Problem 17 - Algebra and Trigonometry

Question

15–24 The system of linear equations has a unique solution. Find the solution using Gaussian elimination or Gauss-Jordan elimination. $$\left\\{\begin{aligned} x+y+z =2 \\ 2 x-3 y+2 z &=4 \\ 4 x+y-3 z &=1 \end{aligned}\right.$$

Step-by-Step Solution

Answer

The solution is \(x = 1\), \(y = 0\), \(z = 1\).

Step 1: Write the Augmented Matrix

The given system of equations can be represented as an augmented matrix. The matrix for the system is:\[\begin{bmatrix}1 & 1 & 1 & | & 2 \2 & -3 & 2 & | & 4 \4 & 1 & -3 & | & 1\end{bmatrix}\]

Step 2: Perform Row Operations to Reach Row Echelon Form

To simplify the matrix, start by eliminating the elements below the first pivot (1 in the first row, first column):- Subtract 2 times the first row from the second row: \\[\begin{bmatrix}2 & -3 & 2 & 4\end{bmatrix} - 2\begin{bmatrix}1 & 1 & 1 & 2\end{bmatrix} \rightarrow \begin{bmatrix}0 & -5 & 0 & 0\end{bmatrix}\]- Subtract 4 times the first row from the third row: \\[\begin{bmatrix}4 & 1 & -3 & 1\end{bmatrix} - 4\begin{bmatrix}1 & 1 & 1 & 2\end{bmatrix} \rightarrow \begin{bmatrix}0 & -3 & -7 & -7\end{bmatrix}\]The matrix is now:\[\begin{bmatrix}1 & 1 & 1 & | & 2 \0 & -5 & 0 & | & 0 \0 & -3 & -7 & | & -7\end{bmatrix}\]

Step 3: Continue Row Operations to Reach Upper Triangular Form

Further simplify the matrix by eliminating elements below the second pivot (-5 in the second row, second column):- Multiply the second row by \(-1/5\) to make the pivot 1: \\[\begin{bmatrix}0 & -5 & 0 & 0\end{bmatrix} \rightarrow \begin{bmatrix}0 & 1 & 0 & 0\end{bmatrix}\]- Add 3 times the new second row to the third row to eliminate the element beneath the pivot: \\[\begin{bmatrix}0 & -3 & -7 & -7\end{bmatrix} + 3\begin{bmatrix}0 & 1 & 0 & 0\end{bmatrix} \rightarrow \begin{bmatrix}0 & 0 & -7 & -7\end{bmatrix}\]The matrix now is:\[\begin{bmatrix}1 & 1 & 1 & | & 2 \0 & 1 & 0 & | & 0 \0 & 0 & -7 & | & -7\end{bmatrix}\]

Step 4: Obtain the Identity Matrix for the Coefficients

Continue simplifying the matrix to achieve identity on the left:- Divide the third row by -7 to make the pivot 1: \\[\begin{bmatrix}0 & 0 & -7 & -7\end{bmatrix} \rightarrow \begin{bmatrix}0 & 0 & 1 & 1\end{bmatrix}\]The matrix now reads:\[\begin{bmatrix}1 & 1 & 1 & | & 2 \0 & 1 & 0 & | & 0 \0 & 0 & 1 & | & 1\end{bmatrix}\]

Step 5: Back Substitute to Find Solutions

Now back substitute to find the solution variables:- From the third row, \(z = 1\).- Substitute \(z = 1\) into the second row to find \(y\): \[ 0 \, y \, + \, 0 \, z \, = 0 \Rightarrow y = 0 \]- Substitute \(y = 0\) and \(z = 1\) in the first row to find \(x\): \[ x \, + \, 0 \, + \, 1 \, = 2 \Rightarrow x = 1 \]The solution is \(x = 1\), \(y = 0\), \(z = 1\).

Step 6: Verify the Solution

Substitute \(x = 1\), \(y = 0\), and \(z = 1\) back into the original equations to confirm:- First equation: \(1 + 0 + 1 = 2\)- Second equation: \(2 \cdot 1 - 3 \cdot 0 + 2 \cdot 1 = 4\)- Third equation: \(4 \cdot 1 + 0 \cdot 1 - 3 \cdot 1 = 1\)All equations are satisfied; this confirms that our solution is correct.

Key Concepts

Linear Equations

Linear equations form the backbone of many mathematical problems and real-world applications. They are equations that express a straight line when plotted on a graph. These equations take on the general form of \(ax + by + cz = d\), where \(x\), \(y\), and \(z\) are variables, and \(a\), \(b\), \(c\), and \(d\) are constants.
Linear equations can be used to describe a wide range of phenomena, from calculating the trajectory of a moving object to determining the relationship between economic factors. The fundamental goal of working with them is to find the values of the variables that make the equation true.

In the provided exercise, you're working with a system of three linear equations. These equations are not isolated; they are connected because they share the same set of variables: \(x\), \(y\), and \(z\). Solving this system means finding values for these variables that satisfy all equations simultaneously. In this case, understanding and using Gaussian elimination effectively helps in achieving that goal.

Augmented Matrix

An augmented matrix is a crucial tool in solving systems of linear equations. It takes all the information from a set of linear equations and arranges it into a matrix form, simplifying the problem.

For instance, in the given exercise, the system of equations is represented as an augmented matrix:

• The coefficients of the variables form the main component of the matrix (the left part).
• The constants from the equations form the rightmost column.

The structure of this augmented matrix allows for systematic application of row operations, making it easier to convert the system into a format suitable for solving. By focusing on the order and relations of the numbers, you're equipped to handle the system efficiently, helping to visually and interactively engage with the data.

Row Echelon Form

Row Echelon Form (REF) is a simplified version of your augmented matrix, achieved by applying various row operations. The goal in attaining REF is to transform the matrix so that it matches a specific pattern:

• Each leading entry (or pivot) in a row is to the right of the leading entry in the previous row.
• Below each pivot, all entries are zeros.

These characteristics make it easy to solve a system of equations through back substitution, as you've identified the hierarchical order in which variables can be calculated.
By reaching the REF in the exercise, we successfully reduced the matrix to three simpler lines, each providing information about one variable in terms of the others. This transformation highlights the power of systematic row operations to transform complexity into simplicity and clarity.

Back Substitution

Back substitution is the final step in solving a system of linear equations once the matrix is in Row Echelon Form. This technique involves using the matrix’s information from the bottom row upwards to determine the variable values one by one.
In the exercise, with your matrix fully simplified, back substitution reveals the values of \(z\), \(y\), and \(x\) in this order due to the nature of the echelon form:

• You start with the last equation, which usually involves a single variable, making it straightforward to solve.
• You then substitute this solution back into the preceding equations, working your way upwards.

This systematic approach ensures no variable is dependent on another unsolved variable, thus efficiently allowing each variable to be determined in sequence. Back substitution turns a daunting system of equations into manageable, sequential steps