Problem 4 - Precalculus: Mathematics for Calculus

Question

The augmented matrix of a system of linear equations is given in reduced row- echelon form. Find the solution of the system. \text { (a) }\left[\begin{array}{llll} 1 & 0 & 0 & 2 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & 3 \end{array}\right] \quad \text { (b) }\left[\begin{array}{llll} 1 & 0 & 1 & 2 \\ 0 & 1 & 1 & 1 \\ 0 & 0 & 0 & 0 \end{array}\right] \quad \text { (c) }\left[\begin{array}{llll} 1 & 0 & 0 & 2 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 0 & 3 \end{array}\right] $$x=$$_____ $$y=$$_____ $$z=$$_____ $$x=$$_____ $$y=$$_____ $$z=$$_____ $$x=$$_____ $$y=$$_____ $$z=$$_____

Step-by-Step Solution

Answer

(a) x=2, y=1, z=3. (b) x=2-t, y=1-t, z=t. (c) No solution.

Step 1: Recognize the Reduced Row Echelon Form (RREF) for (a)

Given the matrix for (a) in RREF: \( \begin{bmatrix} 1 & 0 & 0 & 2 \ 0 & 1 & 0 & 1 \ 0 & 0 & 1 & 3 \end{bmatrix} \). This directly gives the solution to the system of equations, where the columns correspond to variables \(x, y, z\), and the last column is the constants\.

Step 2: Read Solutions from Matrix for (a)

For each row, the solution for each variable is explicitly shown. From the matrix: \(x = 2\), \(y = 1\), \(z = 3\). Therefore: \(x = 2\), \(y = 1\), \(z = 3\)\.

Step 3: Recognize the RREF for (b)

Given the matrix for (b) in RREF: \( \begin{bmatrix} 1 & 0 & 1 & 2 \ 0 & 1 & 1 & 1 \ 0 & 0 & 0 & 0 \end{bmatrix} \). The third row indicates there are infinitely many solutions, as it corresponds to \(0 = 0\)\.

Step 4: Express Solutions for (b)

Define a free variable, \(z = t\). The system becomes: \(x + z = 2\), so \(x = 2 - t\). \(y + z = 1\), so \(y = 1 - t\). The solution is: \(x = 2 - t\), \(y = 1 - t\), \(z = t\)\.

Step 5: Identify Inconsistency in Matrix (c)

For the matrix (c): \( \begin{bmatrix} 1 & 0 & 0 & 2 \ 0 & 1 & 0 & 1 \ 0 & 0 & 0 & 3 \end{bmatrix} \), the last row suggests inconsistency because it translates to \(0 = 3\). This indicates there is no solution to the system\.

Key Concepts

Reduced Row-Echelon Form

When dealing with linear equations, one powerful method to solve them is by using matrices. The reduced row-echelon form (RREF) is a type of matrix that is simplified in such a way that it becomes easy to identify the solutions to a system of linear equations. When a matrix is in RREF, it allows us to directly read the solutions of the variables involved. Here is how a matrix in RREF looks:

• Every leading entry (the first non-zero number in a row) is 1.
• Each leading 1 is the only non-zero entry in its column.
• The leading 1 in each row is to the right of any leading 1s in the rows above it.
• Any rows consisting entirely of zeros are at the bottom.

The main advantage of converting a matrix into RREF is the ease of extracting solutions, as the matrix reveals the relationships between the variables and constants directly.

System of Linear Equations

A system of linear equations consists of multiple equations with multiple variables. Each equation represents a linear relationship. Solving the system means finding values for the variables that satisfy all equations simultaneously.
Systems can have:

• A unique solution: where exactly one set of values fulfills all equations.
• Infinite solutions: where multiple sets of values satisfy the equations.
• No solution: where no set of values can satisfy all equations.

We represent these systems using matrices, and manipulating them can lead us to their solutions. In practical terms, the first step is to write the augmented matrix for the given system and then convert it to RREF to identify the solutions.

Infinite Solutions

In certain systems, you may encounter a scenario where there are infinitely many solutions. This occurs when there is not a unique solution to the system, often due to dependent equations.
In matrix form, a row like \[ \begin{bmatrix} 0 & 0 & 0 & 0 \end{bmatrix} \] indicates that the system is underdetermined; hence, one or more free variables exist.
To express these solutions, we select a free variable, often referred to as a parameter like \( z = t \), and express other variables in terms of this parameter.
For example, if the system is given by the matrix: \[ \begin{bmatrix} 1 & 0 & 1 & 2 \ 0 & 1 & 1 & 1 \ 0 & 0 & 0 & 0 \end{bmatrix} \] we assign \( z = t \) (where \( t \) is any real number), and solve for \( x \) and \( y \) using the equations derived from the matrix.

Inconsistent System

A system of linear equations is inconsistent when there are no solutions that satisfy all the equations. In terms of matrices, this is represented as a row form where one side is zero, and the constant is a non-zero number.
Take for instance a row like \[ \begin{bmatrix} 0 & 0 & 0 & 3 \end{bmatrix} \], which mathematically translates to \( 0 = 3 \). Such a statement is blatantly false, suggesting a contradiction.
When we convert a system to RREF and identify such rows, we can conclude that the system is inconsistent. This means no values for the variables can satisfy all the original equations together.
Understanding this concept helps to highlight systems that need reassessment or redefinition before attempting to find a solution.