Problem 42
Question
In Problems 35-42, each matrix is the reduced echelon matrix for a system with variables \(x_{1}, x_{2}, x_{3}\), and \(x_{4}\). Find the solution set of each system. \(\left[\begin{array}{rrrr:r}1 & 0 & 0 & 0 & 7 \\ 0 & 1 & 0 & 0 & -3 \\ 0 & 0 & 1 & -2 & 5 \\ 0 & 0 & 0 & 0 & 0\end{array}\right]\)
Step-by-Step Solution
Verified Answer
\(x_1 = 7\), \(x_2 = -3\), \(x_3 = 5 + 2x_4\); \(x_4\) is free (any real number).
1Step 1: Understand the Matrix Structure
The given matrix is in reduced echelon form, which provides information about the solution to the system of equations corresponding to the variables \(x_1, x_2, x_3,\) and \(x_4\). Each row represents an equation.
2Step 2: Translate Rows into Equations
The first row translates to the equation \(x_1 = 7\), the second row translates to \(x_2 = -3\), and the third row translates to \(x_3 - 2x_4 = 5\). The fourth row is all zeros and doesn't contribute to any equation.
3Step 3: Solve for Leading Variables
From the first two rows, we directly obtain \(x_1 = 7\) and \(x_2 = -3\). These are the specific solutions for those variables.
4Step 4: Parameterize the Free Variable
The third row \(x_3 - 2x_4 = 5\) can be rewritten as \(x_3 = 5 + 2x_4\). Here, \(x_4\) can be any real number, making it a free variable in the solution set.
5Step 5: Express the Solution Set
The solution set consists of specific values for \(x_1\) and \(x_2\), and an expression for \(x_3\) in terms of \(x_4\). The solution is: \( (x_1, x_2, x_3, x_4) = (7, -3, 5 + 2x_4, x_4) \) where \(x_4\) is any real number.
Key Concepts
Reduced Echelon FormSystem of EquationsFree VariableSolution Set
Reduced Echelon Form
When solving systems of equations, one common representation is the matrix. A special form of this matrix is the reduced echelon form. This form is particularly handy because it makes finding solutions much easier. The reduced echelon form of a matrix has several distinctive features:
- The leading entry in each non-zero row is 1.
- Each leading 1 is the only non-zero entry in its column.
- The rows consisting entirely of zeros are at the bottom.
- The leading 1 of a non-zero row is always to the right of the leading 1 of the previous row.
System of Equations
The core idea behind matrices in the context of linear algebra is to represent systems of equations. Each row of a matrix can be thought of as an equation in a system, with each column corresponding to a variable in those equations. For our example:
- The first row suggests the equation: \(x_1 = 7\).
- The second row corresponds to: \(x_2 = -3\).
- The third row gives us: \(x_3 - 2x_4 = 5\).
- The fourth row, being entirely zeroes, doesn't contribute new information.
Free Variable
In the context of a system of equations, a free variable is one that does not have a leading 1 in any row of the reduced echelon form matrix. This means it can take on any value.For our matrix, the variable \(x_4\) is a free variable because it does not have a unique equation that defines it. Instead, it is used to express other variables, such as in the equation \(x_3 = 5 + 2x_4\). By selecting different values for \(x_4\), we can determine a range of solutions for \(x_3\), which highlights the flexibility provided by free variables.
Solution Set
The solution set is the collection of all possible solutions to a system of equations. It describes all the ways the variables can be values that satisfy all equations in the system simultaneously.In our example, we have determined:
- \(x_1 = 7\), a fixed solution.
- \(x_2 = -3\), also fixed.
- \(x_3 = 5 + 2x_4\), which varies based on \(x_4\).
- \(x_4\) is not fixed, but a free variable that can be any real number.
Other exercises in this chapter
Problem 41
In Problems 35-42, each matrix is the reduced echelon matrix for a system with variables \(x_{1}, x_{2}, x_{3}\), and \(x_{4}\). Find the solution set of each s
View solution Problem 41
Solve each system by using the substitution method. \(\left(\begin{array}{l}\frac{x}{2}-\frac{2 y}{5}=\frac{-23}{60} \\ \frac{2 x}{3}+\frac{y}{4}=\frac{-1}{4}\e
View solution Problem 42
Solve each system by using the substitution method. \(\left(\begin{array}{c}\frac{2 x}{3}-\frac{y}{2}=\frac{3}{5} \\\ \frac{x}{4}+\frac{y}{2}=\frac{7}{80}\end{a
View solution Problem 43
Explain the difference between a matrix and a determinant.
View solution