Problem 29 - College Algebra

Question

Solve each system by Gaussian elimination. $$ \begin{aligned} 6 x-5 y+6 z &=38 \\ \frac{1}{5} x-\frac{1}{2} y+\frac{3}{5} z &=1 \\ -4 x-\frac{3}{2} y-z &=-74 \end{aligned} $$

Step-by-Step Solution

Answer

The solution to the system is \(x = 6\), \(y = 4\), \(z = -\frac{131}{104}\).

Step 1: Write the Augmented Matrix

The first step is to convert the system of equations into an augmented matrix. For the given system: \[\begin{bmatrix} 6 & -5 & 6 & | & 38 \ \frac{1}{5} & -\frac{1}{2} & \frac{3}{5} & | & 1 \ -4 & -\frac{3}{2} & -1 & | & -74 \end{bmatrix}\]

Step 2: Eliminate x from Row 2 and Row 3

To eliminate \(x\) from the second and third row, perform these row operations: - For Row 2: Multiply Row 1 by \(\frac{1}{30}\) and subtract from Row 2.- For Row 3: Multiply Row 1 by \(-\frac{2}{3}\) and add to Row 3.The updated matrix is: \[\begin{bmatrix} 6 & -5 & 6 & | & 38 \ 0 & \frac{19}{30} & -\frac{23}{30} & | & -\frac{19}{30} \ 0 & -\frac{23}{3} & 3 & | & -\frac{56}{3} \end{bmatrix}\]

Step 3: Eliminate y from Row 3

Now eliminate \(y\) from Row 3 by performing the operation: Subtract \(\frac{-23}{3} \div \frac{19}{30}\) times Row 2 from Row 3. The updated matrix is: \[\begin{bmatrix} 6 & -5 & 6 & | & 38 \ 0 & \frac{19}{30} & -\frac{23}{30} & | & -\frac{19}{30} \ 0 & 0 & \frac{416}{19} & | & \frac{-524}{19} \end{bmatrix}\]

Step 4: Solve for z

Since Row 3 now only has \(z\) as the variable, solve for \(z\): \[\frac{416}{19}z = \frac{-524}{19} \Rightarrow z = -\frac{524}{416} = -\frac{131}{104}\]

Step 5: Back-Substitute to Solve for y

Use the value of \(z\) to solve for \(y\) in Row 2: \[\frac{19}{30}y - \frac{23}{30} \left(-\frac{131}{104}\right) = -\frac{19}{30}\] Simplifying: - \(\frac{19}{30}y + \frac{3013}{3120} = -\frac{19}{30}\)- Solve for \(y\) to find \(y = 4\).

Step 6: Back-Substitute to Solve for x

With \(y\) and \(z\) known, substitute into Row 1 to solve for \(x\): \[6x - 5(4) + 6\left(-\frac{131}{104}\right) = 38\] Calculate: - \(6x - 20 - \frac{786}{104} = 38\)- Solve for \(x\) to find \(x = 6\).

Key Concepts

Augmented Matrix

An augmented matrix is a crucial tool when solving systems of linear equations using Gaussian elimination. It is created by converting a system of linear equations into matrix form where each equation's coefficients become a row in the matrix.
This matrix includes an additional column at the end which represents the constants from the right-hand side of the equations. In this way, the augmented matrix compactly represents the entire system:

• The left portion contains the coefficients of the variables.
• The rightmost column represents the constants from each equation.

For example, given the system:\[ 6x - 5y + 6z = 38 \ \frac{1}{5}x - \frac{1}{2}y + \frac{3}{5}z = 1 \ -4x - \frac{3}{2}y - z = -74\]The corresponding augmented matrix is:\[\begin{bmatrix} 6 & -5 & 6 & | & 38 \\frac{1}{5} & -\frac{1}{2} & \frac{3}{5} & | & 1 \-4 & -\frac{3}{2} & -1 & | & -74 \ \end{bmatrix}\]Remember, the augmented matrix serves as a foundation for applying row operations to solve the system.

Row Operations

Row operations are the manipulations we perform on the rows of an augmented matrix to simplify it. These are essential to the Gaussian elimination method, helping us transform the matrix and ultimately solve the system.
There are three main types of row operations:

• Swapping rows: You can exchange the positions of any two rows.
• Multiplying a row by a non-zero scalar: This scales all entries of the row by the same number.
• Adding or subtracting a multiple of one row to another: This is used to create zeros in specific positions to simplify the system.

For example, in the given problem:

• To eliminate \(x\) from Row 2, you might subtract a multiple of Row 1 from Row 2.
• This changes the system step by step towards a simpler form until it's ready for back-substitution.

It's important to apply these operations meticulously while maintaining the matrix's structural integrity.

Back-Substitution

After transforming the augmented matrix into an upper triangular form via row operations, the next step is back-substitution. This is the process used to solve the simplified system of equations.
Back-substitution starts from the bottom row of the matrix and works upwards to find the values of the variables. Here’s a general approach:

• Identify the row with the fewest variables, usually with only one variable term.
• Solve for that variable directly since no other unknowns appear in this equation.
• Substitute the found value into the equations above it to solve for the remaining variables.

In this problem, once the matrix is reduced to have zeros below the diagonal, you solve for \(z\) first, like this:\[\frac{416}{19}z = \frac{-524}{19} \implies z = -\frac{131}{104}\]With \(z\) known, substitute its value into Row 2 to find \(y\) and further into Row 1 to find \(x\). Methodically solving in this reverse order ensures each calculation builds on previously found results.

System of Linear Equations

A system of linear equations consists of multiple linear equations with several variables. Each equation represents a line on a graph, and the solution to the system is the point or set of points where these lines intersect.
In solving systems of equations, Gaussian elimination is a systematic method that involves representing the system in augmented matrix form and applying row operations to simplify. The ultimate goal is to reach a form where it's easy to discern the solution by back-substitution.
For the system:\[6x - 5y + 6z = 38 \\frac{1}{5}x - \frac{1}{2}y + \frac{3}{5}z = 1 \-4x - \frac{3}{2}y - z = -74\]Gaussian elimination helps find the unique set of \((x, y, z)\) values that satisfy all three equations simultaneously.
Understanding the geometry behind the system can also help visualize why certain solutions appear. Each variable corresponds to a dimension, and the system's solution lies where these planes intersect.