Problem 18 - Precalculus with Limits

Question

Performing Row Operations In Exercises 17 and \(18 ,\) perform the row operation and write the equivalent system. Add \(- 2\) times Equation 1 to Equation 3 $$\left\\{ \begin{aligned} x - 2 y + 3 z = 5 & \text { Equation } 1 \\ - x + 3 y - 5 z = 4 & \text { Equation } 2 \\\ 2 x - 3 z = 0 & \text { Equation } 3 \end{aligned} \right.$$ What did this operation accomplish?

Step-by-Step Solution

Answer

The row operation gives the new Equation 3 as \(2y - 3z = -10\). The equivalent system after the operation is \[\begin{align*} x - 2y + 3z &= 5 \-x + 3y - 5z &= 4 \ 2y - 3z &= -10 \end{align*}\]. This operation eliminated 'x' from Equation 3, thus simplifying the system.

Step 1: Apply the Row Operation

First, the operation wants to add \(-2\) times Equation 1 to Equation 3. Perform this by multiplying each term in Equation 1 by \(-2\) and then adding each corresponding term in Equation 3. This gives the new Equation 3 as: \(-2x + 4y - 6z = -10 + 2x -3z = 0\), which simplifies to: \(2y - 3z = -10\).

Step 2: Write the New System

Substitute the new Equation 3 into the original system. The equivalent system after the row operation is: \[\begin{align*} x - 2y + 3z &= 5 \ -x + 3y - 5z &= 4 \ 2y - 3z &= -10 \end{align*}\]

Step 3: Interpret the Result

By performing this operation, the 'x' variable has been eliminated from Equation 3. This operation makes the system simpler and one step closer to being in echelon form, which is a desirable form when using methods like back-substitution to solve the system.

Key Concepts

Row Operations

Row operations are foundational tools in linear algebra used to simplify systems of linear equations. They involve three main types: swapping two rows, multiplying a row by a non-zero constant, and adding a multiple of one row to another row. In exercises involving row operations, like the one above, these actions are performed to make the system easier to solve.

For instance, in the given system, adding \(-2\) times Equation 1 to Equation 3 is a form of simplification. The objective is often to eliminate a variable in one of the equations, making it easier to solve the system step-by-step. This type of operation is useful when you're working towards a triangular form, which is closer to echelon or reduced echelon form. When variables are eliminated correctly, it reduces complexity, paving the way to find solutions.

System of Equations

A system of equations is a set of two or more equations with the same set of unknowns. In this exercise, the system consists of three equations in three variables: \(x, y,\) and \(z\). Solving the system means finding values for these variables that satisfy all equations simultaneously.

These systems can be linear, where each equation represents a line in two dimensions or a plane in three dimensions. Solving the system involves finding where these lines or planes intersect. Row operations are often employed to simplify the system by eliminating variables, finding intersections methodically, and making it easier to interpret the relationships between the equations.

Echelon Form

The echelon form of a matrix or system of equations is a step towards simplifying a problem in linear algebra. It describes a format where each successive row has more leading zeros than the previous one. This triangular shape helps in solving systems using methods like back-substitution.

Transforming the system into this form involves eliminating variables from equations, a task typically accomplished via row operations. In our example, by performing the operation \(-2\) times Equation 1 added to Equation 3, the 'x' variable is removed from Equation 3. This adjustment helps transition the system closer to echelon form, facilitating more straightforward problem-solving.

Back-Substitution

Back-substitution is a method used to solve a system of equations that has been transformed into echelon form. Once a system is in this form, solving begins from the last equation up to the first.

Consider the echelon form you might achieve with the given system. Start from the last equation, which has the fewest variables, and solve for one variable. Then, substitute this value back into previous equations to find other variables. This process repeats until you've found all variable values.

This technique is efficient and practical, especially when used in concert with row operations and transforming a system into echelon form, unveiling solutions in a tiered manner.