Problem 43 - Precalculus: Mathematics for Calculus

Question

Determine whether the system of linear equations is inconsistent or dependent. If it is dependent, find the complete solution. $$\left\\{\begin{aligned} x-y+3 z &=3 \\ 4 x-8 y+32 z &=24 \\ 2 x-3 y+11 z &=4 \end{aligned}\right.$$

Step-by-Step Solution

Answer

The system is dependent; solution: \((x, y, z) = (2t, 5t - 3, t)\).

Step 1: Write the system in matrix form

First, represent the system of equations in augmented matrix form, separating the coefficients from the constants. The matrix form is:\[\begin{bmatrix}1 & -1 & 3 & | & 3 \4 & -8 & 32 & | & 24 \2 & -3 & 11 & | & 4\end{bmatrix}\]

Step 2: Apply Gaussian elimination

Use the Gaussian elimination method to simplify the matrix. Subtract 4 times the first row from the second row:\[\begin{bmatrix}1 & -1 & 3 & | & 3 \0 & -4 & 20 & | & 12 \2 & -3 & 11 & | & 4\end{bmatrix}\]Next, subtract 2 times the first row from the third row:\[\begin{bmatrix}1 & -1 & 3 & | & 3 \0 & -4 & 20 & | & 12 \0 & -1 & 5 & | & -2\end{bmatrix}\]

Step 3: Continue to row reduce

Eliminate the leading coefficient in the third row by subtracting \(-\frac{1}{4}\) times the second row from the third:\[\begin{bmatrix}1 & -1 & 3 & | & 3 \0 & -4 & 20 & | & 12 \0 & 0 & 0 & | & 0\end{bmatrix}\]The last row indicates a row of zeros, which suggests a dependent system.

Step 4: Express the system in parametric form

With one row of zeros, the system has free variables. Let \(z = t\) (a parameter). Solve the second row for \(y\):\[-4y + 20t = 12 \implies y = 5t - 3\]Substitute in the first equation to solve for \(x\):\[x - (5t - 3) + 3t = 3 \\implies x - 5t + 3 + 3t = 3 \\implies x = 2t\]The solution in parametric form: \( (x, y, z) = (2t, 5t - 3, t)\).

Step 5: Interpret the results

Since the system reduces to a row of zeros and produces a parametric solution, it is dependent. The solutions can be expressed in terms of the parameter \(t\).

Key Concepts

Gaussian elimination

The process of Gaussian elimination is fundamental when dealing with systems of linear equations. It simplifies the system to make it easier to find solutions. This method involves performing row operations to transform the original augmented matrix into an upper triangular form or even further, into a reduced row echelon form (RREF). Here are the basic steps involved:

• Identify the pivot element in the first column. If needed, rearrange the rows to bring a non-zero element at the top.
• Use this pivot to eliminate the entries below it by subtracting suitable multiples of the pivot row from the rows below.
• Repeat the process for each subsequent column, ensuring a step-by-step reduction towards a simpler form.

The goal is to end up with zeros below the pivot elements, thereby pushing the system closer to a format where solutions can be easily identified. In our exercise, Gaussian elimination is crucial as it reveals the dependent nature of the system.

parametric form solution

A parametric form solution is used to express the solutions of a dependent system of linear equations. When a system is dependent, it has infinitely many solutions, often owing to the existence of free variables.
The general approach to writing a parametric solution involves:

• Identifying the free variables in the system, which can take any real value.
• Expressing the remaining variables in terms of these free variables.

In the given exercise, by using parametric representation, the variable \( z \) is set as the parameter \( t \), thus: \[(x, y, z) = (2t, 5t - 3, t)\]Here, \( t \) can be any real number, providing a clear visualization of the infinite solutions available to the system.

row reduction

Row reduction is a sequence of operations applied to the rows of a matrix to bring it into a more manageable form such as row echelon form (REF) or reduced row echelon form (RREF). These operations are pivotal in simplifying the matrix and are classified as:

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

The objective is to create a staircase-like form in the coefficient matrix. In this exercise, row reduction is key to eliminating the leading coefficients below the diagonal, revealing dependencies between variables. The process results in a simpler form, showing a row of zeros, which confirms a dependent system.

free variables in linear equations

Free variables arise in systems of equations that have more unknowns than equations, or when equations are linearly dependent. They represent the degrees of freedom within the solution.
In solving linear equations, free variables are those that can accept any value. These are not pivot variables and usually appear naturally when the system is expressed in its row-reduced form.

• Free variables are indicated by the column positions that do not contain a leading 1 after row reduction.
• They allow the expression of an infinite number of solutions.
• Systems with free variables are classified as being dependent.

In the exercise, the variable \( z \) is free and can take any value, leading to different solutions for \( x \) and \( y \). This interplay gives students insight into how dependent systems can still display rich structures due to their inherent flexibility.