Problem 61 - Precalculus

Question

Solve the system of linear equations using Gaussian elimination with back- substitution. $$\begin{aligned} 2 x+5 y &=9 \\ x+2 y-z &=3 \\ -3 x-4 y+7 z &=1 \end{aligned}$$

Step-by-Step Solution

Answer

The system of equations has no solution, as it leads to an inconsistent row after elimination.

Step 1: Write the Augmented Matrix

First, we need to write the system of equations as an augmented matrix.The system is: 1) \( 2x + 5y = 9 \) 2) \( x + 2y - z = 3 \) 3) \( -3x - 4y + 7z = 1 \)The augmented matrix is:\[\begin{bmatrix}2 & 5 & 0 & | & 9 \1 & 2 & -1 & | & 3 \-3 & -4 & 7 & | & 1 \\end{bmatrix}\]

Step 2: Perform Row Operations to Obtain an Upper Triangular Matrix

In this step, we aim to form zeros below the first pivot (first element of the first row).- For the second row, subtract half (\( \frac{1}{2} \)) the first row from it to make the element below the first pivot zero: \[ R_2 = R_2 - \frac{1}{2} R_1 \] - For the third row, add "1.5 times the first row" to eliminate \(-3\) and get zero below the first pivot: \[ R_3 = R_3 + 1.5 R_1 \]The new matrix becomes:\[\begin{bmatrix}2 & 5 & 0 & | & 9 \0 & -0.5 & -1 & | & -1.5 \0 & 3.5 & 7 & | & 14.5 \\end{bmatrix}\]

Step 3: Eliminate Below the Second Pivot

Form zeros below the second pivot in the second column:- For the third row, add "7 times the second row" to eliminate \(3.5\) from the third row: \[ R_3 = R_3 + (7)(R_2) \]Newly transformed matrix:\[\begin{bmatrix}2 & 5 & 0 & | & 9 \0 & -0.5 & -1 & | & -1.5 \0 & 0 & 0 & | & 4 \\end{bmatrix}\]

Step 4: Back Substitution

Since the third row is now \([0, 0, 0] | 4\), indicating no solution (infinity x zero cannot equal 4), Gaussian elimination identifies inconsistencies showing that the system has no solutions.

Key Concepts

Linear Equations

Linear equations are equations in which each term is either a constant or a product of a constant and a single variable. In their simplest form, these equations appear as straight lines when plotted on a graph. They involve constants and variables raised only to the first power. In our problem, we deal with the following system of linear equations:\[\begin{aligned}2x + 5y &= 9, \x + 2y - z &= 3, \-3x - 4y + 7z &= 1 \end{aligned} \]Breaking this down, each equation represents a flat plane in three-dimensional space, and solving the system means finding the specific point (or points) where these planes intersect. However, intersection might not always occur, which could lead to no solutions, as in this particular case.

Augmented Matrix

An augmented matrix is a compact way to represent a system of linear equations. It combines the coefficient matrix of the system with the constant terms, making it easier to apply matrix operations.
To write an augmented matrix, you align all the coefficients and constants of each equation:

• If we have the equations\[ 2x + 5y = 9 \],\[ x + 2y - z = 3 \] and\[ -3x - 4y + 7z = 1 \],
• The augmented matrix becomes:

\[\begin{bmatrix}2 & 5 & 0 & | & 9 \1 & 2 & -1 & | & 3 \-3 & -4 & 7 & | & 1 \\end{bmatrix}\]
This matrix beautifully aligns all the information needed to apply row operations and solve the system.

Row Operations

Row operations are the key to transforming an augmented matrix into a simpler form to solve a system of equations. There are three primary row operations:

• Swapping the positions of two rows.
• Multiplying a row by a non-zero constant.
• Adding or subtracting a multiple of one row to another row.

In our example, performing row operations helps convert the matrix into an upper triangular form, where all entries below the pivot (leading non-zero element) in each column are zeros.
First, adjusting the second and third rows relative to the first row eliminates the coefficients beneath the first pivot. Then, a combination of operations on the second row helps achieve zeros below the second pivot. The process exemplifies the structured steps necessary for Gaussian elimination.

Back-Substitution

Back-substitution is a method used in solving systems of linear equations, particularly once the augmented matrix is in an upper triangular form. The goal is to solve for variables starting from the bottom row and moving upwards. Typically, this would involve solving for one variable at a time using already determined values from previous rows.
However, in the provided exercise, after row operations, the matrix reached a state where the final row was\[\begin{bmatrix}0 & 0 & 0 & | & 4 \\end{bmatrix}\]
This suggests a contradiction, as zero cannot equal four, indicating an inconsistency in the system - hence, no solution exists. This showcases that even without finding specific values for variables, the form of the matrix itself can reveal the nature of the solutions.