Problem 47 - 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. $$\begin{aligned} 2 x-y+3 z &=0 \\ x+2 y-z &=5 \\ 2 y+z &=1 \end{aligned}$$

Step-by-Step Solution

Answer

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

Step 1: Construct the Augmented Matrix

Start by writing the system of equations in the form of an augmented matrix. The coefficients of \( x \), \( y \), and \( z \) are the components of the matrix, and the constants are the last column. This gives us: \[\begin{bmatrix}2 & -1 & 3 & 0 \1 & 2 & -1 & 5 \0 & 2 & 1 & 1\end{bmatrix}\]

Step 2: Perform Row Operations to Create Leading Ones

To create a leading one in the first row, divide the entire first row by 2. This changes the augmented matrix to: \[\begin{bmatrix}1 & -0.5 & 1.5 & 0 \1 & 2 & -1 & 5 \0 & 2 & 1 & 1\end{bmatrix}\]

Step 3: Eliminate the First Column Below the Leading One

Subtract row 1 from row 2 to get a zero below the leading one in the first column:\( R_2 = R_2 - R_1 \). This results in:\[\begin{bmatrix}1 & -0.5 & 1.5 & 0 \0 & 2.5 & -2.5 & 5 \0 & 2 & 1 & 1\end{bmatrix}\]

Step 4: Create a Leading One in the Second Row

Divide the entire second row by 2.5 to create a leading one:\[\begin{bmatrix}1 & -0.5 & 1.5 & 0 \0 & 1 & -1 & 2 \0 & 2 & 1 & 1\end{bmatrix}\]

Step 5: Eliminate the Second Column Below the Leading One

Subtract twice the second row from the third row to produce a zero below the leading one in the second column:\( R_3 = R_3 - 2R_2 \). This gives us:\[\begin{bmatrix}1 & -0.5 & 1.5 & 0 \0 & 1 & -1 & 2 \0 & 0 & 3 & -3\end{bmatrix}\]

Step 6: Create a Leading One in the Third Row

Divide the entire third row by 3 to create a leading one:\[\begin{bmatrix}1 & -0.5 & 1.5 & 0 \0 & 1 & -1 & 2 \0 & 0 & 1 & -1\end{bmatrix}\]

Step 7: Back-Substitute to Solve for Variables

Now that we have the rows in row-echelon form, substitute backwards to solve for \( z \), \( y \), and then \( x \). From the third row: \[ z = -1 \]Using the value of \( z \) in the second row:\[ y - (-1) = 2 \rightarrow y + 1 = 2 \rightarrow y = 1 \]Finally, substitute \( y \) and \( z \) into the first row:\[ x - 0.5(1) + 1.5(-1) = 0 \rightarrow x - 0.5 - 1.5 = 0 \rightarrow x = 2 \]

Step 8: Final Solution

The final solution for the system of equations is:\[x = 2, \quad y = 1, \quad z = -1\]

Key Concepts

Augmented Matrix

The concept of an augmented matrix is a handy tool in solving systems of equations using matrix operations. Essentially, an augmented matrix combines the coefficient matrix of a system of linear equations with the constants from the equations. It provides a compact and structured way to represent and manipulate the equations.

In the context of the given exercise, we had the system:

• System of equations: \( 2x - y + 3z = 0 \)
• \( x + 2y - z = 5 \)
• \( 2y + z = 1 \)

The coefficients form the augmented matrix:\[\begin{bmatrix}2 & -1 & 3 & | & 0 \1 & 2 & -1 & | & 5 \0 & 2 & 1 & | & 1 \\end{bmatrix}\]The vertical bar typically separates the coefficients from the constants, emphasizing the augmentation.

System of Equations

A system of equations is a set of two or more equations with the same variables. Solving such a system means finding a set of variable values that satisfy all equations simultaneously. In our context, the variables of interest are \(x\), \(y\), and \(z\), which must satisfy the system:

• \(2x - y + 3z = 0\)
• \(x + 2y - z = 5\)
• \(2y + z = 1\)

There are various methods to solve these systems, such as substitution, elimination, and matrix methods. The introduction of matrix techniques, like using an augmented matrix, offers an organized approach that scales well for larger systems.

Leading One

"Leading One" refers to a concept in linear algebra, specifically used in matrix manipulation techniques like Gaussian elimination. When performing row operations on an augmented matrix, a leading one is the first non-zero number in a row from the left.

Achieving a leading one is a goal when transforming a matrix into row-echelon form. In the current exercise, after dividing row 1 by 2, we converted the first column's entry into a leading one:\[\begin{bmatrix}1 & -0.5 & 1.5 & | & 0 \1 & 2 & -1 & | & 5 \0 & 2 & 1 & | & 1 \\end{bmatrix}\]By continuing similar operations, we systematically create leading ones across the diagonal, simplifying the matrix step-by-step.

Back-Substitution

Back-substitution is the final phase in solving a system of equations once the matrix form is semi-solved, ideally achieving an upper triangular matrix. This process involves starting from the last equation and solving upwards.

For the matrix form achieved from row operations:

• Third row: \( z = -1 \)
• Second row: \( y - z = 2 \rightarrow y = 1 \)
• First row: \( x - 0.5y + 1.5z = 0 \rightarrow x = 2 \)

You work backward, starting with the simplest equation at the bottom, and solve each equation step by step by substituting known values, which are found from the more straightforward equations. This methodical approach ensures all solutions are valid and satisfy all original equations.