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

Step-by-Step Solution

Answer

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

Step 1: Write the Augmented Matrix

First, we represent the system of equations in terms of an augmented matrix. The system \( x + y = -1 \), \( y + z = 4 \), and \( x + z = 1 \) is expressed as: \[ \begin{bmatrix} 1 & 1 & 0 & | & -1 \ 0 & 1 & 1 & | & 4 \ 1 & 0 & 1 & | & 1 \end{bmatrix} \]

Step 2: Swap Rows if Necessary

Check if any row swaps would simplify the process. In this case, swapping the first and last rows helps, resulting in: \[ \begin{bmatrix} 1 & 0 & 1 & | & 1 \ 0 & 1 & 1 & | & 4 \ 1 & 1 & 0 & | & -1 \end{bmatrix} \]

Step 3: Eliminate Through Row Operations

To eliminate \( x \) in the third row, subtract the first row from the third row. \[ R_3 = R_3 - R_1 \] Resulting in:\[ \begin{bmatrix} 1 & 0 & 1 & | & 1 \ 0 & 1 & 1 & | & 4 \ 0 & 1 & -1 & | & -2 \end{bmatrix} \]

Step 4: Simplify the Matrix

Now subtract the second row from the third row to eliminate \( y \) in the third row.\[ R_3 = R_3 - R_2 \]Resulting in:\[ \begin{bmatrix} 1 & 0 & 1 & | & 1 \ 0 & 1 & 1 & | & 4 \ 0 & 0 & -2 & | & -6 \end{bmatrix} \]

Step 5: Solve for Each Variable

- From the third row, solve for \( z \):\[ -2z = -6 \quad \implies \quad z = 3 \]- Substitute \( z \) into the second row to solve for \( y \):\[ y + 3 = 4 \quad \implies \quad y = 1 \]- Substitute \( z \) into the first row to solve for \( x \):\[ x + 3 = 1 \quad \implies \quad x = -2 \]

Step 6: Verify the Solution

Substitute \( x = -2 \), \( y = 1 \), and \( z = 3 \) back into the original equations to ensure they all hold true:- \( x + y = -2 + 1 = -1 \)- \( y + z = 1 + 3 = 4 \)- \( x + z = -2 + 3 = 1 \)All equations are satisfied, confirming the solution is correct.

Key Concepts

Row Operations

Row operations are fundamental tools used in solving systems of equations represented by matrices. These operations allow us to manipulate the rows of a matrix to achieve a desired simpler form. These forms help in finding solutions to the system. There are three types of row operations:

• *Row swapping*: Exchanging two rows in a matrix, which can help simplify calculations or position pivot elements more clearly.
• *Row multiplication*: Multiplying all elements in a row by a non-zero constant to normalize or scale the row.
• *Row addition*: Adding or subtracting rows from each other to eliminate variables and simplify the system.

In the exercise, row operations were used to transform the augmented matrix into a simpler form. Initially, a row swap placed the row with the element 1 as the leading entry at a more strategic position. Subsequent operations eliminated specific variables in different rows, facilitating the back substitution process to find the values of variables.

System of Equations

A system of equations consists of multiple equations that share the same set of variables. The ultimate goal is to find values for these variables that simultaneously satisfy all equations within the system. These systems can be approached using various methods:

• *Graphical method*: Solving by finding the intersection points of the equations' graphs.
• *Substitution method*: Solving one equation for one variable and substituting this into the other equations.
• *Elimination method*: Eliminating variables systematically to solve for the remaining ones.
• *Matrix method*: Using matrices and row operations as seen in this example.

In our exercise, the system of equations was converted into an augmented matrix. By applying row operations, this system was transformed into row-echelon form. This form significantly eases the process of solving for each variable by back substitution.

Algebraic Solution

An algebraic solution involves finding the exact values of variables that satisfy a given set of equations. It moves from a simplified form (like row-echelon or reduced row-echelon form of a matrix) to derive these exact solutions. For the exercise, once the augmented matrix was simplified, the algebraic solution consisted of solving for each variable sequentially. The steps were:

• Finding the value of the last variable by directly solving the simplified form.
• Back-substituting this value into previous equations to find the other variables.

In this case, solving for the values of \(x\), \(y\), and \(z\) involved starting from the simplest equation and working upwards. For instance, after finding \(z = 3\), it was substituted back to find \(y = 1\), and further substituting to find \(x = -2\). Finally, verifying the solution by substituting these values back into the original equations ensures that they hold true, providing confidence in the correctness of the solution.