Problem 38 - College Algebra with Corequisite Support

Question

For the following exercises, solve the system by Gaussian elimination. $$ \begin{aligned} x+y-4 z &=-4 \\ 5 x-3 y-2 z &=0 \\ 2 x+6 y+7 z &=30 \end{aligned} $$

Step-by-Step Solution

Answer

The solution is \( x = -\frac{11}{32}, \ y = \frac{59}{32}, \ z = \frac{11}{8} \).

Step 1: Write the Augmented Matrix

Start by writing the system of equations in an augmented matrix form. The system given is: \[\begin{cases} 1x + 1y - 4z = -4 \5x - 3y - 2z = 0 \2x + 6y + 7z = 30\end{cases} \]The augmented matrix is: \[\begin{bmatrix} 1 & 1 & -4 & | & -4 \5 & -3 & -2 & | & 0 \2 & 6 & 7 & | & 30 \end{bmatrix}\]

Step 2: Perform Row Operations to Obtain Zero Below First Pivot

Use row operations to create zeros below the first pivot in the first column. Subtract 5 times the first row from the second row and 2 times the first row from the third row:\[R2 = R2 - 5R1 \rightarrow \begin{bmatrix} 5 & -3 & -2 & | & 0 \-5 & -5 & 20 & | & 20 \end{bmatrix}\]\[R3 = R3 - 2R1 \rightarrow \begin{bmatrix} 2 & 6 & 7 & | & 30 \-2 & -2 & 8 & | & 8 \end{bmatrix}\]Updated matrix:\[\begin{bmatrix} 1 & 1 & -4 & | & -4 \0 & -8 & 18 & | & 20 \0 & 4 & 15 & | & 38 \end{bmatrix}\]

Step 3: Perform Row Operations to Obtain Zero Below Second Pivot

Divide the second row by -8 to make the leading coefficient 1, then use row operations to create zero below this pivot in the third row:\[R2 = \frac{1}{-8} R2 \rightarrow \begin{bmatrix} 0 & 1 & -\frac{9}{4} & | & -\frac{5}{4} \end{bmatrix}\]Subtract 4 times the new second row from the third row:\[R3 = R3 - 4R2 \rightarrow \begin{bmatrix} 0 & 0 & 24 & | & 33 \end{bmatrix}\]Updated matrix:\[\begin{bmatrix} 1 & 1 & -4 & | & -4 \0 & 1 & -\frac{9}{4} & | & -\frac{5}{4} \0 & 0 & 24 & | & 33 \end{bmatrix}\]

Step 4: Solve for z

Now the matrix is in row echelon form and we can start back substitution. The third row yields the equation:\[ 24z = 33 \]Solving for \( z \) gives:\[ z = \frac{33}{24} = \frac{11}{8} \]

Step 5: Back Substitute to Solve for y

Substitute \( z = \frac{11}{8} \) back into the second row:\[1y - \frac{9}{4}\cdot\frac{11}{8} = -\frac{5}{4} \]Simplifying:\[1y - \frac{99}{32} = -\frac{5}{4} \]Convert \(-\frac{5}{4}\) to \(-\frac{40}{32}\) and solve for \( y \):\[y = -\frac{40}{32} + \frac{99}{32} = \frac{59}{32} \]

Step 6: Back Substitute to Solve for x

Substitute \( y = \frac{59}{32} \) and \( z = \frac{11}{8} \) back into the first row:\[1x + \frac{59}{32} - 4\cdot\frac{11}{8} = -4 \]Simplifying gives:\[1x + \frac{59}{32} - \frac{44}{8} = -\frac{128}{32} \]Combining like terms:\[1x + \frac{59 - 176}{32} = -\frac{128}{32} \]\[1x = -\frac{128}{32} + \frac{117}{32} \]\[x = -\frac{11}{32} \]

Step 7: Conclusion

The solution to the system is:\( x = -\frac{11}{32}, \ y = \frac{59}{32}, \ z = \frac{11}{8} \)

Key Concepts

Augmented Matrix

An augmented matrix is a compact way to capture the structure of a system of linear equations. It helps streamline the process of solving these systems using techniques such as Gaussian elimination. For a linear system, the augmented matrix is constructed by placing the coefficients of each variable from the equations into a matrix, with an additional column for the constants on the right side of each equation.

This matrix essentially becomes a snapshot of all equations with their variables and constants aligned in a structured grid. For example, consider the following system of equations:

• \( x + y - 4z = -4 \)
• \( 5x - 3y - 2z = 0 \)
• \( 2x + 6y + 7z = 30 \)

Converting this into an augmented matrix would give us: \[\begin{bmatrix} 1 & 1 & -4 & | & -4 \5 & -3 & -2 & | & 0 \2 & 6 & 7 & | & 30 \end{bmatrix} \] Notice how each row corresponds to an equation and each column (excluding the augmented part) corresponds to a particular variable’s coefficients, after which the constants follow on the right side.

Row Operations

Row operations are the main tools we use in Gaussian elimination to manipulate an augmented matrix and eventually solve linear systems. These operations are essential for systematically reducing a matrix to a form that is easier to interpret for solutions.

There are three types of row operations:

• Swapping two rows
• Multiplying a row by a non-zero constant
• Adding or subtracting the multiples of rows from one another

Using these operations, you can transform the matrix into simpler forms, such as making zeros below pivot positions. For instance, to clear out the first column below the first pivot (which is the top-leftmost number), you might subtract a suitable multiple of the first row from the other rows. This is how you progress through the steps of Gaussian elimination, transforming the matrix while retaining its solution set intact.

Back Substitution

Back substitution is the final stage of Gaussian elimination used once the matrix is in row echelon form. Essentially, it involves solving for the variables starting from the last equation, which should ideally only involve one variable, then substituting back into previous equations to find the remaining unknowns.

For example, after transforming our matrix, we might end up with a simplified form like: \[\begin{bmatrix} 1 & 1 & -4 & | & -4 \0 & 1 & -\frac{9}{4} & | & -\frac{5}{4} \0 & 0 & 24 & | & 33 \end{bmatrix} \] In this situation, solve the third row for \( z \) to get \( z = \frac{11}{8} \). Next, substitute \( z \) into the second row equation to solve for \( y \), and finally, use the values of \( y \) and \( z \) in the first row to determine \( x \). This process systematically narrows down each variable step-by-step, from the bottom row up.

Row Echelon Form

The row echelon form (REF) of a matrix is an intermediate step in solving systems of linear equations using Gaussian elimination. It represents a staircase-like structure in the matrix where each leading non-zero entry of a row is to the right of the leading entry in the row above. This form is achieved through a series of row operations.

REF is characterized by:

• All non-zero rows are above any rows of all zeros
• The leading entry (pivot) of each non-zero row after the first occurs to the right of the leading entry of the previous row
• All entries in a column below a leading entry are zeros

Once a matrix is in row echelon form, back substitution can be applied to find the solutions of the system. For example, transforming the augmented matrix into this form involves operations such as subtracting multiples of rows from others to clear out entries below the pivots. This reduces the matrix into a more manageable shape that directly correlates to simplified equations within the system.