/*! 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 29 Solving Systems of Equations Usi... [FREE SOLUTION] | 91Ó°ÊÓ

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Solving Systems of Equations Using Matrices. $$\left\\{\begin{array}{l}x+2 y=1 \\\x+5 y=-2\end{array}\right.$$

Short Answer

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

Step by step solution

01

Express the System as a Matrix Equation

We can express the system of equations as a matrix equation as follows: \[ \begin{bmatrix} 1 & 2 \ 1 & 5 \end{bmatrix} \begin{bmatrix} x \ y \end{bmatrix} = \begin{bmatrix} 1 \ -2 \end{bmatrix} \]
02

Find the Inverse of the Coefficient Matrix

We now find the inverse of the coefficient matrix. So, the matrix \(A = \begin{bmatrix} 1 & 2 \ 1 & 5 \end{bmatrix}\). The inverse of a 2x2 matrix \[A = \begin{bmatrix} a & b \ c & d \end{bmatrix}\] is given by the formula: \(A^{-1} = \frac{1}{ad-bc} \begin{bmatrix} d & -b \ -c & a \end{bmatrix}\). By substituting the values from our matrix we get \(A^{-1} = \frac{1}{1*5 - 2*1} \begin{bmatrix} 5 & -2 \ -1 & 1 \end{bmatrix} = \begin{bmatrix} 5/3 & -2/3 \ -1/3 & 1/3 \end{bmatrix}\).
03

Multiply the Inverse Matrix with the Coefficient Matrix

For the final step, we multiply the inverse of the matrix with the matrix of constants to find the solution. \[ \begin{bmatrix} 5/3 & -2/3 \ -1/3 & 1/3 \end{bmatrix} \begin{bmatrix} 1 \ -2 \end{bmatrix} = \begin{bmatrix} 1 \ -1 \end{bmatrix}\]. So, the solution is \(x = 1\) and \(y = -1\).

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Key Concepts

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

Matrix Inversion
Matrix inversion involves converting a given square matrix into another matrix such that when the two are multiplied, the result is the identity matrix. Imagine the identity matrix like the number 1 in regular multiplication, where any number multiplied by 1 remains unchanged. In matrices, multiplying a matrix by its inverse gives the identity matrix, much like multiplying a number by its reciprocal gives 1.
To invert a 2x2 matrix, we use the formula provided in the exercise. This formula involves switching certain elements and changing the signs of others, then dividing by the determinant of the original matrix. The determinant is a special value that helps in understanding the properties of the matrix. Inverting matrices is crucial when solving systems of equations as it allows us to find solutions via matrix multiplication.
Coefficient Matrix
A coefficient matrix is derived from the coefficients of the variables in a system of linear equations. For our specific exercise, this matrix is \[ \begin{bmatrix} 1 & 2 \ 1 & 5 \end{bmatrix} \] and represents the numerical values multiplying the variables in both equations.
By arranging these coefficients in a matrix, we can more easily handle the equations using linear algebra techniques. This matrix becomes the centerpiece of the equation, from which further operations like finding the inverse are performed. Understanding the role of the coefficient matrix is vital because it sets the foundation for manipulating the system as a whole using matrices.
Matrix Equation
A matrix equation is a compact and organized way of representing a system of linear equations. It is commonly expressed in the form \(AX = B\), where \(A\) is the coefficient matrix, \(X\) is the matrix for variables, and \(B\) is the matrix for constants.
In our exercise, the matrix equation simplifies the system of equations into a form that is amenable to algebraic methods. Instead of managing multiple equations, we manage a single tidy matrix equation. This method highlights how algebraic operations can be expanded beyond numbers to handle broader and more complex systems.
Inverse Matrix
The inverse of a matrix serves a pivotal role in solving matrix equations. Just like inverses in elementary arithmetic cancel out numbers to reveal certain results, inverse matrices help "cancel out" the coefficient matrix in the matrix equation, thereby isolating the variables.
When the inverse matrix is multiplied by both sides of the matrix equation, it effectively removes the coefficient matrix from the equation. This operation enables us to directly solve for the variable matrix. It's important to note that not all matrices have inverses, and having an inverse is a condition for being able to solve certain types of matrix equations. The inverse is specific and unique to the matrix from which it was derived, reinforcing the tailored relationship between the given system and its solution.
Matrices in Algebra
Matrices in algebra provide a powerful framework for solving equations and performing operations that would be incredibly complex if done traditionally. They enable compact expressions and manipulations of systems that involve multiple variables and equations. In educational settings, matrices are crucial for simplifying and visualizing large problems.
Operations on matrices such as addition, multiplication, and inversion help deliver solutions to many algebraic problems. In our specific case, using matrices to solve a system of equations illustrates how they simplify and streamline the processes of algebra. Their use extends beyond theoretical exercises, finding applications in diverse fields such as physics, computer science, and economics. Understanding matrices' roles reveals the elegance and efficiency embedded within algebraic operations.

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Most popular questions from this chapter

A couple has \(\$ 10,000\) to invest for their child's wedding. Their accountant recommends placing at least \(\$ 6000\) in a high-yield investment and no more than \(\$ 4000\) in a low-yield investment. (a) Use \(x\) to denote the amount of money placed into the high-yield investment. Use \(y\) to denote the amount of money placed into the low-yield investment. Write a system of linear inequalities that describes the possible amounts the couple could invest in each type of venture. (b) Graph the region that represents all possible amounts the couple could put into each investment if they wish to follow the accountant's advice.

Apply elementary row operations to a matrix to solve the system of equations. If there is no solution, state that the system is inconsistent. \(\left\\{\begin{array}{l}x+4 z=-3 \\ x-5 y=0 \\ z+4 y=2\end{array}\right.\) (Hint: Be careful with the order of the variables.)

Involve positive-integer powers of a square matrix \(A . A^{2}\) is defined as the product \(A A ;\) for \(n \geq 3, A^{n}\) is defined as the product \(\left(A^{n-1}\right) A\) Let \(A=\left[\begin{array}{ll}4 & 1 \\ 3 & 1\end{array}\right] .\) Find the inverses of \(A^{2}\) and \(A^{3}\) without computing the matrices \(A^{2}\) and \(A^{3} .\)

A family owns and operates three businesses. On their income-tax return, they have to report the depreciation deductions for the three businesses separately. In \(2004,\) their depreciation deductions consisted of use of a car, plus depreciation on 5 -year equipment (on which onc-fifth of the original value is deductible per year) and 10-year equipment (on which one- tenth of the original valuc is deductible per year). The car use (in miles) for cach business in 2004 is given in the following table, along with the original value of the depreciable 5 - and 10 year equipment used in each business that year. $$\begin{array}{|c|c|c|c|}\hline & \begin{array}{c}\text { Car } \\\\\text { Use }\end{array} & \begin{array}{c}\text { Original } \\\\\text { Value, 5-Year }\end{array} & \begin{array}{c}\text { Original } \\\\\text { Value, 10-Year }\end{array} \\\\\text { Business } & \text { (miles) } & \text { Equipment (S) } & \text { Equipment (S) } \\\\\hline 1 & 3200 & 9850 & 435 \\\2 & 8800 & 12,730 & 980 \\\3 & 6880 & 2240 & 615\\\\\hline\end{array}$$ The depreciation deduction for car use in 2004 was 37.5 cents per mile. Use matrix multiplication to determine the total depreciation deduction for each business in 2004.

A golf club manufacturer makes a profit of \(\$ 3\) on a driver and a profit of \(\$ 2\) on a putter. To meet dealer demand, the company needs to produce between 20 and 50 drivers and between 30 and 50 putters each day. The maximum number of clubs produced each day by the company is 80. How many of each type of club should be produced to maximize profit?

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