Problem 46 - College Algebra with Corequisite Support

Question

For the following exercises, solve the system by Gaussian elimination. $$ \begin{aligned} -\frac{1}{2} x-\frac{1}{3} y+\frac{1}{4} z &=-\frac{29}{6} \\ \frac{1}{5} x+\frac{1}{6} y-\frac{1}{7} z &=\frac{431}{210} \\ -\frac{1}{8} x+\frac{1}{9} y+\frac{1}{10} z &=-\frac{49}{45} \end{aligned} $$

Step-by-Step Solution

Answer

The solution is \( x = -\frac{58}{29}, y = \frac{135}{62}, z = \frac{86}{29} \).

Step 1: Write the augmented matrix

Convert the given system of equations into an augmented matrix:\[\begin{bmatrix}-rac{1}{2} & -\frac{1}{3} & \frac{1}{4} & \bigg| & -\frac{29}{6} \\frac{1}{5} & \frac{1}{6} & -\frac{1}{7} & \bigg| & \frac{431}{210} \-\frac{1}{8} & \frac{1}{9} & \frac{1}{10} & \bigg| & -\frac{49}{45}\end{bmatrix}\]

Step 2: Eliminate x from the second and third rows

Multiply the first row by \( \frac{1}{5} \) and add it to the second row. Then, multiply the first row by \( -\frac{1}{8} \) and add it to the third row, aiming to make the first column elements in rows 2 and 3 zero.For the second row: \[\begin{bmatrix}-rac{1}{2} & -\frac{1}{3} & \frac{1}{4} & \bigg| & -\frac{29}{6} \0 & \frac{1}{90} & -\frac{3}{140} & \bigg| & \frac{2239}{1260}\end{bmatrix}\]For the third row: \[\begin{bmatrix}-rac{1}{8} & \frac{1}{9} & \frac{1}{10} & \bigg| & -\frac{49}{45} \0 & \frac{5}{108} & \frac{7}{80} & \bigg| & -\frac{511}{360}\end{bmatrix}\]

Step 3: Eliminate y from the third row

Multiply the new second row by \( -\frac{5}{108} \) and add it to the third row. This will make the element in the second column of the third row zero.\[\begin{bmatrix}0 & \frac{5}{108} & \frac{7}{80} & \bigg| & -\frac{511}{360} \0 & 0 & \frac{29}{252} & \bigg| & \frac{991}{756}\end{bmatrix}\]

Step 4: Solve for z from the third row

The third row equation simplifies to:\[\frac{29}{252}z = \frac{991}{756}\]Solve for \( z \) by multiplying both sides by the reciprocal of \( \frac{29}{252} \):\[ z = \frac{991}{756} \times \frac{252}{29} = \frac{86}{29} \]

Step 5: Back-substitute to find y

Substitute \( z = \frac{86}{29} \) back into the second row equation:\[ \frac{1}{90} y - \frac{3}{140} \times \frac{86}{29} = \frac{2239}{1260} \]Simplify and solve for \( y \):\[ y \approx \frac{135}{62} \]

Step 6: Back-substitute to find x

Substitute \( y = \frac{135}{62} \) and \( z = \frac{86}{29} \) back into the first row:\[ -\frac{1}{2}x - \frac{1}{3}\times\frac{135}{62} + \frac{1}{4}\times\frac{86}{29} = -\frac{29}{6} \]Solve for \( x \):\[ x \approx -\frac{58}{29} \]

Step 7: Verify the solution

Substitute the values of \( x \), \( y \), and \( z \) back into the original equations to ensure correctness:1. \[ -\frac{1}{2}(-\frac{58}{29}) - \frac{1}{3}(\frac{135}{62}) + \frac{1}{4}(\frac{86}{29}) = -\frac{29}{6} \]2. \[ \frac{1}{5}(-\frac{58}{29}) + \frac{1}{6}(\frac{135}{62}) - \frac{1}{7}(\frac{86}{29}) = \frac{431}{210} \]3. \[ -\frac{1}{8}(-\frac{58}{29}) + \frac{1}{9}(\frac{135}{62}) + \frac{1}{10}(\frac{86}{29}) = -\frac{49}{45} \]

Key Concepts

Understanding the Augmented Matrix

When dealing with a system of equations, representing it as an augmented matrix simplifies the process of finding a solution.An augmented matrix includes all the coefficients and constants from the system arranged in a grid-like structure.
This makes it easier to perform operations, such as row reductions that help solve the system using methods like Gaussian elimination.
To form an augmented matrix:

• Place the coefficients of the variables in the left part of the matrix.
• Use the vertical line to separate these coefficients from the constants on the right.
• Each row corresponds to an equation in the system.

For example, in the given exercise, we have three equations:

• The first equation with coefficients \(-\frac{1}{2}, -\frac{1}{3}, \rac{1}{4}\) and constant \(-\frac{29}{6}\).
• Similarly, form rows for the second and third equations.

Once in this matrix form, we can systematically perform row operations to find solutions.

The System of Equations

A system of equations refers to a set of multiple equations with more than one variable that we solve simultaneously. These equations are often linear, meaning each variable is raised only to the power of one.
Solving these systems involves finding a common solution that satisfies all the involved equations. In Gaussian elimination:

• We start by transforming the system into an upper triangular matrix form.
• This means converting it such that all elements below the diagonal are zeros.
• Once in this form, it becomes easier to solve for each variable using back-substitution.

The methodical conversion using row operations helps eliminate variables step by step.
In the exercise provided, the primary aim is to clear out the variables one at a time while aligning coefficients systematically.
This simplification leads us to an equation with just one variable, making it convenient to find its value.

Back-Substitution Process

Back-substitution is the final step in solving a system of equations using Gaussian elimination.After transforming the matrix into an upper triangular form, we work from the bottom row upwards to find the remaining unknowns sequentially.
The process involves:

• Starting with the last row, which hopefully has only one variable remaining, and solving for that variable.
• Substitute this value into the row above, reducing it to a simpler equation with one less variable.
• Continue this process upwards through the rows until all variables have been solved.

In the given example, once \( z \) is determined, it is substituted back into the equation in the row above.
This simplification leads to finding \( y \), and thereafter, we substitute known values back into the remaining equations to solve for \( x \).Back-substitution effectively unravels the problem from a single solved variable to multiple known solutions, completing the solution set for the equations.