Problem 50 - A Graphical Approach to College Algebra

Question

Use row operations on an augmented matrix to solve each system of equations. Round to nearest thousandth when appropriate. $$\begin{aligned} 2 x+y-3 z &=1 \\ x+y+2 z &=5 \\ 3 x+2 y-z &=-3 \end{aligned}$$

Step-by-Step Solution

Answer

The solution is \(x = 9.0\), \(y = -12.8\), \(z = 4.4\).

Step 1: Write the Augmented Matrix

Begin by writing the system of equations as an augmented matrix. The given system is:\[\begin{align*}2x + y - 3z &= 1 \x + y + 2z &= 5 \3x + 2y - z &= -3\end{align*}\]This corresponds to the augmented matrix:\[\begin{bmatrix}2 & 1 & -3 & | & 1 \1 & 1 & 2 & | & 5 \3 & 2 & -1 & | & -3\end{bmatrix}\]

Step 2: Eliminate x from the Second and Third Rows

We want to use row operations to eliminate the variable \(x\) from the second and third rows. First, subtract two times the second row from the first row and replace Row 1:\[ R_1 = R_1 - 2R_2 \]\[\begin{bmatrix}1 & -1 & -7 & | & -9 \1 & 1 & 2 & | & 5 \3 & 2 & -1 & | & -3\end{bmatrix}\]Next, eliminate \(x\) from Row 3 by subtracting three times Row 2 from Row 3:\[ R_3 = R_3 - 3R_2 \]\[\begin{bmatrix}1 & -1 & -7 & | & -9 \1 & 1 & 2 & | & 5 \0 & -1 & -7 & | & -18\end{bmatrix}\]

Step 3: Simplify the Second Row

Use the row operations to further simplify the second row. Subtract Row 1 from Row 2:\[ R_2 = R_2 - R_1 \]The resulting matrix is:\[\begin{bmatrix}1 & -1 & -7 & | & -9 \0 & 2 & 9 & | & 14 \0 & -1 & -7 & | & -18\end{bmatrix}\]

Step 4: Solve for y and z

Add Row 2 to Row 3 to eliminate \(y\) from Row 3:\[ R_3 = R_3 + R_2 \]\[\begin{bmatrix}1 & -1 & -7 & | & -9 \0 & 2 & 9 & | & 14 \0 & 1 & 2 & | & -4\end{bmatrix}\]Now Row 3 can be used to simplify for \(y\):Subtract twice Row 3 from Row 2:\[ R_2 = R_2 - 2R_3 \]\[\begin{bmatrix}1 & -1 & -7 & | & -9 \0 & 0 & 5 & | & 22 \0 & 1 & 2 & | & -4\end{bmatrix}\]From this, solve for \(z\):\[ 5z = 22 \Rightarrow z = 4.4 \approx 4.4 \]

Step 5: Back Substitute to Solve for x and y

With \(z = 4.4\), substitute back into the third row to solve for \(y\):\[ y + 2(4.4) = -4 \]\[ y + 8.8 = -4 \]\[ y = -12.8 \approx -12.8 \]Finally, solve for \(x\) using the first equation:\[ x - (-12.8) - 7(4.4) = -9 \]\[ x + 12.8 - 30.8 = -9 \]\[ x = 9 \approx 9.0 \]

Step 6: Present the Solution

After solving all equations, the solution to the system is:\[x = 9.0, \quad y = -12.8, \quad z = 4.4\]

Key Concepts

Row Operations

Row operations are a crucial technique in linear algebra, especially when dealing with systems of equations via augmented matrices. These operations help us manipulate rows of a matrix to simplify and eventually solve the equations. There are three types of row operations:

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

These operations are performed to transform the matrix into a simpler form, typically aiming for a row-echelon form. This makes it easier to solve variables systematically. For instance, in our exercise, we first eliminate certain variables from rows to simplify the solving process. Much like peeling the layers of an onion, row operations gradually reveal the solution to the system.

System of Equations

A system of equations consists of multiple equations that share a set of variables. The goal is to find values for these variables that would satisfy all the given equations simultaneously. In our example, we have three equations involving three variables: \( x \), \( y \), and \( z \).
By representing a system of equations as an augmented matrix, it becomes easier to apply systematic approaches like row operations to find these variable values. These systems are commonly encountered in fields that require modeling and solving real-world problems, such as engineering and economics.
Efficient methods for solving them include substitution, elimination, and using matrices, of which the latter is most powerful in complex larger systems. Linear algebra offers rich techniques for these solutions, making mathematics an essential tool for problem-solving.

Linear Algebra

Linear algebra is a branch of mathematics focused on vectors, matrices, and linear transformations. It provides a potent framework for solving systems of equations, like the one in our exercise.
When working with an augmented matrix, as we did, linear algebra helps us visualize equations in a compact form and utilize matrix operations to streamline solutions. It's widely used because it can handle large and intricate systems, offering neat and elegant methods for finding solutions.
Core principles of linear algebra, such as vector spaces and eigenvalues, offer deeper insights and tools beyond simple matrix operations. Yet, mastering basic row operations and systems of equations is foundational for utilizing these advanced techniques effectively.

Back Substitution

Back substitution is a method used after obtaining a row-echelon form or reduced row-echelon form of a matrix. It involves solving for the variables from the bottom row upwards, hence the term 'back'.
Once we have simplified our augmented matrix, typically through row operations, we start with the last row which usually has the simplest form - involving only one of the variables. Solving for this variable, we substitute its value into the preceding rows. This step-by-step substitution continues until all variables are solved.
In our exercise, once \( z \) was found, it was substituted back to solve \( y \), and finally, \( x \). This technique ensures that each variable is accurately determined based on those already computed. It is an essential step to ensure the solution is both complete and consistent with all original equations.