Problem 19 - Precalculus

Question

In Exercises \(9-26\), put each system of linear equations into triangular form and solve the system if poesible. Classify each system as consistent independent, consistent dependent, or inconsistent. $$ \left\\{\begin{aligned} 2 x-y+z &=1 \\ 2 x+2 y-z &=1 \\ 3 x+6 y+4 z &=9 \end{aligned}\right. $$

Step-by-Step Solution

Answer

The system is consistent and independent, with solution \(x = \frac{1}{3}\), \(y = \frac{2}{3}\), \(z = 1\).

Step 1: Write the System in Augmented Matrix Form

To begin solving the system, we need to write the system of equations as an augmented matrix. The augmented matrix consists of the coefficients of the variables as well as the constants from the right-hand side:\[\begin{bmatrix}2 & -1 & 1 & | & 1 \ 2 & 2 & -1 & | & 1 \ 3 & 6 & 4 & | & 9 \end{bmatrix}\]

Step 2: Perform Row Operations to Achieve Triangular Form

Next, we'll use row operations to convert the augmented matrix into a triangular form.- Subtract Row 1 from Row 2: \[\begin{bmatrix}2 & -1 & 1 & | & 1 \ 0 & 3 & -2 & | & 0 \ 3 & 6 & 4 & | & 9 \end{bmatrix}\]- Subtract \(\frac{3}{2}\) times Row 1 from Row 3: \[\begin{bmatrix}2 & -1 & 1 & | & 1 \ 0 & 3 & -2 & | & 0 \ 0 & \frac{15}{2} & \frac{5}{2} & | & \frac{15}{2} \end{bmatrix}\]

Step 3: Make Further Row Reductions

Continue reducing to simplify:- Subtract \(\frac{5}{3}\) times Row 2 from Row 3: \[\begin{bmatrix}2 & -1 & 1 & | & 1 \ 0 & 3 & -2 & | & 0 \ 0 & 0 & 5 & | & 5 \end{bmatrix}\]

Step 4: Solve the Triangular System

The resulting upper triangular matrix can be used to solve for the variables by back substitution:- From Row 3: \(5z = 5\) implies \(z = 1\).- Substitute \(z = 1\) into Row 2: \(3y - 2(1) = 0\) leading to \(y = \frac{2}{3}\).- Substitute both \(z = 1\) and \(y = \frac{2}{3}\) into Row 1: \(2x - \frac{2}{3} + 1 = 1\) simplifies to \(x = \frac{1}{3}\).

Step 5: Classify the System

Now that we've solved for \(x\), \(y\), and \(z\), we can classify the system. Since the system has a unique solution, it is consistent and independent.

Key Concepts

Triangular Form

In the world of systems of linear equations, triangular form plays a crucial role in solving problems more efficiently. Triangular form means that the system of equations has been rearranged so that each equation (or row) contains fewer variables than the previous one. This makes it easier to solve the system through back substitution.

Imagine solving a mathematical puzzle; each step simplifies the equations further. When an equation system is in triangular form, the last equation usually contains only one variable. Moving upwards through the equations, each preceding equation introduces one more variable. This methodical organization allows us to solve the equations from the bottom up.

For instance, once we have an upper triangular form, the lower rows help eliminate variables from the upper ones, leading us to find individual variable values more straightforwardly. Think of it like peeling an onion, layer by layer, until the core is reached.

Augmented Matrix

An augmented matrix is a handy tool for representing and solving systems of equations. By transforming a set of linear equations into a matrix, we compactly store all the coefficients and constants involved in the system.

The augmented matrix merges the coefficient matrix, which contains the coefficients of variables, with the column matrix of constants found on the right side of the equations. This combination is separated by a vertical line or simply included as additional columns.

For example, a system

• 2x - y + z = 1
• 2x + 2y - z = 1
• 3x + 6y + 4z = 9

can be represented as: \[\begin{bmatrix}2 & -1 & 1 & | & 1 \2 & 2 & -1 & | & 1 \3 & 6 & 4 & | & 9 \end{bmatrix}\]This visualization helps perform operations to simplify and eventually solve the system efficiently.

Row Operations

Row operations are essential tools to alter augmented matrices without changing the solution set of the system they represent. There are three main types of row operations:

• Row swapping: Exchanging two rows in the matrix can help position leading coefficients strategically.
• Row multiplication: Multiplying a row by a non-zero constant can simplify equations or facilitate other operations.
• Row addition: Adding or subtracting a multiple of one row to/from another helps eliminate variables, gradually achieving triangular form.

These operations ensure that the equation's essence stays intact while gradually transforming the matrix towards a more easily solvable form. Think of them as tools for clearing away clutter, focusing more sharply on the ultimate solution.

Effective application of these operations helps to streamline the solution process, making the eventual back-substitution phase more manageable. Skillful manipulation using row operations is vital to mastering linear algebra and solving systems efficiently.

Consistent Independent System

A consistent independent system is a gold standard when it offers a unique solution for all variables. It means the equations intersect at exactly one point in the solution space, providing clear and definitive answers.

When solving a system, identifying whether it is consistent and independent can dictate how we approach the problem and interpret results:

- **Consistent** means that there is at least one solution. - **Independent** indicates that the solution is unique. In our example, solving the triangular matrix resulted in unique values for each variable. This unique solution clarifies that the system is consistent and independent.

The significance of recognizing this type of system lies in its predictability and reliability in applications. Systems that are consistent and independent provide exact solutions, ensuring robust decision-making in fields ranging from engineering to economics.