Problem 3 - Precalculus: Mathematics for Calculus

Question

The following matrix is the augmented matrix of a system of linear equations in the variables \(x, y,\) and \(z\). (It is given in reduced row-echelon form.) \left[\begin{array}{rrrr} 1 & 0 & -1 & 3 \\ 0 & 1 & 2 & 5 \\ 0 & 0 & 0 & 0 \end{array}\right] (a) The leading variables are __________. (b) Is the system inconsistent or dependent?__________ (c) The solution of the system is: $$x=$$__________ , $$y=$$ __________ , $$=$$ __________

Step-by-Step Solution

Answer

The leading variables are \(x\) and \(y\). The system is dependent. \(x = 3 + z\), \(y = 5 - 2z\), \(z = z\).

Step 1: Understanding the Matrix

First, observe that the given matrix is in reduced row-echelon form (RREF). It indicates how the system of linear equations corresponds directly to variables in the system. The matrix can be written as equations: \(x - z = 3\), and \(y + 2z = 5\). The last row \(0=0\) does not provide any new information.

Step 2: Identifying Leading Variables

Leading variables are those that have leading 1s in the row-echelon form. In our matrix, the leading 1 appears in the columns corresponding to \(x\) and \(y\), which means \(x\) and \(y\) are leading variables. The column for \(z\) does not have a leading 1, indicating it is a free variable.

Step 3: Determining Consistency

A system of linear equations is inconsistent if it contains a row like \([0 h0 h0 tb] \eq 0\), which would indicate a contradiction (e.g., \(0 = 1\)). In our case, every row has either 0 equals 0 or valid equations, indicating that this system is consistent. The presence of a free variable (\(z\)) means the system is dependent, leading to infinitely many solutions.

Step 4: Expressing the Solutions of the System

Since \(z\) is a free variable, express the other variables in terms of \(z\): from the first equation \(x = 3 + z\) and from the second equation \(y = 5 - 2z\). Therefore, the solution for the system in terms of \(z\) is: \(x = 3 + z\), \(y = 5 - 2z\), and \(z = z\).

Key Concepts

Augmented Matrix

An augmented matrix is used to represent and solve systems of linear equations. It combines both the coefficients of the variables and the constants from the equations into one single matrix. Imagine you have a system of linear equations with multiple variables. Instead of writing out each equation, you can organize all the information into this compact form.
Consider a system:

• Equation 1: \(a_1x + b_1y + c_1z = d_1\)
• Equation 2: \(a_2x + b_2y + c_2z = d_2\)

This information is presented in matrix form by creating rows for each equation and columns aligning with each variable followed by the constants on the right side: \[\left[ \begin{array}{cccc} a_1 & b_1 & c_1 & d_1\ a_2 & b_2 & c_2 & d_2 \end{array} \right]\]This form allows for ease of applying techniques like row reduction to solve for the variables.

Reduced Row-Echelon Form

Reduced Row-Echelon Form (RREF) is a special type of matrix form that greatly facilitates solving linear equations. It's achieved through row operations and has some important characteristics. A matrix is in RREF if:

• All rows consisting entirely of zeros are at the bottom.
• The first nonzero number from the left (known as a leading 1) in each nonzero row is to the right of the leading 1 in the row above it.
• Each leading 1 is the only nonzero entry in its column.

These properties make it straightforward to back-substitute and find the solutions to a system of equations.
For example, given our augmented matrix:\[\left[ \begin{array}{rrrr}1 & 0 & -1 & 3 \0 & 1 & 2 & 5 \0 & 0 & 0 & 0\end{array} \right]\]This matrix is in RREF. You can see the zeros at the bottom and the pattern of leading 1s, which make it easy to read off the solutions of the system.

Leading Variables

Leading variables in a system of equations are those corresponding to the columns with leading 1s in the reduced row-echelon form of the matrix. These are the variables that are directly solved in terms of other variables, if necessary.
In our reduced row-echelon form matrix example:\[\left[ \begin{array}{rrrr}1 & 0 & -1 & 3 \0 & 1 & 2 & 5 \0 & 0 & 0 & 0\end{array} \right]\]The leading 1 appears in the first and second columns corresponding to variables \(x\) and \(y\), respectively. Therefore, \(x\) and \(y\) are the leading variables in this case. They are determined first and can be expressed in terms of any free variables in the system.

Free Variable

A free variable arises when not every column in the coefficient part of the matrix (excluding the augmented part) contains a leading 1 in the reduced row-echelon form. This means the variable associated with that column is not directly determined by the system and can take multiple values leading to infinitely many solutions.
For our augmented matrix example:\[\left[ \begin{array}{rrrr}1 & 0 & -1 & 3 \0 & 1 & 2 & 5 \0 & 0 & 0 & 0\end{array} \right]\]The column corresponding to \(z\) has no leading 1. Hence, \(z\) is a free variable. You can choose any value for \(z\), and then use the leading variables' equations to find corresponding values for \(x\) and \(y\). This flexibility in the choice of \(z\) results in a dependent system with infinite solutions, all parameterized by \(z\).