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Chapter 12: Problem 26

Use matrices to solve each system of equations. \(\left\\{\begin{array}{l}x+y=3 \\ x-y=-1\end{array}\right.\)

Short Answer

Expert verified
The solution to the system of equations is \( x = 1 \) and \( y = 2 \).

Step by step solution

01

Convert the System into Matrix Form

First, we need to express the given system of equations in matrix form. The system of equations provided is: \( x + y = 3 \) and \( x - y = -1 \). We can write this system in matrix form as \( A \mathbf{x} = \mathbf{b} \), where\[ A = \begin{bmatrix} 1 & 1 \ 1 & -1 \end{bmatrix}, \quad \mathbf{x} = \begin{bmatrix} x \ y \end{bmatrix}, \quad \mathbf{b} = \begin{bmatrix} 3 \ -1 \end{bmatrix} \].
02

Find the Inverse of Matrix A

To solve for \( \mathbf{x} \), we first need to find the inverse of matrix \( A \). For a 2x2 matrix \( A = \begin{bmatrix} a & b \ c & d \end{bmatrix} \), the inverse is given by \( A^{-1} = \frac{1}{ad-bc} \begin{bmatrix} d & -b \ -c & a \end{bmatrix} \).Here, \( a = 1, b = 1, c = 1, \) and \( d = -1 \). So, the determinant \( ad-bc = 1(-1) - 1(1) = -1 - 1 = -2 \). Therefore, the inverse is:\[ A^{-1} = \frac{1}{-2} \begin{bmatrix} -1 & -1 \ -1 & 1 \end{bmatrix} = \begin{bmatrix} \frac{1}{2} & \frac{1}{2} \ \frac{1}{2} & -\frac{1}{2} \end{bmatrix} \]
03

Solve for \( \mathbf{x} \) by Multiplying \( A^{-1} \times \mathbf{b} \)

Now, we find \( \mathbf{x} = A^{-1} \mathbf{b} \) by multiplying the inverse of \( A \) with \( \mathbf{b} \).\[ \mathbf{x} = \begin{bmatrix} \frac{1}{2} & \frac{1}{2} \ \frac{1}{2} & -\frac{1}{2} \end{bmatrix} \begin{bmatrix} 3 \ -1 \end{bmatrix} = \begin{bmatrix} \frac{1}{2}(3) + \frac{1}{2}(-1) \ \frac{1}{2}(3) - \frac{1}{2}(-1) \end{bmatrix} = \begin{bmatrix} \frac{3}{2} - \frac{1}{2} \ \frac{3}{2} + \frac{1}{2} \end{bmatrix} = \begin{bmatrix} 1 \ 2 \end{bmatrix} \].
04

Interpret the Solution

The solution matrix \( \mathbf{x} = \begin{bmatrix} 1 \ 2 \end{bmatrix} \) gives us the values of \( x \) and \( y \). Thus, \( x = 1 \) and \( y = 2 \). These values solve the given system of equations.

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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 a set of variables that you want to find solutions for. In our given problem, we have a system with two equations and two variables:
  • Equation 1: \( x + y = 3 \)
  • Equation 2: \( x - y = -1 \)
The objective here is to determine the values of \( x \) and \( y \) that satisfy both equations at the same time.
Systems of equations can have different types of solutions:
  • A single solution.
  • No solution, which happens when the system is inconsistent, meaning the lines are parallel.
  • Infinite solutions, when equations represent the same line.
To solve a system of linear equations efficiently, we often convert the equations into matrix form, which simplifies the process of finding the solution using matrix methods like matrix inversion. This provides a clearer computational path compared to traditional substitution or elimination methods.
Matrix Inversion
Matrix inversion is a powerful tool for solving systems of equations. To find the solution for the variables in a system like the one provided, we can convert the system into a matrix equation, \( A \mathbf{x} = \mathbf{b} \). Here, \( A \) is the coefficients matrix, \( \mathbf{x} \) is the variables vector, and \( \mathbf{b} \) is the constants vector.In terms of matrices, an inverse matrix \( A^{-1} \) is like a reciprocal of a number:
  • If \( A \cdot A^{-1} = I \), where \( I \) is the identity matrix (similar to 1 for numbers), then \( A^{-1} \) is the inverse of \( A \).
  • The inverse is useful because multiplying both sides of the matrix equation \( A \mathbf{x} = \mathbf{b} \) by \( A^{-1} \) gives\( \mathbf{x} = A^{-1} \mathbf{b} \).
However, not all matrices have an inverse. A matrix can be inverted only if its determinant is non-zero. If the determinant is zero, the matrix is singular and does not have an inverse, making direct matrix inversion impossible.
Determinant Calculation
The determinant is a special number that can be calculated from a matrix. It offers vital information about the matrix's properties, particularly whether or not a matrix has an inverse. For a 2x2 matrix \( A = \begin{bmatrix} a & b \ c & d \end{bmatrix} \), the determinant is calculated as:
  • \( \,a \times d - b \times c\)

In our problem, with the matrix \( A = \begin{bmatrix} 1 & 1 \ 1 & -1 \end{bmatrix} \), the determinant calculation is as follows:
  • \( (1) \times (-1) - (1) \times (1) = -1 - 1 = -2 \)
  • Thus, \( A\) is invertible, as its determinant is not zero.
Understanding the determinant helps us to know if a matrix can be inverted, which is critical when using the matrix inversion method to solve systems of equations.

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Most popular questions from this chapter

Solve each system by graphing, if possible. If a system is inconsistent or if the equations are dependent, state this. (Hint: Several coordinates of points of intersection are fractions.) $$ \left\\{\begin{array}{l} x=\frac{1-3 y}{4} \\ y=\frac{12+3 x}{2} \end{array}\right. $$

Solve each system by elimination (addition). $$ \left\\{\begin{array}{l} \frac{1}{3} x+\frac{1}{2} y=\frac{31}{6} \\ 0.3 x+0.2 y=3.9 \end{array}\right. $$

Solve each system, if possible. If a system is inconsistent or if the equations are dependent, state this. $$ \left\\{\begin{array}{l} 3 x-y=5 \\ 21 x=7(y+5) \end{array}\right. $$

Solve each system by any method. If a system is inconsistent or if the equations are dependent, so indicate. $$ \left\\{\begin{array}{l} \frac{3}{2} p+\frac{1}{3} q=2 \\ \frac{2}{3} p+\frac{1}{9} q=1 \end{array}\right. $$

Use Cramer's rule to solve system of equations. If a system is inconsistent or if the equations are dependent, so indicate. \(\left\\{\begin{array}{l}2 x+5 y-13=0 \\ -2 x+13=5 y\end{array}\right.\)
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