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Given that the augmented matrix in row-reduced form is equivalent to the augmented matrix of a system of linear equations, (a) determine whether the system has a solution and (b) find the solution or solutions to the system, if they exist. \(\left[\begin{array}{rrrr|r}1 & 0 & 0 & 0 & 2 \\ 0 & 1 & 0 & 0 & -1 \\ 0 & 0 & 1 & 1 & 2 \\ 0 & 0 & 0 & 0 & 0\end{array}\right]\)

Short Answer

Expert verified
The given system of linear equations has infinitely many solutions, represented as: \[x1 = 2, x2 = -1, x3 = 2 - x4, x4 ∈ R\]

Step by step solution

01

Analyze the given matrix for inconsistencies

To determine whether the given system of linear equations has a solution, we need to check for any inconsistencies within the row-reduced matrix. If there is a row of the form [0 0 0 .. 0 | k], where k is a non-zero constant, then the system is inconsistent and has no solution. Analyzing the given row-reduced matrix: \[\left[\begin{array}{rrrr|r}1 & 0 & 0 & 0 & 2 \\\ 0 & 1 & 0 & 0 & -1 \\\ 0 & 0 & 1 & 1 & 2 \\\ 0 & 0 & 0 & 0 & 0\end{array}\right]\] We can see there is no inconsistent row present.
02

Determine whether the system has a solution

We have determined that there is no inconsistency within the given row-reduced matrix. This means that the system is consistent, which implies that it has a solution or infinitely many solutions. In our case, there are 4 variables and 3 non-zero rows, so the system will have infinitely many solutions.
03

Find the solution (free variables and basic variables)

To find the solution for the given system of equations, we need to identify the free variables and the basic variables. In our row-reduced matrix, columns 1, 2, and 3 correspond to the basic variables, and column 4 corresponds to the free variable. Basic variables are x1, x2, and x3, while the free variable is x4.
04

Express the basic variables in terms of free variable

Now, express the basic variables x1, x2, and x3 in terms of the free variable x4. The given matrix corresponds to the following system of linear equations: \(x1 = 2\), \(x2 = -1\), \(x3 + x4 = 2\). From those equations, we have: \(x1 = 2\), \(x2 = -1\), \(x3 = 2 - x4\). In conclusion, the solution to the given system of linear equations is: \[x1 = 2, x2 = -1, x3 = 2 - x4, x4 ∈ R\]

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Key Concepts

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

Row-Reduced Matrices
The row-reduced form of a matrix, also known as row-reduced echelon form (RREF), is crucial in solving systems of linear equations. It simplifies the matrix associated with the system so that we can easily read off the solutions or determine characteristics of the system, such as consistency and the number of solutions.

For a matrix to be in RREF, it must satisfy the following conditions:
  • Any zero rows are at the bottom of the matrix.
  • The leading entry (first non-zero number from the left) in a non-zero row is 1, known as a pivot.
  • This pivot is to the right of the pivot in the row above it.
  • All entries directly above and below a pivot are zeros.
  • If a column contains a pivot, every other entry in that column is zero.
Understanding row-reduced matrices enables us to quickly move forward and determine whether a system has one solution, no solution, or infinitely many solutions.
Inconsistent Systems
An inconsistent system of equations is one that has no solution. This occurs when the equations represent parallel lines that never intersect, which, in matrix terms, is indicated by a row of all zeroes except for the last entry in an augmented matrix.

During the process of row reduction, if we encounter a row that has the form \(\left[\begin{array}{ccc|c}0 & \cdots & 0 & k\end{array}\right]\) where \(k\) is not zero, this row tells us that the system is inconsistent because it suggests an impossible solution, such as \(0x_1 + 0x_2 + \ldots = k\) where \(k\) is non-zero. It's like saying 'nothing equals something,' which is clearly a contradiction.
Infinitely Many Solutions
A system of equations has infinitely many solutions when there are more unknowns than there are independent equations. In these cases, some variables do not have a unique value and are called free variables.

A telltale sign of this in a row-reduced matrix is the presence of at least one non-pivot column (a column without a leading one), indicating that the corresponding variable can take on multiple values. When you have free variables, you can express the basic (pivot) variables in terms of the free variables, leading to an infinite set of solutions or a solution space, typically described with parameterized variables.
Free Variables and Basic Variables
In the context of solving linear systems, basic variables correspond to columns with pivots in the row-reduced matrix; these variables are dependent on the other variables and can be directly solved for. Free variables are associated with non-pivot columns and represent the set of variables that can be assigned arbitrary values; their values are 'free' to be chosen.

In the step-by-step solution of our exercise, \(x1, x2, \text{ and } x3\) are the basic variables, determined by the columns with pivots in the row-reduced matrix. Variable \(x4\) is a free variable, as its column in the matrix lacks a pivot. This means that \(x4\) can take any real value, and we can express \(x3\) in terms of \(x4\) to give us a family of solutions. The system has infinitely many solutions because there's an entire dimension of values that \(x4\) can take, each leading to a different set of values for the basic variables.

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Most popular questions from this chapter

Find the inverse of the matrix, if it exists. Verify your answer. \(\left[\begin{array}{rrr}1 & 2 & 0 \\ -3 & 4 & -2 \\ -5 & 0 & -2\end{array}\right]\)

Compute the indicated products. \(4\left[\begin{array}{rrr}1 & -2 & 0 \\ 2 & -1 & 1 \\ 3 & 0 & -1\end{array}\right]\left[\begin{array}{rrr}1 & 3 & 1 \\ 1 & 4 & 0 \\ 0 & 1 & -2\end{array}\right]\)

Compute the indicated products. \(\left[\begin{array}{rr}1 & 3 \\ -1 & 2\end{array}\right]\left[\begin{array}{lll}1 & 3 & 0 \\ 3 & 0 & 2\end{array}\right]\)

Jackson Farms has allotted a certain amount of land for cultivating soybeans, corn, and wheat. Cultivating 1 acre of soybeans requires 2 labor-hours, and cultivating 1 acre of corn or wheat requires 6 labor-hours. The cost of seeds for 1 acre of soybeans is \(\$ 12\), for 1 acre of corn is \(\$ 20\), and for 1 acre of wheat is \(\$ 8\). If all resources are to be used, how many acres of each crop should be cultivated if the following hold? a. 1000 acres of land are allotted, 4400 labor-hours are available, and \(\$ 13,200\) is available for seeds. b. 1200 acres of land are allotted, 5200 labor-hours are available, and \(\$ 16,400\) is available for seeds.

(a) write each system of equations as a matrix equation and (b) solve the system of equations by using the inverse of the coefficient matrix. \(\begin{aligned} 2 x_{1}+x_{2}+x_{3}=b_{1} \\ x_{1}-3 x_{2}+4 x_{3}=b_{2} \\\\-x_{1}+x_{3}=b_{3} \\ \text { where } & \text { (i) } b_{1}=1, b_{2}=4, b_{3}=-3 \\ \text { and } & \text { (ii) } b_{1}=2, b_{2}=-5, b_{3}=0 \end{aligned}\)

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