Problem 46 - 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. $$\begin{aligned} x+y &=1 \\ 2 x-z &=0 \\ y+2 z &=-2 \end{aligned}$$

Step-by-Step Solution

Answer

Solution: \(x = -1\), \(y = 2\), \(z = -2\).

Step 1: Write the Augmented Matrix

Convert the given system of equations into an augmented matrix. The system given is: \[\begin{aligned} x+y &=1 \2x-z &=0 \y+2z &=-2 \end{aligned}\]The augmented matrix is: \[\begin{bmatrix} 1 & 1 & 0 & | & 1 \2 & 0 & -1 & | & 0 \0 & 1 & 2 & | & -2 \end{bmatrix}\]

Step 2: Simplify the First Row

Let's make the first row have a leading 1 in the first column (it already has it), so we'll leave it unchanged for now: \[\begin{bmatrix} 1 & 1 & 0 & | & 1 \2 & 0 & -1 & | & 0 \0 & 1 & 2 & | & -2 \end{bmatrix}\]

Step 3: Eliminate First Column of the Second Row

Subtract 2 times the first row from the second row to eliminate the 2 in the second row, first column:\[R_2 = R_2 - 2R_1\]This changes the matrix to:\[\begin{bmatrix} 1 & 1 & 0 & | & 1 \0 & -2 & -1 & | & -2 \0 & 1 & 2 & | & -2 \end{bmatrix}\]

Step 4: Clear First Column of the Third Row

There is already a zero in the first column of the third row, so no changes are needed. The matrix remains:\[\begin{bmatrix} 1 & 1 & 0 & | & 1 \0 & -2 & -1 & | & -2 \0 & 1 & 2 & | & -2 \end{bmatrix}\]

Step 5: Simplify the Second Row

We'll divide the second row by -2 to simplify it:\[R_2 = \frac{1}{-2}R_2\]The matrix becomes:\[\begin{bmatrix} 1 & 1 & 0 & | & 1 \0 & 1 & \frac{1}{2} & | & 1 \0 & 1 & 2 & | & -2 \end{bmatrix}\]

Step 6: Eliminate Second Column of the Third Row

Subtract the second row from the third row to make the second column entry of the third row zero:\[R_3 = R_3 - R_2\]Resulting in:\[\begin{bmatrix} 1 & 1 & 0 & | & 1 \0 & 1 & \frac{1}{2} & | & 1 \0 & 0 & \frac{3}{2} & | & -3 \end{bmatrix}\]

Step 7: Simplify Third Row

Divide the third row by \(\frac{3}{2}\) to get a leading 1 in the third column:\[R_3 = \frac{1}{\frac{3}{2}}R_3\]This results in:\[\begin{bmatrix} 1 & 1 & 0 & | & 1 \0 & 1 & \frac{1}{2} & | & 1 \0 & 0 & 1 & | & -2 \end{bmatrix}\]

Step 8: Back Substitution for the Second Row

Substitute the value of \(z = -2\) from the third row into the second row:\[ y + \frac{1}{2}(-2) = 1 \]Simplifying gives:\[ y - 1 = 1 \y = 2 \]

Step 9: Back Substitution for the First Row

Substitute \(y = 2\) and \(z = -2\) into the first row:\[ x + 2 = 1 \]Giving:\[ x = -1\]

Step 10: Verify the Solution

Substitute \(x = -1\), \(y = 2\), and \(z = -2\) back into the original equations to verify the solution satisfies all:\[\begin{aligned} -1 + 2 &= 1, \2(-1) - (-2) &= 0, \2 + 2(-2) &= -2 \end{aligned}\]All equations are satisfied.

Key Concepts

Row Operations

Row operations are fundamental tools used on matrices to solve systems of equations. These operations transform matrices in a way that simplifies them without altering the solution set they represent. There are three basic types of row operations:

• Swapping two rows
• Multiplying a row by a nonzero scalar
• Adding or subtracting a multiple of one row to another row

By using these row operations, an augmented matrix can be transformed into a simpler form that is easier to work with. This often takes the form of a matrix in row echelon form or reduced row echelon form. In our example, row operations are used systematically to zero out entries below the leading coefficients, paving the way for easy back substitution.

System of Equations

A system of equations is a collection of two or more equations with a set of unknowns. Solving a system means finding values for the unknowns that satisfy every equation in the system. Systems of equations can be represented algebraically as well as graphically or in matrix form. In the given exercise:

• The system consists of three equations with three variables, namely, \( x, y, \text{ and } z \).
• These equations can be transformed into an augmented matrix, which provides a compact representation that is optimal for matrix manipulation.

Using matrices and row operations, such a system can be effectively solved by transforming it to a form where the solution can be directly read off or easily derived using back substitution.

Back Substitution

Back substitution is a method used to solve systems of linear equations organized in an upper triangular matrix form. This means starting from the last equation and moving backward to solve for each unknown one at a time. Once the augmented matrix is in its simplified form, back substitution helps find the exact values of the variables:

• First, solve the last equation for the corresponding variable.
• Substitute this value back into the preceding equations to solve for the other variables.
• Continue this process until all variables have been determined.

In this example, after row operations are performed, the matrix forms an upper triangular shape, allowing us to use back substitution to find \( z = -2 \), then \( y = 2 \), and finally, \( x = -1 \).

Gaussian Elimination

Gaussian elimination is a systematic method used for solving systems of linear equations. It involves a series of operations to transform this system's matrix into an upper triangular form. The steps in Gaussian elimination are:

• Converting the system of equations into an augmented matrix.
• Using row operations to zero out the entries below the pivot positions, hence simplifying the matrix to row echelon form.
• Applying back substitution to find the solutions.

The augmented matrix in our exercise is transformed via Gaussian elimination, leveraging row operations to systematically reach a simplified form, where back substitution can be effectively applied. This method efficiently provides the solution set \((x, y, z) = (-1, 2, -2)\) for the given system of equations.