Problem 69 - Precalculus

Question

Solve the system of linear equations using Gaussian elimination with back- substitution. $$\begin{array}{rr} 3 x_{1}-2 x_{2}+x_{3}+2 x_{4}= & -2 \\ -x_{1}+3 x_{2}+4 x_{3}+3 x_{4}= & 4 \\ x_{1}+x_{2}+x_{3}+x_{4}= & 0 \\ 5 x_{1}+3 x_{2}+x_{3}+2 x_{4}= & -1 \end{array}$$

Step-by-Step Solution

Answer

The system has no solution because it leads to a contradiction (0 = -3).

Step 1: Write the Augmented Matrix

Convert the given system of equations into an augmented matrix form.\[\begin{bmatrix}3 & -2 & 1 & 2 & | & -2 \-1 & 3 & 4 & 3 & | & 4 \1 & 1 & 1 & 1 & | & 0 \5 & 3 & 1 & 2 & | & -1\end{bmatrix}\]

Step 2: Perform Gaussian Elimination (Forward Elimination)

The goal in Gaussian elimination is to convert the matrix into an upper triangular form. We do this by performing row operations.- Use row 1 to eliminate the elements below in column 1: - R2 = R2 + \(\frac{1}{3}\) * R1. - R3 = R3 - \(\frac{1}{3}\) * R1. - R4 = R4 - \(\frac{5}{3}\) * R1.Result:\[\begin{bmatrix}3 & -2 & 1 & 2 & | & -2 \0 & \frac{5}{3} & \frac{11}{3} & \frac{11}{3} & | & \frac{10}{3} \0 & \frac{5}{3} & \frac{2}{3} & \frac{1}{3} & | & \frac{2}{3} \0 & \frac{19}{3} & -\frac{2}{3} & -\frac{1}{3} & | & \frac{1}{3}\end{bmatrix}\]

Step 3: Continue Gaussian Elimination

Now, we need to create zeros below the leading coefficient in column 2.- Use row 2 to eliminate elements below in column 2: - R3 = R3 - R2. - R4 = R4 - \(\frac{19}{5}\) * R2.Result:\[\begin{bmatrix}3 & -2 & 1 & 2 & | & -2 \0 & \frac{5}{3} & \frac{11}{3} & \frac{11}{3} & | & \frac{10}{3} \0 & 0 & -3 & -2 & | & -2 \0 & 0 & -12 & -8 & | & -11\end{bmatrix}\]

Step 4: Finalize Upper Triangular Form

Perform final row operation to make the matrix strictly upper triangular.- Use row 3 to make element below leading one zero: - R4 = R4 - 4 * R3.Result:\[\begin{bmatrix}3 & -2 & 1 & 2 & | & -2 \0 & \frac{5}{3} & \frac{11}{3} & \frac{11}{3} & | & \frac{10}{3} \0 & 0 & -3 & -2 & | & -2 \0 & 0 & 0 & 0 & | & -3\end{bmatrix}\]

Step 5: Check for Inconsistencies

The last row implies that 0 = -3, which is a contradiction, indicating that the system has no solution.

Key Concepts

Linear Equations

Linear equations are mathematical expressions that describe a straight line when plotted on a graph. These equations are pivotal in modeling a variety of real-world situations.
In the given exercise, each equation is in a linear form and involves several variables. Specifically, these variables are denoted as \(x_1, x_2, x_3,\) and \(x_4\).
The solution to these equations can help in finding values for these variables such that all the equations hold true simultaneously. This is often done by combining these equations in a way to simplify and solve them, which is particularly useful in fields like physics, engineering, and economics.

Augmented Matrix

An augmented matrix is a compact and convenient way to represent a system of linear equations. It combines the coefficients of the variables and the constants of the equations into one matrix form.
For the given system, the augmented matrix looks like this:

• The first part represents the coefficients of variables \(x_1, x_2, x_3,\) and \(x_4\).
• The vertical bar "|" separates these coefficients from the constants on the right side of the equations.

The purpose of forming an augmented matrix is to perform systematic calculations known as Gaussian elimination. This helps to transform the system into a simpler form, making it easier to solve.

Back-substitution

Back-substitution is a process used in solving linear systems after applying Gaussian elimination. It is used once the system of equations is converted into an upper triangular form.
When the matrix is in this form, each equation can be solved starting from the last row and moving upwards to the first row.
During back-substitution, you solve for one variable at a time by plugging the values of already found variables back into the previous equations. This sequential method ensures that you have all the variable values in the end.
Unfortunately, in this exercise, we never reach the step of back-substitution because we identify an inconsistency during Gaussian elimination.

Inconsistent System

An inconsistent system is one in which no solution satisfies all of the equations simultaneously.
In this example, the final result of Gaussian elimination gave us a row that could be interpreted as \(0 = -3\), a clear impossibility in mathematics.
This sort of result indicates that there is a contradiction among the equations themselves. Thus, either there was an error in the equations or they describe a system where no common solution exists.

• An inconsistent system typically means the lines represented by the equations do not intersect at any point.
• Identifying an inconsistency early can save time since it indicates further calculations are unnecessary.

Recognizing these cases is vital for efficiently handling problems involving linear equations.