Problem 21 - College Algebra

Question

For the following exercises, solve each system by Gaussian elimination. $$ \begin{aligned} 2 x+3 y-4 z &=5 \\\\-3 x+2 y+z &=11 \\\\-x+5 y+3 z &=4 \end{aligned} $$

Step-by-Step Solution

Answer

The solution is \(x = -\frac{1649}{539},\ y = \frac{191}{77},\ z = -\frac{30}{7}\).

Step 1: Write the Augmented Matrix

The first step in solving this system by Gaussian elimination is to write the augmented matrix. The matrix represents the coefficients of the variables along with the constants from the right of the equation.\[\begin{bmatrix}2 & 3 & -4 & | & 5 \-3 & 2 & 1 & | & 11 \-1 & 5 & 3 & | & 4 \\end{bmatrix}\]

Step 2: Perform Row Operations to Reach Upper Triangular Form

We need to zero out the elements below the first pivot (first row, first column). Start by adding \(\frac{3}{2}\) times the first row to the second row to eliminate the first element in the second row, and \(\frac{1}{2}\) times the first row to the third row.New second row: \[0 \ & \frac{11}{2} \ & -\frac{5}{2} \ & \left | \ & \frac{43}{2} \\]New third row:\[0 \ & \frac{11}{2} \ & 1 \ & \left | \ & \frac{13}{2} \\]Now the matrix:\[\begin{bmatrix}2 & 3 & -4 & | & 5 \0 & \frac{11}{2} & -\frac{5}{2} & | & \frac{43}{2} \0 & \frac{11}{2} & 1 & | & \frac{13}{2} \\end{bmatrix}\]

Step 3: Simplify to Reduce Further

Subtract the third row from the second row to zero out its first column.New third row:\[0 \ & 0 \ & -\frac{7}{2} \ & \left | \ & 15 \\]Updated matrix:\[\begin{bmatrix}2 & 3 & -4 & | & 5 \0 & \frac{11}{2} & -\frac{5}{2} & | & \frac{43}{2} \0 & 0 & -\frac{7}{2} & | & 15 \\end{bmatrix}\]

Step 4: Back-Substitute to Find Solutions

Now we back-substitute from the bottom row upward to find the variables. Start with the equation from the last row:\[-\frac{7}{2}z = 15\]Solving for \(z\), we find:\[z = -\frac{30}{7}\]Substitute \(z\) into the second row:\[\frac{11}{2}y - \frac{5}{2}\left(-\frac{30}{7}\right) = \frac{43}{2}\]Solve for \(y\), we get:\[y = \frac{191}{77}\]Finally, substitute \(y\) and \(z\) into the first row:\[2x + 3\left(\frac{191}{77}\right) - 4\left(-\frac{30}{7}\right) = 5\]Solving for \(x\), we find:\[x = -\frac{1649}{539}\]

Step 5: Present the Final Solution

The solution to the system of equations is:\[\begin{aligned}x &= -\frac{1649}{539}\y &= \frac{191}{77}\z &= -\frac{30}{7}\\end{aligned}\]These values are obtained by reducing the system to an upper triangular matrix and then back-substituting.

Key Concepts

Augmented Matrix

The first step in the process of Gaussian Elimination is forming an augmented matrix. This matrix derives its name from including not just the coefficients of the variables in the equations but also the constants on the right-hand side of the equations. Think of it as a compact table that encapsulates the entire system of linear equations.For the given system of equations:- \(2x + 3y - 4z = 5\)- \(-3x + 2y + z = 11\)- \(-x + 5y + 3z = 4\)The augmented matrix representation is:\[\begin{bmatrix}2 & 3 & -4 & | & 5 \-3 & 2 & 1 & | & 11 \-1 & 5 & 3 & | & 4\end{bmatrix}\]Here, each row corresponds to one of the equations, while the columns contain the coefficients of the variables \(x, y,\) and \(z\), along with the constants on the right-hand side. This matrix forms the foundation for subsequent operations aimed at simplifying the system.

Row Operations

Once the augmented matrix is set up, the next step involves using row operations to simplify it. Row operations are the tools that help us manipulate the rows of the matrix to eventually solve the system.There are three main types of row operations:

• Swapping two rows
• Multiplying a row by a nonzero constant
• Adding or subtracting multiples of one row to another

In this exercise, the primary goal was to transform the matrix into an upper triangular form. This involves making strategic changes to zero out certain elements in the matrix, specifically below the pivot (the leading coefficient of each row).For example, to eliminate the \(-3\) and \(-1\) in the first column of the second and third rows, we used:- Adding \(\frac{3}{2}\) times the first row to the second row.- Adding \(\frac{1}{2}\) times the first row to the third row.Through these operations, the focus is on appending zeros strategically, which will facilitate easier computation in later stages.

Upper Triangular Form

The upper triangular form is a matrix where all the entries below the main diagonal are zeros. Converting an augmented matrix to this form is a crucial milestone in solving linear systems using Gaussian elimination. This form lays the groundwork for easily solving the variables from the bottom up.For our exercise, after performing row operations, the matrix obtained looks like this:\[\begin{bmatrix}2 & 3 & -4 & | & 5 \0 & \frac{11}{2} & -\frac{5}{2} & | & \frac{43}{2} \0 & 0 & -\frac{7}{2} & | & 15\end{bmatrix}\]Here, the lower-left part of the matrix contains all zeros below the pivots, simplifying further calculation. This format is crucial because, with zeros below the diagonal, it's straightforward to start back-substitution from the bottom of the matrix upward to solve for the variables \(x, y, \) and \(z\). The transition to an upper triangular form facilitates finding concrete values for these variables.

Back-Substitution

The method of back-substitution is a backwards approach used after the augmented matrix has been converted into an upper triangular form. This is the point where we solve for each variable step-by-step, starting with the last equation.In our upper triangular form matrix, the bottommost equation appears as:\[-\frac{7}{2}z = 15\]From this equation, solving for \(z\) is straightforward, resulting in \(z = -\frac{30}{7}\). With \(z\) known, we move up to the previous row to solve for \(y\):\[\frac{11}{2}y - \frac{5}{2}(-\frac{30}{7}) = \frac{43}{2}\]Solving this provides \(y = \frac{191}{77}\). Lastly, it's time to solve for \(x\) by substituting both \(y\) and \(z\) into the first row equation:\[2x + 3(\frac{191}{77}) - 4(-\frac{30}{7}) = 5\]Completing this calculation reveals \(x = -\frac{1649}{539}\). This phase encapsulates the last steps of the elimination method, providing definitive solutions for the variables by methodically backtracking through our triangular matrix.