Question

The augmented matrix of a system of linear equations has been carried to the following by row operations. In each case solve the system. a. \(\left[\begin{array}{rrrrrr|r}1 & 2 & 0 & 3 & 1 & 0 & -1 \\ 0 & 0 & 1 & -1 & 1 & 0 & 2 \\ 0 & 0 & 0 & 0 & 0 & 1 & 3 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0\end{array}\right]\) b. \(\left[\begin{array}{rrrrrr|r}1 & -2 & 0 & 2 & 0 & 1 & 1 \\ 0 & 0 & 1 & 5 & 0 & -3 & -1 \\ 0 & 0 & 0 & 0 & 1 & 6 & 1 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0\end{array}\right]\) c. \(\left[\begin{array}{rrrrr|r}1 & 2 & 1 & 3 & 1 & 1 \\ 0 & 1 & -1 & 0 & 1 & 1 \\ 0 & 0 & 0 & 1 & -1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0\end{array}\right]\) d. \(\left[\begin{array}{rrrrr|r}1 & -1 & 2 & 4 & 6 & 2 \\ 0 & 1 & 2 & 1 & -1 & -1 \\ 0 & 0 & 0 & 1 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 & 0\end{array}\right]\)

Short answer

Solutions involve assigning parameters due to more variables than equations.

Step-by-step solution

Analyze Structure - Part a

The given matrix is\[\begin{array}{cccccc|c}1 & 2 & 0 & 3 & 1 & 0 & -1 \0 & 0 & 1 & -1 & 1 & 0 & 2 \0 & 0 & 0 & 0 & 0 & 1 & 3 \0 & 0 & 0 & 0 & 0 & 0 & 0\end{array}\]This matrix indicates a system of equations with more variables than equations, leading to a system with parameters. The third row gives that the 6th variable must equal 3.

Solution Identification - Part a

From the augmented row form:- From row 3: \( x_6 = 3 \)- From row 2: \( x_3 - x_4 + x_5 = 2 \)- From row 1: \( x_1 + 2x_2 + 3x_4 + x_5 = -1 \)Choose parameters:\( x_2 = s, x_4 = t, x_5 = u \).- Solve: \( x_3 = 2 + t - u \)- Solve: \( x_1 = -1 - 2s - 3t - u \)Thus, the solution is:\( x_1 = -1 - 2s - 3t - u \)\( x_2 = s \)\( x_3 = 2 + t - u \)\( x_4 = t \)\( x_5 = u \)\( x_6 = 3 \)

Analyze Structure - Part b

The given matrix is:\[\begin{array}{cccccc|c}1 & -2 & 0 & 2 & 0 & 1 & 1 \0 & 0 & 1 & 5 & 0 & -3 & -1 \0 & 0 & 0 & 0 & 1 & 6 & 1 \0 & 0 & 0 & 0 & 0 & 0 & 0\end{array}\]This indicates a system with three equations and six variables, meaning the system is underdetermined. The solution involves parameters.

Solution Identification - Part b

From the augmented row form:- From row 3: \( x_5 + 6x_6 = 1 \)- From row 2: \( x_3 + 5x_4 - 3x_6 = -1 \)- From row 1: \( x_1 - 2x_2 + 2x_4 + x_6 = 1 \)Choose parameters: \( x_2 = s, x_4 = t, x_6 = u \).- Solve: \( x_5 = 1 - 6u \)- Solve: \( x_3 = -1 - 5t + 3u \)- Solve: \( x_1 = 1 + 2s - 2t - u \)Thus, the solution is:\( x_1 = 1 + 2s - 2t - u \)\( x_2 = s \)\( x_3 = -1 - 5t + 3u \)\( x_4 = t \)\( x_5 = 1 - 6u \)\( x_6 = u \)

Analyze Structure - Part c

The given matrix is:\[\begin{array}{ccccc|c}1 & 2 & 1 & 3 & 1 & 1 \0 & 1 & -1 & 0 & 1 & 1 \0 & 0 & 0 & 1 & -1 & 0 \0 & 0 & 0 & 0 & 0 & 0\end{array}\]This system has 5 variables and three equations, leading to a solution with parameters.

Solution Identification - Part c

From the augmented row form:- From row 3: \( x_4 - x_5 = 0 \Rightarrow x_4 = x_5 \).- From row 2: \( x_2 - x_3 + x_5 = 1 \).- From row 1: \( x_1 + 2x_2 + x_3 + 3x_4 + x_5 = 1 \).Choose parameters: \( x_3 = s, x_5 = t \).- Solve: \( x_4 = t \)- Solve: \( x_2 = 1 + s - t \)- Solve: \( x_1 = 1 - 2(1+s-t) - s - 3t - t \)Thus, the solution is:\( x_1 = -1 - s - 4t \)\( x_2 = 1 + s - t \)\( x_3 = s \)\( x_4 = t \)\( x_5 = t \)

Analyze Structure - Part d

The given matrix is:\[\begin{array}{ccccc|c}1 & -1 & 2 & 4 & 6 & 2 \0 & 1 & 2 & 1 & -1 & -1 \0 & 0 & 0 & 1 & 0 & 1 \0 & 0 & 0 & 0 & 0 & 0\end{array}\]This system presents a situation where there are more variables than equations, involving undetermined components and parameters.

Solution Identification - Part d

From the augmented row form:- From row 3: \( x_4 = 1 \).- From row 2: \( x_2 + 2x_3 + x_4 - x_5 = -1 \).- From row 1: \( x_1 - x_2 + 2x_3 + 4 \times 1 + 6x_5 = 2 \).Choose parameters: \( x_3 = s, x_5 = t \).- Solve for knowns: \( x_4 = 1 \)- Solve: \( x_2 = -1 - 2s - 1 + t = -2s + t - 2 \)- Solve for \( x_1 \): \( 2 - 4 - x_2 - 2s - 6t \)Thus, the solution is:\( x_1 = -2s + 3t + 4 \)\( x_2 = -2s + t - 2 \)\( x_3 = s \)\( x_4 = 1 \)\( x_5 = t \)

Key concepts

Augmented Matrix

An augmented matrix is a powerful tool in linear algebra that combines the coefficients of a system of linear equations with the constants in the equations. This matrix acts as a compact representation of the system which simplifies the process of finding solutions using row operations.

To construct an augmented matrix, take each equation of the system and align them in rows, placing all the coefficients of each variable into the main part of the matrix. The constants are placed in an additional column, often referred to as the "augmenting" column.

This arrangement permits systematic operations to solve the system, which is especially useful for systems that may lead to solutions with parameters, such as those with infinitely many solutions or no solutions at all. Using row reduction techniques, we can convert the augmented matrix to a simpler form to evaluate possible solutions.

System of Linear Equations

A system of linear equations consists of multiple equations that share several variables. The goal is to determine the values of these variables that satisfy all equations simultaneously.

In the example given in the exercise, each scenario represented by an augmented matrix can have:

• Unique solution: When the number of independent equations matches the number of unknowns.
• Infinitely many solutions: When there are more variables than independent equations.
• No solution: An inconsistent system where no set of variables satisfy all equations.

Studying different systems requires interpreting the augmented matrix and using row operations to simplify each system to understand its characteristics and possible solutions.

Row Operations

Row operations are essential for simplifying augmented matrices and solving system of equations. There are three types of row operations:

• Swapping two rows in the matrix.
• Multiplying a row by a nonzero scalar.
• Adding or subtracting a multiple of one row to another.

Using these operations systematically helps bring the matrix closer to a form called reduced row-echelon form (RREF).

RREF is a matrix form where each leading coefficient is 1, and zeroes are below and above each leading coefficient in its column. Achieving RREF makes it easier to identify the solutions and understand the relationship between variables, particularly in discovering if parameters play a role in the solution.

Solution with Parameters

When solving systems of linear equations, you'll often encounter solutions with parameters. Such systems generally have more variables than independent equations, leading to infinitely many solutions.

In these solutions, we express one or more variables in terms of other "free" variables. These free variables act as parameters that you can choose freely. Assigning different values to these parameters provides different specific solutions to the system.

For example, in our step-by-step solutions:

• The variable choices like setting some variables to parameters, such as \(x_2 = s\), \(x_4 = t\), or \(x_5 = u\), help establish a general form that solves the system for all possible values of the parameters.

This approach allows flexibility and demonstrates the nature of underdetermined systems where parametrization is key.