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Solving Systems of Equations Using Matrices. $$\left\\{\begin{array}{rr}2 x+4 y= & 5 \\\\-4 x-8 y= & -10\end{array}\right.$$

Short Answer

Expert verified
The solutions to the system of equations are \(x = 2.5 - 2y\) for any real number \(y\).

Step by step solution

01

Represent System of Equations as a Coefficient Matrix

The given system of equations can be written as a coefficient matrix as follows: \( \left[ \begin{array}{ c c | c } 2 & 4 & 5 \\-4 & -8 & -10 \end{array} \right] \).
02

Simplifying Coefficient Matrix

The next step is reducing the matrix to row-echelon form. First, let's simplify the second row by multiplying every element of the row by -1. The matrix now reads \( \left[ \begin{array}{ c c | c } 2 & 4 & 5 \4 & 8 & 10 \end{array} \right] \). Further simplification reveals the two equations are linearly dependent, equivalent, actually, and that this is an underdetermined system. The equivalent system, in matrix form, is \( \left[ \begin{array}{ c c | c } 1 & 2 & 2.5 \0 & 0 & 0 \end{array} \right] \).
03

Solve the System of Equations

Upon deciphering the simplified matrix, it becomes apparent that y can be any real number. Conversely, x's value depends on y's and can be solved for as follows, x = 2.5 - 2y.
04

Verification

To verify that these solutions make both original equations true, plug x and y back into the original equations, one finds that they hold true, which affirms the validity of the solutions derived.

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Key Concepts

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

Coefficient Matrix
In the realm of linear algebra, translating a system of linear equations into a matrix can provide a clear pathway to finding the solution. The coefficient matrix is a rectangular array that encapsulates the coefficients of the variables in the system. For instance, for the equations

\[2x + 4y = 5\]
\[-4x - 8y = -10,\]

the coefficient matrix is

\[\begin{bmatrix}2 & 4 \-4 & -8d{bmatrix}\]

where the first column contains the coefficients of x and the second the coefficients of y. This compact representation is crucial as it forms the basis for various methods of solving systems of equations, including the use of row operations to transform the matrix into a more manageable form.
Row-Echelon Form
The row-echelon form (REF) of a matrix is achieved through a series of row operations that streamline the process of solving systems of equations. A matrix is in REF when all nonzero rows are above any rows of all zeros, each leading coefficient (also known as a pivot) of a nonzero row is to the right of the leading coefficient of the row above it, and all entries in a column below a leading coefficient are zeros.

Transforming a coefficient matrix into REF makes it straightforward to use back substitution to find the solutions to the system. In our exercise, REF was achieved, revealing an essential characteristic of the system that we will touch upon in the next section.
Linearly Dependent Equations
In a system of equations, when one equation can be derived from another by multiplication or addition, the equations are said to be linearly dependent. This dependency implies that they do not provide unique information about the system's solution. In our exercise example, when the second row is multiplied by -1, we discover that both equations in the system are essentially the same, making them linearly dependent.

In terms of the matrix representation, this results in a row of zeros after we simplify, which is an indicator of dependency. The presence of linearly dependent equations in a system has significant implications for the number of solutions the system might have, as we'll explore in the final concept.
Underdetermined System
An underdetermined system of equations is one where there are more variables than independent equations, leading to an infinite number of possible solutions. After simplification and realisation of linear dependency in our example, we are left with what is essentially one equation with two variables, creating an underdetermined system.

This is evident in the matrix row of zeros, which indicates that the second equation does not limit the possible values of the variables. Consequently, one variable can be assigned any value, and the other variable can be determined in terms of this free variable. In the exercise, setting y to be a free variable allows x to be expressed as a function of y. Underdetermined systems are common in real-world scenarios where insufficient information is available to pinpoint a single solution.

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Most popular questions from this chapter

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.

Find the decoding matrix for each encoding matrix. $$\left[\begin{array}{ll}1 & -3 \\\1 & -2\end{array}\right]$$

If \(A=\left[\begin{array}{ll}0 & 1 \\ a & 0\end{array}\right]\) and \(B=\left[\begin{array}{ll}0 & a \\ 1 & 0\end{array}\right],\) for what value(s) of \(a \operatorname{does} A B=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right] ?\)

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 \(\left(A^{n-1}\right) A\) Find \(\left(A^{3}\right)^{-1}\) and \(\left(A^{-1}\right)^{3},\) where \(A=\left[\begin{array}{rr}-5 & -1 \\ 4 & 1\end{array}\right] .\) What do you observe?

The sum of the squares of two positive integers is \(85 .\) If the squares of the integers differ by 13 find the integers.

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