Problem 46 - Algebra for College Students

Question

For Problems \(45-50\), change each augmented matrix of the system to reduced echelon form and then indicate the solutions of the system. \(\left(\begin{array}{c}x+3 y-2 z=-1 \\ 2 x-5 y+7 z=4\end{array}\right)\)

Step-by-Step Solution

Answer

The solution is \((x, y, z) = (-\frac{13}{11} - t, t - \frac{6}{11}, t)\), where \(t\) is any real number.

Step 1: Write the Augmented Matrix

The given system of equations is \(x + 3y - 2z = -1\) and \(2x - 5y + 7z = 4\). Write this system as an augmented matrix: \[\begin{bmatrix} 1 & 3 & -2 & | & -1 \2 & -5 & 7 & | & 4 \end{bmatrix}\]

Step 2: Make the Leading Entry of First Row 1

The leading entry of the first row is already 1, so no changes are necessary. The matrix remains: \[\begin{bmatrix} 1 & 3 & -2 & | & -1 \2 & -5 & 7 & | & 4 \end{bmatrix}\]

Step 3: Eliminate the First Column Below Leading 1

Subtract 2 times the first row from the second row to create a zero below the leading 1 in the first row:\[R_2 = R_2 - 2R_1\]Making the matrix:\[\begin{bmatrix} 1 & 3 & -2 & | & -1 \0 & -11 & 11 & | & 6 \end{bmatrix}\]

Step 4: Make the Leading Entry of Second Row 1

Divide the second row by -11 to make the leading coefficient 1:\[R_2 = \frac{1}{-11}R_2\]This transforms the matrix to:\[\begin{bmatrix} 1 & 3 & -2 & | & -1 \0 & 1 & -1 & | & -\frac{6}{11} \end{bmatrix}\]

Step 5: Eliminate the Second Column of the First Row

Subtract 3 times the second row from the first row to eliminate the y-term from the first row:\[R_1 = R_1 - 3R_2\]Resulting in the matrix:\[\begin{bmatrix} 1 & 0 & 1 & | & -\frac{13}{11} \0 & 1 & -1 & | & -\frac{6}{11} \end{bmatrix}\]

Step 6: Write the System from the Reduced Row Echelon Form

Translate the reduced matrix back into equations:1. \(x + z = -\frac{13}{11}\)2. \(y - z = -\frac{6}{11}\)

Step 7: Solve the System for Free Variable

Assuming \(z = t\) (where \(t\) is any real number), solve for \(x\) and \(y\):1. \(x = -\frac{13}{11} - t\)2. \(y = t - \frac{6}{11}\)

Step 8: Write the Solution

The solution can be expressed as: \[(x, y, z) = (-\frac{13}{11} - t, t - \frac{6}{11}, t)\]where \(t\) is any real number. This represents a line of solutions depending on parameter \(t\).

Key Concepts

Augmented Matrix

The concept of an augmented matrix is crucial in solving systems of equations using matrices. An augmented matrix is an extension of a typical coefficient matrix. It includes a column of constants from the right-hand side of a system of linear equations. This additional column allows us to represent the entire system compactly in matrix form.
Consider the system of equations given in the problem: \(x + 3y - 2z = -1\) and \(2x - 5y + 7z = 4\). To create an augmented matrix from this system, we align the coefficients of each variable into rows, and separate the constant terms that follow the equal sign with a vertical line.
The augmented matrix for our example is:

• First row: [1, 3, -2 | -1]
• Second row: [2, -5, 7 | 4]

It represents all the necessary information to apply matrix operations with the aim of solving the system. By using this matrix format, we can utilize various row operations to simplify and ultimately solve the system.

Systems of Equations

At its core, a system of equations involves finding values for variables that satisfy all given equations simultaneously. In our problem, we deal with a system of linear equations with three variables: \(x, y, z\). Our goal is to find values of these variables that satisfy both equations in the system.
This is commonly represented as:

• Equation 1: \(x + 3y - 2z = -1\)
• Equation 2: \(2x - 5y + 7z = 4\)

By converting these into an augmented matrix, we apply a process known as row reduction - performing operations to simplify the matrix to reduced row echelon form (RREF). When an augmented matrix is reduced, it can easily be translated back into a system of equations with each equation having a straightforward relation among the variables, thus simplifying solving further to find unique or parametric solutions.

Parametric Solution

In systems of equations, especially those that are linear and do not have a unique solution, a parametric solution is often used. This type of solution expresses some variables in terms of others, usually involving a parameter, often denoted as \(t\).
In our exercise, after transforming the augmented matrix to reduced row echelon form and translating it back to equations, we have:

• \(x + z = -\frac{13}{11}\)
• \(y - z = -\frac{6}{11}\)

Assuming \(z = t\), where \(t\) is a real number, the system can be solved as:

• \(x = -\frac{13}{11} - t\)
• \(y = t - \frac{6}{11}\)

Thus, the solution is expressed as a parametric representation: \( (x, y, z) = (-\frac{13}{11} - t, t - \frac{6}{11}, t) \), which defines a line in three-dimensional space. Each value of \(t\) gives a specific point on this line, providing insight into the system's infinite nature of solutions.