Problem 4 - Algebra and Trigonometry

Question

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

The augmented matrix of a system represents the coefficients and constant terms of a system of linear equations. Each row corresponds to one equation, and each column corresponds to a variable or the constant term on the right side of the equation.

Step 2: Identify Variables for Part (a)

The matrix for part (a) is \[\begin{bmatrix} 1 & 0 & 0 & 2 \ 0 & 1 & 0 & 1 \ 0 & 0 & 1 & 3 \end{bmatrix}\]This matrix corresponds to the system of equations:1. \( x = 2 \)2. \( y = 1 \)3. \( z = 3 \)

Step 3: Identify Variables for Part (b)

The matrix for part (b) is \[\begin{bmatrix} 1 & 0 & 1 & 2 \ 0 & 1 & 1 & 1 \ 0 & 0 & 0 & 0 \end{bmatrix}\]This corresponds to the system with:1. \( x + z = 2 \)2. \( y + z = 1 \)3. The third row implies no new information as it's a zero row. Choose \( z \) as a parameter: let \( z = t \). Then the solutions are:\( x = 2 - t \), \( y = 1 - t \), \( z = t \).

Step 4: Identify Variables for Part (c)

The matrix for part (c) is \[\begin{bmatrix} 1 & 0 & 0 & 2 \ 0 & 1 & 0 & 1 \ 0 & 0 & 0 & 3 \end{bmatrix}\]This implies the equations:1. \( x = 2 \)2. \( y = 1 \)3. The last row is inconsistent because it implies \( 0 = 3 \), which is impossible. Thus, there is no solution for the system.

Key Concepts

Reduced Row-Echelon Form

Understanding the reduced row-echelon form (RREF) of a matrix is crucial when dealing with systems of linear equations. An augmented matrix in RREF makes it easy to identify solutions directly by transforming a system into a simpler representation. The characteristics of this form are:

• Each leading entry in a row is 1 (also known as a pivot), and it's the only non-zero entry in its column.
• The leading 1 of any row is to the right of a leading 1 in the previous row.
• Rows consisting entirely of zeros are at the bottom of the augmented matrix.

This simplified structure allows us to straightforwardly read off the solutions for each variable when feasible. For example, part (a) of our exercise directly indicates solutions for variables: \[ x = 2, y = 1, z = 3. \] Each variable has its own "pivot" or leading 1, culminating in unique solutions.

System of Linear Equations

A system of linear equations is simply a set of equations with multiple variables. The equations are linear, meaning each equation graphs as a straight line in coordinate space. Solving such systems involves finding common points of intersection, representing solutions to the equations. An augmented matrix is a compact way to represent these equations using just the coefficients of the variables and the constant terms. You arrange these coefficients in rows, with each row corresponding to an individual equation. This form is particularly useful in matrix algebra, especially in methods like Gaussian elimination to work towards finding solutions, as demonstrated in part (a) of our exercise.

Parametric Equations

Parametric equations express the solution of a system of linear equations where not all variables can be solved directly without an arbitrary component. This occurs when the system has infinitely many solutions due to having fewer independent equations than variables, leading to free variables. In part (b) of our exercise, we use a parameter \( t \) to express the solutions as:

• \( x = 2 - t \)
• \( y = 1 - t \)
• \( z = t \)

This representation shows how one variable, \( z \), is free to vary, while the other variables depend on this parameter. Such solutions convey how the system unfolds along a line or plane in space.

Inconsistent System

An inconsistent system of equations is one that does not have a solution. This occurs when there is a conflicting equation in the system, leading to a contradiction. In part (c) of our exercise, observe the last row in the matrix: \[ 0 = 3 \] This result indicates a clear contradiction, as zero cannot equal three. Therefore, the entire system of equations fails due to this impossibility. No solution exists because there is no set of values for the variables that can satisfy all the equations simultaneously. Identifying such inconsistencies is crucial in determining when a system does not have any feasible solution.