Problem 12
Question
In Exercises \(9-12\), the reduced row echelon form of the augmented matrix of a system of equations is given. Find the solutions of the system. $$\left(\begin{array}{cccc} 1 & 0 & 0 & -2 \\ 0 & 1 & 0 & 3 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right)$$
Step-by-Step Solution
Verified Answer
Answer: The solution is $x_1 = -2$, $x_2 = 3$, and $x_3 = 0$.
1Step 1: Identify the variables and write the system of linear equations
From the reduced row echelon form of the matrix, we can see that there are 3 variables and 3 non-zero rows. Based on this information, we can write down the corresponding system of linear equations:
$$\begin{cases}
x_1 = -2 \\
x_2 = 3 \\
x_3 = 0
\end{cases}
$$
2Step 2: Read the solution
Each equation corresponds to one of the variables. The solution of the system of linear equations is:
$$
\begin{cases}
x_1 = -2 \\
x_2 = 3 \\
x_3 = 0
\end{cases}
$$
Key Concepts
System of Linear EquationsAugmented MatrixReduced Row Echelon Form
System of Linear Equations
A system of linear equations is a collection of one or more linear equations involving the same set of variables. For example, two equations that have variables such as x and y are part of a system if they must be solved together. In real-world situations, these systems can represent various scenarios, from simple budgeting problems to complex network optimization.
In the given exercise, the system of linear equations has been derived from a reduced row echelon form matrix. This matrix representation is beneficial as it allows students to easily identify solutions without having to perform any further algebraic manipulations. One way to improve understanding for students would be to relate the matrix to a visual representation, such as a graph or a diagram, where each equation can be seen as a line, and the intersection points represent the solutions to the system.
In the given exercise, the system of linear equations has been derived from a reduced row echelon form matrix. This matrix representation is beneficial as it allows students to easily identify solutions without having to perform any further algebraic manipulations. One way to improve understanding for students would be to relate the matrix to a visual representation, such as a graph or a diagram, where each equation can be seen as a line, and the intersection points represent the solutions to the system.
Augmented Matrix
An augmented matrix is a powerful tool in linear algebra that combines the coefficients of variables in a system of linear equations with the constants on the right side of each equation, creating a single matrix. It essentially represents the entire system in a compact form. To interpret an augmented matrix:
- Look at the columns: each represents the coefficients of a particular variable across all equations.
- The last column represents the constants from the right-hand side of the equations.
- Row operations can be applied to simplify this matrix without changing the solutions of the system.
Reduced Row Echelon Form
Reduced row echelon form (RREF) is the final, simplified version of an augmented matrix that is achieved through a series of row operations like row addition, row multiplication, and row switching. The distinguishing features of an RREF matrix include:
- Leading entries in each row are '1' (also termed as pivot).
- Each leading '1' is the only non-zero number in its column.
- The rows with all zero elements are at the bottom of the matrix.
- The leading '1' moves to the right as you move down the rows.
Other exercises in this chapter
Problem 11
Use substitution to solve the system. $$\begin{aligned}&r+s=0\\\&r-s=5\end{aligned}$$
View solution Problem 11
Find AB. $$A=\left(\begin{array}{rrr} 2 & 0 & -1 \\ 1 & 1 & 2 \\ 0 & 2 & -3 \\ 2 & 3 & 0 \end{array}\right), \quad B=\left(\begin{array}{llll} 1 & 0 & 1 & 1 \\
View solution Problem 12
Use substitution to solve the system. $$\begin{aligned}&t=7 u-4\\\&t=5 u+6\end{aligned}$$
View solution Problem 13
In Exercises \(13-16,\) use Gaussian elimination to solve the system. $$\begin{aligned} -x+3 y+2 z &=0 \\ 2 x-y-z &=3 \\ x+2 y+3 z &=0 \end{aligned}$$
View solution