Problem 20 - Precalculus Mathematics for Calculus

Question

Find the complete solution of the linear system, or show that it is inconsistent. $$\left\\{\begin{aligned} x+y+z &=0 \\ -x+2 y+5 z &=3 \\ 3 x-y &=6 \end{aligned}\right.$$

Step-by-Step Solution

Answer

The solution is \((x, y, z) = (1, -3, 2)\).

Step 1: Write the system as an augmented matrix

First, we need to translate the given system of equations into an augmented matrix. The system: \(\begin{aligned} x + y + z &= 0 \ -x + 2y + 5z &= 3 \ 3x - y &= 6 \end{aligned}\) becomes: \[ \begin{bmatrix} 1 & 1 & 1 & | & 0 \ -1 & 2 & 5 & | & 3 \ 3 & -1 & 0 & | & 6 \end{bmatrix} \]

Step 2: Perform Gaussian Elimination

We will use row operations to transform this matrix into row-echelon form. 1. Make the first element of the first column a pivot (it is already 1).2. Add row 1 to row 2 to eliminate the -1 under the first pivot: \[ R_2 = R_2 + R_1 \] 3. Subtract 3 times row 1 from row 3: \[ R_3 = R_3 - 3R_1 \]This gives us the new matrix: \[ \begin{bmatrix} 1 & 1 & 1 & | & 0 \ 0 & 3 & 6 & | & 3 \ 0 & -4 & -3 & | & 6 \end{bmatrix} \]

Step 3: Continue to Simplify

Continue the elimination process on the new matrix: 1. To make the second row start with a simpler pivot, divide the second row by 3:\[ R_2 = \frac{1}{3}R_2 \] Resulting in: \[ \begin{bmatrix} 1 & 1 & 1 & | & 0 \ 0 & 1 & 2 & | & 1 \ 0 & -4 & -3 & | & 6 \end{bmatrix} \] 2. To eliminate the -4 in the third row, add 4 times the second row to the third row: \[ R_3 = R_3 + 4R_2 \] This reduces to: \[ \begin{bmatrix} 1 & 1 & 1 & | & 0 \ 0 & 1 & 2 & | & 1 \ 0 & 0 & 5 & | & 10 \end{bmatrix} \]

Step 4: Solve the triangular system

We now have the matrix in row-echelon form, which corresponds to the following system of equations: \(\begin{aligned} x + y + z &= 0 \ y + 2z &= 1 \ 5z &= 10 \end{aligned}\) Solve the third equation for \(z\): \[ z = \frac{10}{5} = 2 \] Substitute \(z = 2\) into the second equation to solve for \(y\): \[ y + 2(2) = 1 \implies y = 1 - 4 = -3 \] Finally, substitute \(y = -3\) and \(z = 2\) into the first equation to solve for \(x\): \[ x - 3 + 2 = 0 \implies x = 1 \]

Step 5: Write the final solution

The solution to the system is the ordered triple \((x, y, z) = (1, -3, 2)\). This means the system of equations is consistent and has a unique solution.

Key Concepts

Gaussian Elimination

Gaussian Elimination is a powerful method in linear algebra used to solve systems of linear equations. The main idea is to transform the original system into an equivalent system that is easier to solve.

This process involves a series of steps which include:

• Interchanging the rows.
• Multiplying a row by a nonzero scalar.
• Adding or subtracting the rows to eliminate variables.

These steps manipulate the equations to ultimately reach a form where back substitution can be easily applied.

In our linear system example:

• We first identified our pivot positions, which are the leading coefficients in the rows.
• Then, systematically zeroed out the numbers below each pivot.
• Finally, achieved an upper triangular matrix ready for back substitution.

Gaussian Elimination is systematic and often the preferred method when dealing with complex systems.

Augmented Matrix

An augmented matrix is a compact way to represent a system of linear equations. It includes the coefficients of the variables and the constants from the equations, all snugly arranged into one tidy matrix.

For the given system of equations:

• The rows correspond to the equations.
• The columns except the last one, hold the coefficients of the variables.
• The last column contains the constants from the right side of the equations.

This form allows for easier manipulation compared to handling each equation separately.

The augmented matrix for our system was: \[ \begin{bmatrix} 1 & 1 & 1 & | & 0 \ -1 & 2 & 5 & | & 3 \ 3 & -1 & 0 & | & 6 \ \end{bmatrix} \] The vertical line separates the coefficients from the constants, providing a clear visual structure that helps in performing Gaussian elimination efficiently.

Row Echelon Form

Reaching Row Echelon Form (REF) is a milestone in solving a system of equations using Gaussian Elimination. In REF, the matrix has a stair-like shape where each leading nonzero entry of a row is to the right of the leading entry of the row above it.

The features of Row Echelon Form include:

• Any rows that are completely zero are located at the bottom of the matrix.
• The leading entry of a nonzero row is 1, often called a leading 1.
• Each leading 1 is the only nonzero entry in its column above it.

In the given problem, the matrix was finally manipulated into the following form: \[ \begin{bmatrix} 1 & 1 & 1 & | & 0 \ 0 & 1 & 2 & | & 1 \ 0 & 0 & 5 & | & 10 \ \end{bmatrix} \] At this point, the system is ready for back substitution to determine the solution. REF provides a streamlined way of systematically approaching solutions.

Consistent System

A consistent system of linear equations is one that has at least one solution. In contrast, an inconsistent system has no solutions, usually evidenced by an absurd result like 0 = 1 being derived during Gaussian Elimination.

Throughout the solution processes of linear systems:

• If the matrix process results in no contradictory equation, the system is consistent.
• It might have one or infinitely many solutions.

In the example problem, Gaussian Elimination uncovers a unique solution for the variables, making the system consistent.

The solution found, \((x, y, z) = (1, -3, 2)\), demonstrates how a properly manipulated system, through effective use of matrices and row operations, provides clear insights into the existence of solutions.