Problem 41 - 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. $$\begin{aligned} &x+y=-1\\\ &y+z=4\\\ &x+z=1 \end{aligned}$$

Step-by-Step Solution

Answer

The solution is \(x = -2\), \(y = 1\), \(z = 3\).

Step 1: Write the Augmented Matrix

First, express the given system of equations as an augmented matrix. The system is:\[\begin{aligned} &x+y=-1\ &y+z=4\ &x+z=1\end{aligned}\] The augmented matrix corresponding to the system is:\[\begin{bmatrix} 1 & 1 & 0 & |-1 \ 0 & 1 & 1 & |4 \ 1 & 0 & 1 & |1\end{bmatrix}\]

Step 2: Use Row Operations to Simplify

The objective is to simplify the matrix into row-echelon form using elementary row operations. Start by subtracting the first row from the third row to eliminate the first column variable in the third row:\[\begin{bmatrix} 1 & 1 & 0 & |-1 \ 0 & 1 & 1 & |4 \ 0 & -1 & 1 & |2\end{bmatrix}\]

Step 3: Eliminate y from Third Row

Add the second row to the third row to eliminate the y variable in the third row:\[\begin{bmatrix} 1 & 1 & 0 & |-1 \ 0 & 1 & 1 & |4 \ 0 & 0 & 2 & |6\end{bmatrix}\]

Step 4: Solve for z

Divide the third row by 2 to solve for z:\[z = \frac{6}{2} = 3\]. Now, rewrite the third row as:\[\begin{bmatrix} 1 & 1 & 0 & |-1 \ 0 & 1 & 1 & |4 \ 0 & 0 & 1 & |3\end{bmatrix}\]

Step 5: Back-Substitute to Find y

Use the second row to solve for y, given \(1(y) + 1(z) = 4\) with \(z = 3\). Substitute \(z = 3\) in the second row:\[y + 3 = 4\] Which gives: \[y = 1\]

Step 6: Back-Substitute to Find x

Use the first row to solve for x, given \(x + y = -1\) with \(y = 1\). Substitute \(y = 1\) in the first row:\[x + 1 = -1\] Which gives: \[x = -2\]

Step 7: Verify the Solution

Substitute \(x = -2\), \(y = 1\), \(z = 3\) back into the original equations to verify correctness. All equations are satisfied, confirming the solution.

Key Concepts

Understanding Row Operations

Row operations are a set of techniques applied to an augmented matrix to simplify and ultimately solve a system of equations. These operations help transform the matrix into a more manageable form, often row-echelon form, which then makes it easier to solve for each variable.

There are three types of elementary row operations:

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

By strategically applying these operations, you can systematically eliminate variables from certain rows, leading you towards a solution. Ensuring accuracy in these operations is vital, as a mistake can lead to an incorrect solution.
When performing row operations, it's crucial to maintain balance across the matrix, keeping the arithmetic consistent.

The Role of an Augmented Matrix

In solving systems of equations using matrix operations, the augmented matrix plays a crucial role. An augmented matrix condenses a system of equations into a single matrix by tagging the constants to the end of each equation's row.

For example, the system:\[\begin{aligned}&x+y=-1\&y+z=4\&x+z=1\end{aligned}\] is converted into an augmented matrix:\[\begin{bmatrix}1 & 1 & 0 & |-1 \0 & 1 & 1 & |4 \1 & 0 & 1 & |1\end{bmatrix}\]This matrix combines both the coefficients of the variables and the constants from the right side of each equation. By using this structure, row operations can be applied more efficiently to systematically reach a straightforward solution of the system.
It is vital to correctly set up the augmented matrix to ensure the integrity of subsequent row operations.

Solving a System of Equations

A system of equations is essentially a set of two or more equations with the same set of variables. The goal is to find values of these variables that satisfy all the equations simultaneously.

In the given exercise:\[\begin{aligned}&x + y = -1 \&y + z = 4 \&x + z = 1\end{aligned}\]the solution involves identifying values of \(x\), \(y\), and \(z\) such that all equations are satisfied.
By translating these equations into an augmented matrix and applying row operations, we methodically simplify the matrix to solve for variables one by one.Substituting back each found variable into the equations provides a way to check the consistency of the solution. This exercise's method is both systematic and logical, ensuring that all possible solutions are considered and verified.