Problem 29 - Algebra and Trigonometry

Question

Find the complete solution of the linear system, or show that it is inconsistent. $$ \left\\{\begin{aligned} x+3 y-2 z &=0 \\ 2 x+& 4 z=4 \\ 4 x+6 y &=4 \end{aligned}\right. $$

Step-by-Step Solution

Answer

The system has infinitely many solutions and can be expressed using a free parameter \(t\).

Step 1: Write the System in Matrix Form

First, translate the given equations into an augmented matrix. For the system \[ \begin{align*} x + 3y - 2z &= 0 \ 2x + 0y + 4z &= 4 \ 4x + 6y + 0z &= 4 \end{align*} \], the augmented matrix is: \[\begin{bmatrix} 1 & 3 & -2 & | & 0 \ 2 & 0 & 4 & | & 4 \ 4 & 6 & 0 & | & 4 \end{bmatrix}\]

Step 2: Apply Row Operations to Achieve Row-Echelon Form

Use row operations to simplify the matrix. First, let's subtract 2 times the first row from the second row:\[R_2 \rightarrow R_2 - 2R_1: \begin{bmatrix} 1 & 3 & -2 & | & 0 \ 0 & -6 & 8 & | & 4 \ 4 & 6 & 0 & | & 4 \end{bmatrix}\]Next, subtract 4 times the first row from the third row:\[R_3 \rightarrow R_3 - 4R_1: \begin{bmatrix} 1 & 3 & -2 & | & 0 \ 0 & -6 & 8 & | & 4 \ 0 & -6 & 8 & | & 4 \end{bmatrix}\]

Step 3: Simplify Further to Identify Unique Solutions or Inconsistency

Subtract the second row from the third row:\[R_3 \rightarrow R_3 - R_2: \begin{bmatrix} 1 & 3 & -2 & | & 0 \ 0 & -6 & 8 & | & 4 \ 0 & 0 & 0 & | & 0 \end{bmatrix}\]Since the last row is essentially zero, there is no inconsistency. The system is consistent with at least one free variable.

Step 4: Back Substitute to Solve for Each Variable

From the second row, \[0x - 6y + 8z = 4\]: solve for \(y\): \[\begin{align*} -6y + 8z &= 4 \ y &= \frac{8}{6}z - \frac{4}{6} \ y &= \frac{4}{3}z - \frac{2}{3} \end{align*}\]From the first row, substitute \(y\) to solve for \(x\):\[\begin{align*} x + 3\left(\frac{4}{3}z - \frac{2}{3}\right) - 2z &= 0 \ x + 4z - 2 - 2z &= 0 \ x &= -2z + 2 \end{align*}\]Thus, \(x = -2z + 2\), \(y = \frac{4}{3}z - \frac{2}{3}\), and \(z = z\) (free variable).

Step 5: Write the Solution in Parametric Form

The solution of the system can be expressed in parametric form: - \(x = -2t + 2\) - \(y = \frac{4}{3}t - \frac{2}{3}\) - \(z = t\) where \(t\) is a free parameter that can take any real number. Therefore, the system has infinitely many solutions.

Key Concepts

Augmented Matrix

An augmented matrix is a compact way to represent a system of linear equations. It integrates both the coefficients of the variables and the constants from each equation into a single matrix. In the given system of equations:

• \(x + 3y - 2z = 0\)
• \(2x + 0y + 4z = 4\)
• \(4x + 6y + 0z = 4\)

The augmented matrix looks like this: \[ \begin{bmatrix} 1 & 3 & -2 & | & 0 \ 2 & 0 & 4 & | & 4 \ 4 & 6 & 0 & | & 4 \end{bmatrix} \]Here, each row corresponds to one equation, and the vertical bar separates the coefficients from the right-hand side constants. This format is useful for performing row operations and finding solutions efficiently.

Row Operations

Row operations are essential techniques we use to simplify an augmented matrix. The goal is to transform it into row-echelon form or reduced row-echelon form, which makes it easier to identify solutions.

• **Swapping Rows:** This operation allows you to exchange any two rows.
• **Multiplying a Row by a Non-Zero Scalar:** You can multiply all elements in a row by the same non-zero number.
• **Adding or Subtracting Rows:** You can add or subtract a multiple of one row to or from another.

Use these operations to simplify the matrix, ensuring that each leading entry (the first non-zero number from the left in a row) is a 1 and that zeros are below all leading entries.

Parametric Form

The parametric form of a solution represents the solution set of a linear system using parameters.In our linear system, after reduction, we found expressions for all variables in terms of a free variable \(z\):

• \(x = -2z + 2\)
• \(y = \frac{4}{3}z - \frac{2}{3}\)
• \(z = z\)

To generalize, we let \(z = t\), where \(t\) is a parameter that can be any real number. This gives us:

• \(x = -2t + 2\)
• \(y = \frac{4}{3}t - \frac{2}{3}\)
• \(z = t\)

This form clearly shows that the system has infinitely many solutions, with \(t\) determining each specific solution.

Free Variable

In many systems of equations, particularly those with infinite solutions, some variables do not correspond to a pivot column after row reduction. These variables are free variables.In our system, the variable \(z\) was not bound to a particular value. Instead, it was expressed in terms of itself, \(z = z\), allowing it to take any real number. This liberty makes \(z\) a free variable.

• Free variables imply flexibility in the solution and are essential for expressing parametric solutions.
• They allow for multiple solutions, illustrating the variety of ways a linear system can be satisfied.

Consistent System

A linear system is consistent if there is at least one set of values for the variables that satisfies all of the equations. In our case, after performing the row operations and achieving a row-echelon form:

• The presence of a row where all elements including the constant are zero (like \(0 = 0\)) indicates consistency, not inconsistency.
• An inconsistent system would have a row equivalent to \(0 = c\), where \(c\) is a non-zero constant.

Thus, since the last row of the reduced matrix was all zeros, our system is consistent and has infinitely many solutions, characterized by the free variable \(z\).