Problem 31 - A Graphical Approach to Precalculus with Limits

Question

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

Step-by-Step Solution

Answer

The solution is x = 2 and y = 3.

Step 1: Write the Augmented Matrix

First, translate the given system of equations into an augmented matrix form. The system is:\[\begin{align*}x + y &= 5\x - y &= -1\end{align*}\]The augmented matrix is:\[\begin{bmatrix}1 & 1 & | & 5 \1 & -1 & | & -1\end{bmatrix}\]

Step 2: Use Row Operations to Obtain a Row Echelon Form

Our goal is to transform this matrix into row-echelon form. Start with eliminating the first element of the second row by subtracting the first row from the second row.Subtract Row 1 from Row 2:\[R_2 = R_2 - R_1\]\[\begin{bmatrix}1 & 1 & | & 5 \0 & -2 & | & -6\end{bmatrix}\]

Step 3: Simplify to Identity Matrix Form

Transform the matrix to get the leading coefficient of the second row as 1. Divide the second row by -2:\[R_2 = \frac{1}{-2} R_2\]\[\begin{bmatrix}1 & 1 & | & 5 \0 & 1 & | & 3\end{bmatrix}\]

Step 4: Perform Back Substitution to Solve for x

Subtract 1 times Row 2 from Row 1 to isolate the value of \(x\):\[R_1 = R_1 - R_2\]\[\begin{bmatrix}1 & 0 & | & 2 \0 & 1 & | & 3\end{bmatrix}\]The matrix corresponds to the solutions \(x = 2\) and \(y = 3\).

Step 5: Final Step: Write the Solution

The solution to the system of equations is \(x = 2\) and \(y = 3\).

Key Concepts

Row Operations

Row operations are key moves we perform on rows of a matrix to help solve systems of equations in a straightforward manner. There are three main types of row operations:

• Row swapping: This changes the position of two rows in the matrix, which is sometimes necessary for easier calculations.
• Row scaling: This involves multiplying all entries of a row by a non-zero constant to get a desired number.
• Addition or subtraction of rows: Here, you either add or subtract one row from another, often used to eliminate variables.

Row operations can be performed without changing the solution set of the system, allowing us to manipulate a matrix into a simpler form that is easier to solve. They are fundamental in transforming a matrix into row-echelon form, which serves as an essential step towards finding solutions.

System of Equations

A system of equations is formed when two or more equations with the same set of variables are considered together. These systems can be solved to find the variable values that satisfy all equations simultaneously.
In our example, the given system is:

• \(x + y = 5\)
• \(x - y = -1\)

The goal is to find values for \(x\) and \(y\) that make both of these equations true at the same time. The challenge lies in the interplay between the equations, where solving them together gives us a unique solution, no solution, or infinitely many solutions, depending on the conditions defined by the equations.

Row-Echelon Form

The row-echelon form of a matrix is a type of matrix form that is simpler to work with when solving systems of equations. To be in row-echelon form, a matrix must satisfy the following criteria:

• The leading non-zero entry in each row, called the leading coefficient, is 1.
• Each leading coefficient is to the right of the leading coefficient of the row above it.
• All entries in a column below a leading coefficient are zero.

In our solution, the augmented matrix was transformed using row operations to:\[\begin{bmatrix} 1 & 1 & | & 5 \ 0 & 1 & | & 3 \end{bmatrix}\]This form makes it easy to perform the final steps of solving the system, as it neatly lines up equations for straightforward back substitution, where each equation represents a simple expression for one of the variables.

Back Substitution

Back substitution is the final process in solving a system of equations using matrices once it has been transformed to row-echelon form. With this method, solutions are derived starting from the last equation moving upward.
For the matrix at:\[\begin{bmatrix} 1 & 0 & | & 2 \ 0 & 1 & | & 3 \end{bmatrix}\]we see that the second row gives \(y = 3\). Knowing the value of \(y\), we substitute back into the first row to solve for \(x\):

• From \(1x + 0y = 2\) yields \(x = 2\).

This method ensures that all equations are considered correctly and efficiently, starting with the simplest equations and working backwards to solve for each variable. It works hand in hand with row operations and matrix transformations to provide clear solutions to complex systems.