Problem 18 - Intermediate Algebra

Question

For Problems \(1-22\), solve each of the systems and use matrices as we did in the examples of this section. $$ \left(\begin{array}{rl} x-3 y+2 z & =0 \\ 2 x-4 y-3 z & =19 \\ -3 x-y+z & =-11 \end{array}\right) $$

Step-by-Step Solution

Answer

The solution is \(x = \frac{133}{45}\), \(y = \frac{67}{15}\), \(z = \frac{7}{3}\).

Step 1: Write the System as an Augmented Matrix

The first step is to express the system of equations as an augmented matrix. For the given system:\[\begin{align*}x - 3y + 2z &= 0 \2x - 4y - 3z &= 19 \-3x - y + z &= -11\end{align*}\]The augmented matrix is:\[\left(\begin{array}{ccc|c}1 & -3 & 2 & 0 \2 & -4 & -3 & 19 \-3 & -1 & 1 & -11 \\end{array}\right)\]This matrix contains the coefficients of the variables and the constants from each equation.

Step 2: Apply Row Operations to Obtain Row Echelon Form

The goal is to transform the augmented matrix into a form where the variables can easily be solved back. First, manipulate the rows to have leading ones.- Swap Row 1 and Row 3 to have a 1 in the top-left position.- The matrix becomes:\[\left(\begin{array}{ccc|c}-3 & -1 & 1 & -11 \2 & -4 & -3 & 19 \1 & -3 & 2 & 0 \\end{array}\right)\]- Add 3 times Row 1 to Row 3, and 6 times Row 1 to Row 2 to eliminate the x terms in Row 2 and Row 3.- The matrix becomes:\[\left(\begin{array}{ccc|c}-3 & -1 & 1 & -11 \0 & -10 & -1 & -47 \0 & -10 & 5 & -33 \\end{array}\right)\]

Step 3: Further Simplify the Matrix

To further reduce the matrix:- Divide Row 1 by -3 to make the leading coefficient of Row 1 a 1.\[\left(\begin{array}{ccc|c}1 & \frac{1}{3} & -\frac{1}{3} & \frac{11}{3} \0 & -10 & -1 & -47 \0 & -10 & 5 & -33 \\end{array}\right)\]- Subtract Row 2 from Row 3 to eliminate the y term from Row 3.- The matrix becomes:\[\left(\begin{array}{ccc|c}1 & \frac{1}{3} & -\frac{1}{3} & \frac{11}{3} \0 & -10 & -1 & -47 \0 & 0 & 6 & 14 \\end{array}\right)\]

Step 4: Solve for Variables using Back Substitution

Now that the matrix is in row-echelon form, we can solve for variables starting from the bottom:1. From the third row: \(6z = 14\). Solve for \(z\): \[z = \frac{14}{6} = \frac{7}{3}\]2. Substitute \(z = \frac{7}{3}\) into the second row equation: \(-10y - z = -47\). Solve for \(y\): \[ -10y - \frac{7}{3} = -47 \ -10y = -47 + \frac{7}{3} \ -10y = -\frac{141}{3} + \frac{7}{3} \ -10y = -\frac{134}{3} \ y = \frac{134}{30} = \frac{67}{15}\]3. Substitute \(y = \frac{67}{15}\) and \(z = \frac{7}{3}\) into the first row equation: \(x + \frac{1}{3}y - \frac{1}{3}z = \frac{11}{3}\), and solve for \(x\): \[ x + \frac{1}{3}\left(\frac{67}{15}\right) - \frac{1}{3}\left(\frac{7}{3}\right) = \frac{11}{3} \ x + \frac{67}{45} - \frac{7}{9} = \frac{11}{3} \ x + \frac{67}{45} - \frac{35}{45} = \frac{165}{45} \ x + \frac{32}{45} = \frac{165}{45} \ x = \frac{165}{45} - \frac{32}{45} = \frac{133}{45}\]

Step 5: Conclusion

The solution to the system of equations is:\[x = \frac{133}{45}, \quad y = \frac{67}{15}, \quad z = \frac{7}{3}\]

Key Concepts

System of Equations

A system of equations is a set of two or more equations that we solve simultaneously. Each equation contains multiple variables thatare shared among the equations. The goal is to find values for each variable that satisfy all the equations at once.
In matrix algebra, systems of equations can be conveniently represented as matrices, which helps organize the coefficients and constants. In the exercise above, a system of three equations with three variables—\(x\), \(y\), and \(z\)—is presented as an augmented matrix.
The matrix encapsulates the coefficients (the numbers multiplying the variables) in the main part and the constants on the augmented side, separated by a vertical line.

• Each row represents an equation.
• Each column represents a variable up to the constants.

This organized layout makes it easier to implement systematic approaches like elimination and substitution to solve for the unknowns.

Row Echelon Form

The row echelon form is a stage in the solution process where the augmented matrix is transformed into a simpler form. This form facilitates easy back substitution to find variable values. It features leading coefficients of one and organized zeros, particularly below the leading one in each row.
The given solution transforms the matrix into row-echelon form by using row operations. These operations include swapping rows, multiplying rows by scalars, and adding multiples of rows to other rows. The primary objective is to create a triangular format where:

• The first non-zero number in each row is a 1 (pivot element).
• All elements below a pivot are zeros.

Once in this form, you can effectively pinpoint the variables starting from the last row, in a systematic and straightforward manner, setting the stage for back substitution.

Back Substitution

Back substitution is the final phase of solving a system of equations after achieving the row echelon form. It involves solving for variables starting from the bottom row of the matrix upward. In the reduced matrix, each row represents a simplified equation.
For example, in the exercise provided:

• You begin with the last row, which is simplest, solving for the bottom-most variable.
• Next, substitute this variable’s solved value into the row above it, simplifying and solving that equation.
• Continue this process, rising row by row, substituting solved values back into previous rows, until all variables are determined.

This technique allows one to solve complex systems in a structured manner. It ensures each step builds on previous solutions, making the solving process both logical and efficient.