Problem 35 - College Algebra with Corequisite Support

Question

For the following exercises, solve each system by Gaussian elimination. $$ \begin{array}{l} -\frac{1}{4} x-\frac{5}{4} y+\frac{5}{2} z=-5 \\ -\frac{1}{2} x-\frac{5}{3} y+\frac{5}{4} z=\frac{55}{12} \\ -\frac{1}{3} x-\frac{1}{3} y+\frac{1}{3} z=\frac{5}{3} \end{array} $$

Step-by-Step Solution

Answer

x = \frac{21}{4}, y = \frac{38}{11}, z = \frac{6}{5}.

Step 1: Write the Augmented Matrix

First, write the given system of equations as an augmented matrix.\[\begin{bmatrix}-rac{1}{4} & -rac{5}{4} & rac{5}{2} & | & -5 \-rac{1}{2} & -rac{5}{3} & rac{5}{4} & | & rac{55}{12} \-rac{1}{3} & -rac{1}{3} & rac{1}{3} & | & rac{5}{3}\end{bmatrix}\]

Step 2: Eliminate Element Below First Pivot

To create zeros below the first pivot \(-\frac{1}{4}\) in column 1, we perform row operations.Multiply row 1 by 2 and add to row 2:\[\begin{bmatrix}-rac{1}{4} & -rac{5}{4} & rac{5}{2} & | & -5 \0 & -rac{11}{6} & rac{15}{4} & | & rac{5}{6} \-rac{1}{3} & -rac{1}{3} & rac{1}{3} & | & rac{5}{3}\end{bmatrix}\]

Step 3: Eliminate Second Element in Third Row

Multiply row 1 by 4/3 and add to row 3 to eliminate the first element.\[\begin{bmatrix}-rac{1}{4} & -rac{5}{4} & rac{5}{2} & | & -5 \0 & -rac{11}{6} & rac{15}{4} & | & rac{5}{6} \0 & -rac{5}{3} & rac{5}{2} & | & rac{25}{12}\end{bmatrix}\]

Step 4: Eliminate Element Below Second Pivot

Multiply row 2 by \(-\frac{5}{3}\) and add to row 3 to create a zero below the second pivot.\[\begin{bmatrix}-rac{1}{4} & -rac{5}{4} & rac{5}{2} & | & -5 \0 & -\frac{11}{6} & \frac{15}{4} & | & \frac{5}{6} \0 & 0 & \frac{25}{12} & | & \frac{5}{4}\end{bmatrix}\]

Step 5: Solve for z

The third row now reads: \[ \frac{25}{12}z = \frac{5}{4} \]Solve for \(z\) by multiplying both sides by \(\frac{12}{25}\).\[z = 1.2 = \frac{6}{5}\]

Step 6: Back Substitution to Find y

Substitute \(z = \frac{6}{5}\) into row 2:\[-\frac{11}{6} y + \frac{15}{4} \frac{6}{5} = \frac{5}{6}\]Simplify and solve for \(y\):\[-\frac{11}{6} y + \frac{9}{2} = \frac{5}{6}\]\[-\frac{11}{6} y = -\frac{38}{6}\]Thus,\[y = \frac{38}{11} = \frac{38}{11}\]

Step 7: Back Substitution to Find x

Substitute \(z = \frac{6}{5}\) and \(y = \frac{38}{11}\) back into row 1:\[-\frac{1}{4}x - \frac{5}{4}\frac{38}{11} + \frac{5}{2}\frac{6}{5} = -5\]Simplify and solve for \(x\):\[-\frac{1}{4}x -\frac{95}{22} + 3 = -5\]\[-\frac{1}{4}x = -\frac{21}{4}\]So, \(x = \frac{21}{4}\).

Key Concepts

Augmented Matrix

An augmented matrix is an essential tool in solving systems of equations using Gaussian elimination. It combines the coefficients and constants of a system of linear equations into a single, more manageable matrix form. By doing so, it allows us to use matrix operations instead of directly handling complex equations. For the given system of equations: \[\begin{align*}-\frac{1}{4}x - \frac{5}{4}y + \frac{5}{2}z &= -5\-\frac{1}{2}x - \frac{5}{3}y + \frac{5}{4}z &= \frac{55}{12}\-\frac{1}{3}x - \frac{1}{3}y + \frac{1}{3}z &= \frac{5}{3}\end{align*}\] We write the augmented matrix as: \[\begin{bmatrix}-\frac{1}{4} & -\frac{5}{4} & \frac{5}{2} & | & -5 \-\frac{1}{2} & -\frac{5}{3} & \frac{5}{4} & | & \frac{55}{12} \-\frac{1}{3} & -\frac{1}{3} & \frac{1}{3} & | & \frac{5}{3}\end{bmatrix}\] This representation simplifies the analysis and manipulation of the system, making steps like row reduction and back substitution much easier to handle.

Row Operations

Row operations are fundamental actions applied to the rows of an augmented matrix, which help implement the Gaussian elimination process. The goal is to transform the matrix into a row-echelon form, where we have zeros below the pivot positions (the leading non-zero entry in each row). These operations include:

• Swapping two rows
• Multiplying a row by a non-zero scalar
• Adding or subtracting a multiple of one row from another row

In the given solution, row operations are performed to create zeros below each pivot, progressively simplifying the matrix. For example, multiplying the first row by 2 and adding it to the second row eliminates the \(-\frac{1}{2}\) term in the lower row. These operations maintain the equivalence of the system while reshaping it for easier solving.

Back Substitution

Back substitution is the final step after reaching the row-echelon form through Gaussian elimination. When the matrix is in this form, you typically have an upper triangular matrix where all entries below the pivot elements are zeros. This form allows you to solve for the variables starting from the bottom row upwards.In the given example, the third row ends with a simple equation in \(z\): \[\frac{25}{12}z = \frac{5}{4}\] Solving it gives \(z = \frac{6}{5}\). Next, back substitute \(z\) into the second equation to find \(y\). After finding \(y\), both solutions are substituted back into the first equation to find \(x\). This orderly process ensures that each variable is solved in terms of the known quantities, leading to a complete solution set for the system of equations.

System of Equations

A system of equations is a collection of two or more equations involving the same set of variables. Solving a system means finding all possible values of the variables that satisfy each equation in the system simultaneously. Systems can be:

• Consistent, having at least one solution
• Inconsistent, having no solution
• Dependent, having infinitely many solutions

The given exercise involves solving a system of three linear equations in three variables. Gaussian elimination is a systematic method that transforms these equations into an upper triangular matrix form (through the use of row operations), eventually ensuring that we can isolate each variable. This methodology effectively narrows down the solution of a potentially complex system into a series of straightforward arithmetic equations, providing clarity and build-up to the equations' solutions.