin{bmatrix} 3 \ 6 \end{bmatrix} = \begin{bmatrix} 1 \ 2 \end{bmatrix} \]. Therefore, \( x = 1 \) and \( y = 2 \).
05

Verify the Solution

To confirm the solution is correct, substitute \( x = 1 \) and \( y = 2 \) back into the original equations. For the first equation, \( 2(1) - 2 = 0 \), which is true. For the second equation, \( 1 + 2 = 3 \), which is also true. Hence, \( (x, y) = (1, 2) \) is the correct solution.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

System of Equations
A system of equations is a collection of two or more equations with the same set of unknowns. The main goal when solving a system of equations is to find the values of the unknown variables that satisfy all equations simultaneously. For example, consider the equations:
- \( 2x - y = 0 \) - \( x + y = 3 \) Here we have two equations with two variables, \( x \) and \( y \). To solve these, we can use various methods, such as substitution, elimination, or matrices. Matrices offer a structured way of finding solutions, especially for larger systems.In matrix form, these equations become:
- Coefficient matrix, \( A = \begin{bmatrix} 2 & -1 \ 1 & 1 \end{bmatrix} \)- Variable matrix, \( \mathbf{x} = \begin{bmatrix} x \ y \end{bmatrix} \)- Constant matrix, \( \mathbf{b} = \begin{bmatrix} 0 \ 3 \end{bmatrix} \)This provides a neat alternative to handling systems, capturing all information in a crisp, mathematical framework.
Inverse Matrix
An inverse matrix is like a secret key in matrix algebra. It allows us to reverse the effect of a matrix. If you multiply a matrix by its inverse, you get the identity matrix, much like multiplying a number by its reciprocal gives you one. In terms of solving systems of equations, if we have a system written as \( A\mathbf{x} = \mathbf{b} \), we can find \( \mathbf{x} \) using the inverse matrix \( A^{-1} \). This is done by multiplying both sides by \( A^{-1} \):
\[ \mathbf{x} = A^{-1}\mathbf{b} \]However, not all matrices have an inverse. A matrix must be square (same number of rows and columns) and have a non-zero determinant to be invertible. For our coefficient matrix \( A = \begin{bmatrix} 2 & -1 \ 1 & 1 \end{bmatrix} \), the determinant is non-zero, hence \( A \) has an inverse. Calculating the inverse involves swapping the diagonal elements, changing the signs of the off-diagonal elements, and dividing by the determinant.Knowing how to find an inverse is essential for solving systems efficiently, particularly in higher dimensions.
Determinant of a Matrix
The determinant of a matrix is a special number that can tell us a lot about the matrix. For square matrices, it is a scalar value that can show us if a matrix is invertible and if a system of equations has a unique solution.For a 2x2 matrix \( A = \begin{bmatrix} a & b \ c & d \end{bmatrix} \), the determinant is calculated as:
\[ \det(A) = ad - bc \]In the given exercise, the coefficient matrix \( A = \begin{bmatrix} 2 & -1 \ 1 & 1 \end{bmatrix} \) has a determinant of \( 3 \). This non-zero determinant is what tells us the matrix is invertible, meaning there's a unique solution to the system of equations.The determinant is crucial in determining if there are any solutions:
  • If the determinant is zero, there might be no solution or infinitely many solutions.
  • A non-zero determinant indicates a single, unique solution.
Knowing and calculating determinants is vital for matrix algebra and solving linear systems.

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